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• alyafey22
Re: Advanced Integration techniquesIn summary, this technique is used to solve various integrals. Differentiation with respect to y is used if the function f has partial continuous derivative on a chosen interval.f

#### alyafey22

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MHB
Hello

Integration is one of the most interesting topics in Mathematics. It has a wide area of applications in very different aspects of engineering and science. There are numerous ways to tackle integration problems. For elementary ones please refer to the following http://mathhelpboards.com/calculus-10/integral-calculus-tutorial-1711.html for basic introduction to integral calculus .

For the most part, I don't want you to be afraid by the word ''advanced'' the most important factor is practice and practice as long as you have basic knowledge of elementary integral calculus you should not get afraid. Some exercises here will require basic knowledge of certain properties of special functions which I will introduce before tackling certain problems (I will not focus on the proof) .Furthermore , basic background of complex variables will be such a great help here. I will try to leave certain problems for the reader with a final answer to try in your leisure time .

Finally , I will try to post an exercise every day so if you find any mistake don't hesitate to inform me . Furthermore , if you have any comments , face any problem understanding something or you want to show me your work just send me a pm or post it http://www.mathhelpboards.com/f49/commentary-advanced-integration-techniques-4219/#post19090and I will try to respond as soon as possible .

[new] I created a pdf for my tutorials here.

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1. Differentiation under the integral sign : (Leibniz_integral)

This is one of the most commonly used techniques to solve numerous numbers of questions I will be using this technique to solve many other exercises in the coming posts

Assume that we have the following function of two variables :

$$\displaystyle \int ^{b}_a \, f(x,y) \, dx$$​

Then we can differentiate it with respect to y provided that f has partial continuous derivative on a chosen interval.

$$\displaystyle F'(y) = \int^{b}_{a} f_{y}(x,y)\,dx$$​

Now using this in many problems is not that clear you have to think a lot to get the required answer because many integral questions are just in one variable so you add the second variable and assume it is a function of two variables .

Assume we want to solve the following integral :

$$\displaystyle \int ^{1}_{0} \, \frac{x^2-1}{\ln (x) }$$​

Now that seems very difficult to solve but using this technique we can solve it easily no matter how much power is x raised to . So the crux move is to decide ,where to put the second variable ! So the problem with the integral is that we have a logarithm in the denominator which makes the problem so difficult to tackle !

Remember that we can get a natural logarithm if we differentiate exponential functions i.e $F(a) = 2^a \Rightarrow \,\, F'(a) = \ln(a) \cdot 2^a$

Applying this to our problem

$$\displaystyle F(a)=\int ^{1}_{0} \, \frac{x^a-1}{\ln (x) }$$

Differentiate with respect to $a$

$$F'(b)=\frac{d}{d a}\int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx$$

We see that the integral is a function with two variables $f(x,a)$

So we take the partial derivative with respect to a ..

$$F'(a)=\int^{1}_{0} \frac{\partial }{\partial a}\left(\frac{x^a-1}{\ln(x)}\right) \,\,dx$$

$$F'(a)=\int^{1}_{0} x^a\,\,dx$$

$$F'(a)=\frac{x^{a+1}}{a+1} \bigl]^1_0$$

$$F'(a)=\frac{1}{a+1}$$

Now integrate wrt to a to get $$\displaystyle F(a)$$

$$F(a)=\ln{(a+1)}+C$$

$$F(0)=\ln{1}+C\,\,\text{hence : } C=0$$

$$\int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx=\ln{(a+1)}$$

By this powerful rule we were not only able to solve the integral we also found a general formula for some $$\displaystyle a$$ .

Now to solve our original integral put $$\displaystyle a=2$$

$$\int^{1}_{0} \frac{x^2-1}{\ln(x)}\,\,dx=\ln{(2+1)}=\ln(3)$$

[HW]
$$\int^{\infty}_{0}\frac{\sin (x) }{x}\,dx$$

To Be Continued ...

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1.Differentiation under the integral sign (continued)

It is not always easy to find the function of two variables in which to differentiate. So that requires insight and ability to foresee the function.

Here is an example , find the following integral

$$\displaystyle \int_0^{\frac{\pi}{2}} \frac{x}{\tan x}dx$$​

So where do we put the variable $$\displaystyle a$$ here , that doesn't seem to be straight forward , how do we proceed ?

Well, take a deep breath , as I said it is not always easy to deduce such a function , it might not be so straight forward .

Let us try the following

$$F(a)= \int_0^{\frac{\pi}{2}} \frac{\arctan(a\tan(x))}{\tan (x)}dx$$

Now differentiate with respect to $$\displaystyle a$$ :

$$F'(a)= \int_0^{\frac{\pi}{2}} \frac{1}{1+(a\tan(x) )^2}dx$$

We will show later that

$$\int_0^{\frac{\pi}{2}} \frac{1}{1+(a\tan(x) )^2}dx= \frac{\pi}{2(1+a)}$$

$$F'(a) = \frac{\pi}{2(1+a)}$$

Now Integrate both sides

$$F(a)= \frac{\pi}{2}\ln(1+a) +C$$

Now we need to substitute $a=0$ in order to find $C = 0$

$$\int_0^{\frac{\pi}{2}} \frac{\arctan(a\tan(x))}{\tan (x)}\,dx = \frac{\pi}{2}\ln(1+a)$$

Put $a =1$ in order to get our original integral

$$\int_0^{\frac{\pi}{2}} \frac{x}{\tan (x)}\,dx = \frac{\pi}{2}\ln(2)$$

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Differentiation under the integral sign (continued):

I already gave this as a practice problem I will solve it now to check your solution :

We are given the following :

$$\int^{\infty}_0 \frac{\sin(x) }{x}\, dx$$

This problem can be solved by many ways , but here we will try to solve it by differentiation .

So as I described earlier in the previous examples it is generally not so easy to find the function with two variables .

Actually this step might require trial and error (Tmi) techniques until we get the desired result , so don't just give up if an approach merely doesn't work !.

$$F(a)=\int^{\infty}_0 \frac{\sin(ax) }{x}\, dx$$

Let us try this one :

If we differentiated with respect to a we get the following :

$$F'(a)=\int^{\infty}_0 \cos(ax) \, dx$$

But unfortunately this integral doesn't converge , so this is not the correct one .

Well, that seemed hopeless , but you should benefit from mistakes (Yes). The previous integral will converge if there is an exponential (This is merely the Laplace transform which I will illustrate later ... (Time) ).

So let us try the following :

$$F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx$$

Take the derivative to get :

$$F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx$$

Now this is easy to solve we can use integration by parts twice to get the following :

$$F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}$$

Now integrate both sides :

$$F(a)=-\arctan(a)+C$$

Now to find the value of the constant will take the limit as a grows very large :

$$C=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}$$

So we get our F(a) as the following:

$$F(a)=-\arctan(a)+ \frac{\pi}{2}$$

Now for the particular value of a = 0 we have :

$$\int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}$$

Please if anything not clear you can send me a pm ...

In the next post I will start Hyperbolic Integration.

2.Hyperbolic Integration : hyperbolic functions

Hyperbolic functions are very interesting , they are used to solve several integration problems. The most interesting thing about them is their relation to geometric functions through the see here.

It is really power full that you can switch between trigonometric functions and hyperbolic functions and vice versa , this requires a basic knowledge of complex numbers .

We know from the general definition of hyperbolic functions that :

$$\sinh(x) = \frac{e^x-e^{-x}}{2}$$

$$\cosh(x) = \frac{e^x+e^{-x}}{2}$$

Well, that seems interesting as it seems very close to the Euler's formula for sine and cosine :

$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$

$$\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$

It can be deduced easily that we can use the following to convert from trigonometric to hyperbolic functions :

$$\cos(ix)= \cosh(x)$$

$$\sin(ix)= i\sinh(x)$$

Woo , that is really fabulous so we can apply this to several problems to get rid of the complex i which is really annoying .

Here is an interesting problem :

$$\int - \left( \frac{\pi \csc\left(\pi (\frac{-1-ix}{2})\right)}{4}+\frac{\pi \csc\left(\pi (\frac{-1+ix}{2})\right)}{4}\right)\, dx$$

That seems very very very ... complicated (Yawn) , but wait don't be cheated by the complex view of a problem . Think about our hyperbolic friend;).

Here we can apply basic trigonometric simplifications :

$$\int \frac{-\pi}{4}\left(\frac{1}{\sin\left(\frac{-\pi}{2})\cos(\frac{i\pi x}{2}\right)}+\frac{1}{\sin\left(\frac{-\pi}{2}\right)\cos\left(\frac{i\pi x}{2}\right)} \right) \, dx$$

Now we see how to apply our magic trick

$$\int \frac{\pi}{4}\left(\frac{1}{\cosh\left(\frac{\pi x}{2}\right)}+\frac{1}{\cosh\left(\frac{\pi x}{2}\right)} \right) \, dx$$

This can be further simplified to get the following :

$$\frac{\pi}{2}\int \text{sech}\left(\frac{\pi x}{2}\right)\,dx=2\arctan \left( \tanh\left(\frac{\pi x}{4}\right) \right)+C$$

Ooh , wait , hang on How is that ?!

Well, differentiate the right side to see what is going on :

$$2\frac{\frac{\pi}{4} \text{sech}^2 \left(\frac{\pi x}{4}\right)}{1+\tanh^2 \left(\frac{\pi x}{4}\right)}$$

$$\frac{\pi}{2}\left(\frac{1}{\cosh^2\left(\frac{\pi x}{4}\right) +\sinh^2\left(\frac{\pi x}{4}\right)} \right)$$

Further simplification will do the task :

$$\frac{\pi}{2}\left(\frac{1}{\cosh^2(\frac{\pi x}{4}) +\cosh^2(\frac{\pi x}{4})-1}\right)=\frac{\pi}{2}\left(\frac{1}{\cosh(\frac{\pi x}{2})}\right)= \frac{\pi}{2}\text{sech}\left(\frac{\pi x}{2}\right)$$

To be continued ...

Hyperbolic Integration (continued)

As we have seen in the previous post hyperbolic functions can be so helpful as they help us simplify a lot of operations in a way that is really neat. Also, they help us handle complex variables to get a result that is completely free of complex numbers. This is basically because of the ability to convert from trigonometric functions though Euler formula.

Working with hyperbolic functions for the most part requires a lot of practice. It might not be so clear that the function can be integrated , this will help us in complex integration in the future (Happy).

Let us have the following example :

prove that :

$$I=\int -\frac{1}{2} \left( \csc(-\alpha-ix)+\csc(-\alpha+ix)\right)= \arctan\left(\csc(\alpha) \sinh(x) \right) +C$$

$$\frac{1}{2}\int \csc(\alpha+ix)+\csc(\alpha-ix) \,dx$$

$$\frac{1}{2}\int \frac{1}{\sin(\alpha+ix)}+\frac{1}{\sin(\alpha-ix)}\, dx$$

Applying basic geometry identities :

$$=\frac{1}{2}\int \left(\frac{1}{\sin(\alpha)\cosh(x) +i\sinh(x) \cos(\alpha)}+\frac{1}{\sin(\alpha)\cosh(x) -i\sinh(x) \cos(\alpha)}\right)\, dx$$

Notice that the denominators are the complex conjugate of each other :

$$\frac{1}{2}\int \left(\frac{2\sin(\alpha)\cosh(x)}{\sin^2(\alpha) \cosh ^2(x) +\sinh^2(x) \cos^2(\alpha)}\right)\, dx=\, \int \left(\frac{\sin(\alpha)\cosh(x)}{\sin^2(\alpha)(1+\sinh^2(x)) +\sinh^2(x) (1-\sin^2(\alpha))}\right)\, dx$$

$$\int \left(\frac{\sin(\alpha)\cosh(x)}{\sin^2(\alpha) +\sinh^2(x)}\right)\, dx= \int \frac{\csc(\alpha)\cosh(x)}{ 1+\csc^2(\alpha)\sinh^2(x)}\, dx$$

$$I= \arctan\left(\csc(\alpha) \sinh(x) \right) +C$$

I will start Laplace transform in the next post ...
(Movie)

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3.Integration by Laplace Transform (Laplace)

3.1.Basic Introduction :

Laplace transform is a very powerful transform. It can be used in many applications . For example, it can be used to solve Differential Equations and its rules can be used to solve integration problems .

The basic definition of Laplace transform :

$$F(s) = \mathcal{L}(f(t)) = \int ^{\infty}_0 e^{-st}f(t) \, dt\, \text{ ----(1)}$$

This integral will converge when $$\text{Re}(s)>a \text{ for } |f(t)| \leq Me^{at}$$ (Exponential Type )

Let us see the Laplace transform for some functions f(t) :

1.f(t) =1 Let us find F(s) :

This is elementary $$F(s) =\int ^{\infty}_0 e^{-st} \, dt= \frac{1}{s}$$

2.Also in general $f(t) = t^n \text{ where }n \geq 0$

We can prove by parts that $$F(s) =\int ^{\infty}_0 e^{-st} t^n \, dt=\frac{n!}{s^{n+1}}\text{ ----(2)}$$ ( Try to prove it (Smirk))

3.What about $f(t)= \cos(at)$

Also by parts we can prove that $$F(s) =\int ^{\infty}_0 e^{-st} \cos(at) \, dt=\frac{s}{s^2+a^2}\text{ ----(3)}$$

You can assistant a table for Laplace transform ...

One of the most interesting results of the Laplace :

$$(f * g)(t)= \int^{t}_0 f(s)g(t-s)\,ds\text{ ----(4)}$$ (Convolution)

$$\mathcal{L}\left((f * g)(t)\right)= \mathcal{L}(f(t)) \mathcal{L}(g(t)) \text{ ----(5)}$$

You can see the proof http://www.mathhelpboards.com/f34/problem-week-42-january-14th-2013-a-3000/.

Let us see some examples on integration:

Find the following integral :

$$\int^{\infty}_{0}e^{-2t}t^3\,dt$$

We can directly use the above formula

$$\int ^{\infty}_0 e^{-st}t^n \, dt=\frac{n!}{s^{n+1}}$$

here we have s= 2 , and n =3 , so directly we get :

$$\int^{\infty}_{0}e^{-2t}t^3\,dt= \frac{3!}{2^{3+1}}=\frac{3}{8}$$

So , we see that is becoming easy to find ...

Now let us think about the Laplace Inverse :

So, basically you are given F(s) and we want to get f(t) this is denoted by :

$$\mathcal{L}^{-1}(F(s))= f(t)\text{ or }\mathcal{L}^{-1}(\mathcal{L}(f(t)))= f(t)$$

We are given a function in the variable s and we want to transform it into another in the variable t .

Suppose we have the following examples :

Find the Laplace inverse of the following :

1) $$\frac{1}{s^3}$$

2) $$\frac{s}{s^2+4}$$

1- we can use the result in (2) so we have

$$\mathcal{L}(t^2)= \frac{2!}{t^3}\,\, \Rightarrow \,\, \frac{1}{2}\mathcal{L}(t^2)= \frac{1}{s^3}$$

Now take the inverse to both sides :

$$\frac{t^2}{2}= \mathcal{L}^{-1}\left(\frac{1}{s^3}\right)$$

2- we can use the result in (3) so we have

$$\cos(2t)=\mathcal{L}^{-1}\left(\frac{s}{s^2+4}\right)$$

[HW.1] Find the Laplace of $\sin(at)$

[HW.2] Find the inverse Laplace of $$\frac{1}{s^{n+1}}$$

To be continued ...

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Integration by Laplace Transform (continued)

3.2.1.Interesting results :

Prove the following :

$$B(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}$$ B is Beta function and $\Gamma$ is Gamma function

Note : Just as a basic introduction to special functions, you can think of gamma function as the following : $n!=\Gamma{(n+1)}$ we can extend n to exist in the whole complex plane {except n is a negative integer }.

The beta function is so interesting , we will explain some of its basic properties later (Time) .It can be used to solve many integrals.

For the proof :

We have the defnition (4) in the previous post about convolution .

Let us choose some functions f , and g :

$f(t) = t^{x} \,\, , \, g(t) = t^y$

By substituting in (4) we get the following :

$$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds \text{ ---- (*)}$$

We have from definition (5) that :

$\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y )$

Now we can use the result (2) to deduce

$$\mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$$

Notice that we need to find the inverse of Laplace $\mathcal{L}^{-1}$

$$\mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{-1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$

So we have the following :

$$(t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$

Now substitute in --(*) we get :

$$t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds$$

This looks good , put t=1 we get :

$$\frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$$

By using that $n! = \Gamma{(n+1)}$

we arrive happily to our formula :

$$\int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}$$

To be continued ...

Integration by Laplace Transform (continued)

3.2.2.Interesting results :

Prove the following :

$$\int^{\infty}_0 \frac{f(t)}{t}\,dt=\, \int^{\infty}_0 \, \mathcal{L}(f(t))\, ds\text{ ----(6)}$$

This is a very powerful rule it is basically saying that if we have a function divided by its independent variable and we integrated it in the half positive plane then we can transform it into an integral wrt to the Laplace transform of the function .

So Let us prove it :

we will start from the right hand side of (6)

we know from the definition --(1) that :

$$\int^{\infty}_0 \, \mathcal{L}(f(t))\, ds= \int^{\infty}_0 \,\left( \int^{\infty}_0e^{-st}\, f(t) \,dt\right)\, ds$$

Now by the Fubini's theorem we can rearrange* the double integral :

$$\int^{\infty}_0 \, f(t) \left(\int^{\infty}_0 e^{-st}\,ds\right)\,dt \text{ ----(**)}$$

The integral inside the parenthesis :

$$\int^{\infty}_0 e^{-st}\,ds$$

I know this is elementary ,but you may realize this is the Laplace transform but now the function is f(s) so basically we are just evaluating F(t) when f(s) =1 .

$$\int^{\infty}_0 e^{-st}\,ds = \mathcal{L}(1)= \frac{1}{t}$$

Now substitute this value in the integral in (**)

$$\int^{\infty}_0 \, \frac{f(t)}{t}\,dt$$

which is the left hand side of ----(6) as just required ...

This is just a fabulous rule , it can be used to simplify many computations ...

Let us find the following integral :

$$\int^{\infty}_0 \frac{\sin(t)}{t}\,dt$$

This is not the first time we see this integral and not the last (Happy). We have seen that we can find it using differentiation under the integral sign .

But you are just about to see the power of definition (6) [Doe that look like a movie trailer :D] , never mind .

Back to our integral , so the integral is of the type $\frac{f(t)}{t}$ so we can easily find it .

$$\int^{\infty}_0 \frac{sin(t)}{t}\,dt= \int^{\infty}_{0}\mathcal{L}(\sin(t))\,ds$$

Now what is the Laplace transform of $\sin(t)$ , did you do your homework ?

$$\mathcal{L}(\sin(at)) = \frac{a^2}{s^2+a^2}$$

So putting a =1 we get :

$$\mathcal{L}(\sin(t)) = \frac{1}{s^2+1}$$

Just substitute in our integral :

$$\int^{\infty}_0 \frac{ds}{1+s^2}= \tan^{-1}(s)|_{s=\infty}-\tan^{-1}(s)|_{s=0}=\frac{\pi}{2}$$

Just as expected ...;)

In the next post , I will explain the Gamma function .

* We must prove that we can rearrange the double integral , I didn't go into the details.

4.Integration using special functions

4.1.Gamma Function :

Gamma function is really interesting it is used to solve many interesting integrals, here we try to define some basic properties , prove some of them and take some examples.

Definition :

$$\Gamma{(x+1)} = \int^{\infty}_0 e^{-t}\, t^{x}\, dt \text{ ---(1)}$$

For the first glance that just looks like the Laplace Transform , actually they are closely related .

So let us for simplicity assume that x=n where $$n\geq 0 \text{ , n }$$ (is an integer )and substitute in (1) we get :

$$\Gamma{(n+1)} = \int^{\infty}_0 e^{-t}\, t^{n}\, dt$$

Well, we can use the Laplace transform - you might revise if you forgot -

$$\int^{\infty}_0 e^{-t}\, t^{n}= \frac{n!}{s^n+1}|_{s=1}= n!$$

So we see that there is a relation between the gamma function and the factorial ..

We will assume for the time being that the gamma function is defined as the following

$$n!= \Gamma{(n+1)}$$.

By this definition n\geq 0 where n is any positive integer which is pretty limited but surely this definition will be soon replaced by a stronger one.

Let us have some Examples :

Find the following integrals :

$$\int^{\infty}_0 \, e^{-t}t^4\, dt$$

By definition (1) this can be replaced by

$$\int^{\infty}_0 \, e^{-t}t^4\, dt =\Gamma(4+1)= 4! = 24$$

Pretty good let us continue with some more examples :

1) $$\int^{\infty}_0 \, e^{-t^2}t \, dt$$

2) $$\int^{1}_{0}\ln (t) \, t^2 \, dt$$

1) For the first one clearly we need a substitution before we go ahead :

so let us start by putting $$x=t^2$$ so the integral becomes :

$$\frac{1}{2}\int^{\infty}_0 \, e^{-x}x^{\frac{1}{2}}\cdot x^{-\frac{1}{2}}\, dt =\frac{1}{2}\Gamma(1+0)=\frac{1}{2}$$

2) For the second we use the substitution $$t=e^{-\frac{x}{2}}$$

$$\int^{1}_{0}\ln (t) \, t^2 \, dt = -\frac{1}{4}\int^{\infty}_{0}e^{-\frac{3x}{2}}\cdot x\,dx$$

Using another substitution $$t=\frac{3x}{2}$$ (Back to my favourite symbol )

$$\frac{-1}{9}\int^{\infty}_0 e^{-x}\, x\, dx =\frac{- \Gamma(2)}{9}= \frac{-1}{9}$$

It is an important thing to get used to the symbol $\Gamma$. I am sure that you are saying (boring...) that this seems elementary , but my main aim here is to let you practice the new symbol and get used to solving some problems using it .

[HW] prove $$\frac{\Gamma(5)\cdot \Gamma(2)}{\Gamma(7)}=\frac{1}{30}$$

[HW] Find the integral : $$\int^{\infty}_{0}e^{-\frac{1}{60}t}\, t^{20}\, dt$$ in terms of the gamma function .

In the next post I should extend the idea of a factorial ... you know this is only get exciting ...

To be continued.

4.Integration using special functions (continued)

4.1.Gamma Function (continued) :

For simplicity we assumed int the gamma definition (1) (look the previous post) that gamma only works for positive integers . This definition was so helpful as we assumed the relation between gamma and factorials . Actually ,this restricts the gamma function so much , we want to exploit the real strength of this very nice function .

Hence, we must extend the gamma function to work for all real numbers except for some values. Actually we will see soon that we can extend it to work for all complex numbers except where the function has poles .

There are many representations for the gamma function :

$$\Gamma(z) = \lim_{n=\infty}\frac{n! \, \, n^z}{z(z+1)(z+2) \, . \, . \, . \, (z+n)}\text{ ---- (2) }$$

$$\Gamma(z)\,= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}\text{ ----(3)}$$

Let us focus for the time being on the first representation :

This tells us a lot about the analyticity of gamma function , we notice that the gamma function is analytic everywhere except for z={0,-1,-2,-3, ... } , so zero and the negative integers are excluded and the function is not defined in this set .

We know have extended the gamma function and we want to evaluate interesting values .

Let us start by the following :

Find the integral :

$$\int^{\infty}_{0}\, \frac{e^{-t}}{\sqrt{t}}\, dt$$

Now according to definition (1) this is equal to $$\Gamma\left(\frac{1}{2}\right)$$ but we want to find a value for this if it exists ?

Laplace transform will not help here since we don't know what is the Laplace of a root !

Let us first make a substitution $$\sqrt{t}=x$$

so we have the integral as : $$2\int^{\infty}_{0}\, e^{-x^2}\, dx$$

Now to find this integral we need to do a simple trick ... (Happy)

$$\left(\int^{\infty}_{0}\, e^{-x^2}\, dx \right)^2$$

we will first try to find the upper integral ...

$$\left( \int^{\infty}_{0}\, e^{-x^2}\, dx\right)^2=\left( \int^{\infty}_{0}\, e^{-x^2}\, dx\right)\cdot \left(\int^{\infty}_{0}\, e^{-x^2}\, dx\right)$$

Now this is the crux movement (stay seated , don't shout, it is all ok )

$$\left(\int^{\infty}_{0}\, e^{-x^2}\, dx\right)\cdot \left(\int^{\infty}_{0}\, e^{-y^2}\, dy\right)$$

So we have done nothing just changed the second x by sub .

Now since they are two independent variables we can do the following :

$$\int^{\infty}_0\int^{\infty}_{0}\, e^{-(x^2+y^2)}\, dy \, dx$$

Now by polar substitution we can do the following :

$$\int^{\frac{\pi}{2} }_0\int^{\infty}_{0}\, e^{-r^2}\, r\,dr \, d\theta$$

so this becomes elementary :

$$\int^{\frac{\pi}{2}}_0\,\frac{1}{2}\, d\theta= \frac{\pi}{4}$$

So we have $$\left(\int^{\infty}_{0}\, e^{-x^2}\, dx \right)^2=\frac{\pi}{4}$$

Take the square root to both sides :

$$\int^{\infty}_{0}\, e^{-x^2}\, dx =\frac{\sqrt{\pi}}{2}$$

$$2\int^{\infty}_{0}\, e^{-x^2}\, dx =\sqrt{\pi}$$

So we have our result $$\fbox{[tex] \Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}$$ }[/tex]

This result is very interesting and will use it to solve many integrals .

To be continued ...

4.Integration using special functions (continued)

4.1.Gamma Function (continued) :

There are still lots of properties of gamma function one of which is related to the

factorial function . we know that $$n!=n(n-1)(n-2)\, ... \, 4\cdot 3\cdot 2\cdot 1$$.

For gamma function we have the following property :

$$\Gamma(x+1) = x\Gamma(x)$$

This is interesting because we can find some values for certain inputs for gamma .

Assume that we want to find
$$\Gamma\left(\frac{3}{2}\right)$$:

If we used this property we get :

$$\Gamma\left(1+\frac{1}{2}\right)= \frac{1}{2}\cdot\Gamma\left(\frac{1}{2}\right)=$$$$\frac{\sqrt{\pi}}{2}$$

Not all the time the result will be reduced to a simpler form as the previous example. For example we don't know how to express $$\Gamma(\frac{1}{4})$$ in a simpler form but we can approximate its value :

$$\Gamma(\frac{1}{4})\approx 3.6256 \, ...$$

so sometimes we just solve some integrals in terms of gamma function since we don't know a simpler form ...

For example solve the integral :

$$\int^{\infty}_0 e^{-t}t^{\frac{1}{4}}$$

we know by definition --- (1) of gamma function that this reduces to:

$$\int^{\infty}_0 e^{-t}t^{\frac{1}{4}}=\Gamma\left(\frac{5}{4}\right)=$$$$\frac{\Gamma\left(\frac{1}{4}\right)}{4}$$

We have seen that $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$ but what about $$\Gamma\left(\frac{-1}{2}\right)$$ ?

The question is elementary since we know that $$\Gamma\left(1-\frac{1}{2}\right)=\frac{-1}{2}\Gamma\left(\frac{-1}{2}\right)$$

so we have that $$\Gamma\left(-\frac{1}{2}\right)=-2\sqrt{\pi}$$

Then we can prove that any fraction with the denomenator =2 and the numerator is odd can be reduced into

$$\Gamma\left(\frac{2n+1}{2}\right)= C \,\Gamma\left(\frac{1}{2}\right)$$ where $$C\in \mathbb {Q}\text{ and n}\in \mathbb{Z}$$

ِAs a practice for Gamma find :

$$\int_{0}^{\infty}\frac{e^{-t}\cosh(a\sqrt{t})}{\sqrt{t}}dt$$

We have learned that gamma function has an exponent muliplied by a polynomial but here we have a hyperoblic function ! (Headbang)

Hange on we know that we can expand cosh using power series :

$$\cosh(x) = \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$

put $x=a\sqrt{t}$ we have
:

$$\cosh(a\sqrt{t}) = \sum_{n=0}^{\infty}\frac{a^{2n}\cdot t^n}{(2n)!}$$

subsituting back in the integral we have :

$$\int_{0}^{\infty}e^{-t}\sum_{n=0}^{\infty}\frac{a^{2n}\cdot t^n}{(2n)!\, \sqrt{t}}dt$$

Now since the series is always positive we can swap the integral and the series :

$$\sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!}\left[ \int_{0}^{\infty}e^{-t} t^{n-\frac{1}{2}}dt\right]$$

so we have using gamma :

$$\sum_{n=0}^{\infty}\frac{a^{2n}\cdot \Gamma(n+\frac{1}{2})}{2n!}$$

But we don't know how to simplify further because we need another property which will help us find the solution .

To be continued ...

Last edited:

4.Integration using special functions (continued)

4.1.Gamma Function (continued) :

In the previous post we were stuck trying to simplify $$\Gamma(n+\frac{1}{2})$$ but ,you know, the following property will make our life just easier.

Legendre Duplication Formula: (LDF)*

$$\Gamma\left(\frac{1}{2}+n\right) = {(2n)! \over 4^n n!} \sqrt{\pi}$$

The proof of this identity requires some manipulation of beta function .

So , we have just got our free gift to continue our example , so we were stuck at :

$$\sum^{\infty}_{n=0} \frac{a^{2n}\,\Gamma\left(\frac{1}{2}+n\right)}{(2n!)}$$

so using LDF we get the following :

$$\sum^{\infty}_{n=0}\frac{a^{2n}\,\left({(2n)! \over 4^n n!} \sqrt{\pi}\right)}{(2n!)}$$

Further simplification we get :

$$\sqrt{\pi}\,\sum^{\infty}_{n=0} \frac{{a^{2n}\over 4^n }}{n!}$$

Now that looks familiar since we know that :

$$\sum^{\infty}_{n=0}\frac{z^n}{n!}=e^z$$

Putting $$z={a^2\over 4 }$$ and multiplying by $\sqrt{\pi}\,$ we get :

$$\sqrt{\pi}\sum^{\infty}_{n=0} \frac{\left({a^2\over 4 }\right)^n}{n!}=\sqrt{\pi}\,e^{{a^2\over 4 }}$$

So we have finally that :

$$\int_{0}^{\infty}\frac{e^{-t}\cosh(a\sqrt{t})}{\sqrt{t}}dt\,=\,\sqrt{\pi}e^{{a^2 \over 4}}$$

Finally we have the interesting property Euler's Reflection Formula : (ERF)*

$$\Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \frac \pi {\sin \left({\pi z}\right)} \,\,\forall z \notin \mathbb{Z}$$

Let us have some examples on that :

Simplify the following expressions :

1. $$\Gamma\left(\frac{3}{4}\right)\,\Gamma\left(\frac{1}{4}\right)$$

2. $$\Gamma{\left(\frac{1+i}{2}\right)} \, \Gamma{\left(\frac{1-i}{2}\right)}$$

Solution :

1. we can rewrite as $$\Gamma\left(1-\frac{1}{4}\right)\,\Gamma\left(\frac{1}{4}\right)$$

Now by ERF we have the following :

$$\Gamma\left(1-\frac{1}{4}\right)\,\Gamma\left(\frac{1}{4}\right)=\frac \pi {\sin \left({\frac{\pi}{4} }\right)}=\sqrt{2}\,\pi$$

2. We can rewrite as $$\Gamma{\left(\frac{1+i}{2}\right)} \, \Gamma{\left(1-\frac{1+i}{2}\right)}$$

This expression simplifies to:

$$\frac{\pi}{\sin\left(\frac{\pi(1+i)}{2}\right)}= \frac{\pi}{\cos \left( \frac{i \pi}{2}\right)}$$

By geometry to hyperbolic conversions we get :

$$\frac{\pi}{\cosh\left(\frac{\pi}{2}\right)}=\pi \, \text{sech}\left(\frac{\pi}{2}\right)$$

LDF and ERF will be so much beneficial when we discuss the beta function in the next post .

* I will use these shortcuts when ever calling these formulas.

4.Integration using special functions (continued)

4.2.Beta Function :

General definition :

$$\int^1_0 t^{x-1}(1-t)^{y-1} \, dt= B(x,y)$$

This is actually my favourite function , and what is so nice about it , that it has many representations. Also it is related to Gamma function :

$$B(x,y)\,=\,{\Gamma(x)\Gamma(y) \over \Gamma(x+y)}$$

We have proved this identity earlier when we discussed convolution.

We shall realize the symmetry of beta function that is to say $$B(x,y) = B(y,x)$$.

Beta function has many other representations all can be deduced through substitution :

$$B(x,y) \,=\, \int ^{\infty}_0 \frac{t^{x-1}}{(1+t)^{x+y}}\, dt$$

$$B(x,y) \,=\,2\,\int^{\frac{\pi}{2}}_0 \cos^{2x-1}(t) \sin^{2y-1}(t)\,dt$$

The proofs are left to the reader as practice ...

Let us discuss some examples :

Find the following integral :

$$\int^{\infty}_0 \frac{1}{(x^2+1)}\, dx$$

we know that $$\int^{\infty}_0 \frac{1}{(x^2+1)}=\frac{\pi}{2}$$

Let us try to solve it by beta ,First put $x=\sqrt{t}$

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)}\,dt$$

Now we know the second representation of beta so we have :

$$x-1={-1 \over 2} \,\, \Rightarrow \,\, x={1\over 2}$$

$$x+y={1} \,\, \Rightarrow \,\, y={1\over 2}$$

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)}\,dt=\frac{B({1 \over 2},{1 \over 2})}{2}=\frac{\Gamma({1 \over 2})\, \Gamma({1\over 2})}{2}= \frac{\sqrt{\pi} \cdot \sqrt{\pi}}{2}=\frac{\pi}{2}$$

Now let us try to solve another similar integral :

$$\int^{\infty}_0 \frac{1}{(x^2+1)^2}\, dx$$

Using the same substitution as the previous example we get :

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^2}\,dt$$

$$x-1={-1 \over 2} \,\, \Rightarrow \,\, x={1\over 2}$$

$$x+y={2} \,\, \Rightarrow \,\, y={3\over 2}$$

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^2}\,dt=\frac{B({1 \over 2},{3 \over 2})}{2}=\frac{\Gamma({1 \over 2})\, \Gamma({3\over 2})}{2}= \frac{\Gamma({1 \over 2})\, \Gamma({1\over 2})}{4}=\frac{\pi}{4}$$

We might try to generalize :

$$\int^{\infty}_0 \frac{1}{(x^2+1)^n}\, dx \,\,\, \forall\,\, n> \frac{1} {2}$$

Using the same substitution again and again ..(boring) ..

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^n}\,dt$$

$$x-1={-1 \over 2} \,\, \Rightarrow \,\, x={1\over 2}$$

$$x+y={n} \,\, \Rightarrow \,\, y=n-{1\over 2}$$

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^n}\,dt=\frac{\Gamma ({1\over 2})\cdot \Gamma\left(n-{1\over 2}\right)}{2\Gamma(n)}$$

Now how to simplify $$\Gamma\left(n-{1\over 2}\right)$$ ?

But remember that by LDF that $$\Gamma\left(k+{1\over2}\right) =\frac{(2k)!\sqrt{\pi}}{4^k\,k!}$$

Now let $$k=n-1$$ so we have $$\Gamma\left(n-{1\over2}\right) =\frac{(2n-2)!\sqrt{\pi}}{4^{n-1}\,(n-1)!}=\frac{\Gamma\left(2n-1\right)\sqrt{\pi}}{4^{n-1}\,\Gamma(n)}$$

Substituting in our integral we have the following :

$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^n}\,dt=\frac{2\pi \,\cdot \Gamma\left(2n-1 \right)}{4^n \cdot \Gamma^2(n)}$$

$$\int^{\infty}_0 \frac{1}{(x^2+1)^n}\, dx \,=\frac{\pi \,\cdot \Gamma\left(2n-1 \right)}{2^{2n-1}\cdot \Gamma^2(n)}$$

It is easy to say that for $n\in \mathbb{Z}^+ \,$ we get a $\pi$ multiplied by some rational number ...

[HW] Verify the previous exercises using the formula ...

To be continued ...

4.Integration using special functions (continued)

4.2.Beta Function (continued):

Prove that :

$$\int_0^1\frac{x^n}{\sqrt{1-x}}\,dx=2\,\frac{(2n)!}{(2n+1)!}$$

Where the double factorial ! is defined as the following :

$$n! = \begin{cases}n\cdot(n-2)... 5\cdot 3\cdot 1&; \mbox{if }n>0 \, \text{odd} \\n\cdot(n-2)... 6\cdot 4\cdot 2&; \mbox{if } n>0 \, \text{even}\\1 &; \mbox{if } n=0,-1\end{cases}$$

Solution :

Now , we shall be familiar by just looking that this is the beta function :

$$\int_0^1 x^n\cdot (1-x)^{-\frac{1}{2}}\, dx$$

$$\fbox{[tex]x-1 = n \,\, \Rightarrow \,\, x=n+1\,\,\,\,$$

$$y-1= \frac{-1}{2}\,\, \Rightarrow \,\, y=\frac{1}{2}$$}[/tex]

$$\int_0^1 x^n\cdot (1-x)^{-\frac{1}{2}}\, dx=B(n+1,\frac{1}{2})$$

$$B\left( n+1,\frac{1}{2} \right)=\frac{\Gamma \left( \frac{1}{2} \right)\Gamma(n+1)}{\Gamma \left( n+\frac{3}{2}\right)}=\frac{2 \sqrt{\pi}\, \Gamma(n+1)}{(n+\frac{1}{2})\Gamma \left( n+\frac{1}{2} \right)}$$

Now you shall realize that we must use LDF:

$$\frac{\sqrt{\pi}\Gamma(n+1)}{(n+\frac{1}{2})\Gamma\left(n+\frac{1}{2}\right)}=\frac{2\sqrt{\pi}\, n!}{(2n+1)\sqrt{\pi}\, \frac{(2n)!}{4^n\, n!}}$$

$$\frac{2\, 2^{2n} (n!)^2}{(2n)!}=2\frac{2^{2n}(n\cdot(n-1)\,...\,3\cdot 2\cdot 1)^2}{(2n+1)\cdot 2n\cdot(2n-1)\, ... \, 3\cdot 2\cdot 1}$$

Now we should separate odd and even terms in the denomenator :

$$2\frac{2^{2n}(n\cdot(n-1)\,...\,3\cdot 2\cdot 1)^2}{(2n\cdot(2n-2)\, ... \, 4\cdot 2)((2n+1)\cdot (2n-1)\, ... \, 3\cdot 1)}$$

$$2\frac{(2n\cdot(2n-2)\,...\,6\cdot 4\cdot 2)^2}{(2n \cdot (2n-2)\, ... \, 4\cdot 2)((2n+1)\cdot (2n-1)\, ... \, 3\cdot 1)}=2\frac{(2n)!}{(2n+1)!}$$

Find the following integral:

$$\int^{\infty}_{-\infty}\, \left(1+\frac{x^2}{n-1}\right)^{-\frac{n}{2}}\,dx$$

Solution :

First we shall realize the evenness of the integral :

$$2\int^{\infty}_{0}\, \left(1+\frac{x^2}{n-1}\right)^{-\frac{n}{2}}\,dx$$

We should think of a substitution to get the second beta integral ...

Let $$t=\frac{x^2}{n-1}\,\,\, \Rightarrow \,\,\, dt =\frac{2x}{n-1}\,dx$$

$$\sqrt{n-1}\int^{\infty}_{0}\frac{t^{-\frac{1}{2}}}{(1+t)^{\frac{n}{2}}}\, \,dx$$

Now we see that our integral becomes so familiar :

$$\sqrt{n-1}\, B\left(\frac{1}{2},\frac{n-1}{2}\right)= \frac{\sqrt{\pi(n-1)}\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}$$

[HW] Prove $$\int^{\infty}_{0}\frac{x^{2m+1}}{(ax^2+c)^n} dx\,=\frac{m!\,(n -m -2)!}{2(n-1)!\, a^{m+1}\,c^{n-m-1}}$$

To be continued ...

4.Integration using special functions (continued)

4.2.Beta Function (continued):

Find the following integral:

$$\int^{\infty}_0\, \frac{x^{-p}}{x^3+1}\,dx$$

Solution :

We should foresee that this is the second beta integral.

Let us do the substitution :
$$x^3\,=t\,$$

$$\frac{1}{3}\, \int^{\infty}_0\, \frac{t^{-\frac{p+2}{3}}}{t+1}\,dt$$

Now we should find x, y :

$$x =\frac{1-p}{3}$$

$$y+x =1\,\,\, \Rightarrow \,\, y = 1-\frac{1-p}{3}$$

so we have our beta representation of the integral :

$$\frac{B\left(\frac{1-p}{3}, \frac{1-p}{3}\right)}{3}= \frac{\Gamma\left(\frac{1-p}{3}\right) \Gamma\left(1-\frac{1-p}{3}\right)}{3}$$

Now we should use the
ERF :

$$\frac{\Gamma \left( \frac{1-p}{3}\right) \Gamma\left(1-\frac{1-p}{3}\right)}{3}= \frac{\pi}{3\, \sin\left(\frac{\pi(1-p)}{3}\right)}= \, \frac{\pi}{3}\,\csc\left(\frac{\pi-\pi p}{3}\right)$$

Now let us try to find :

$$\int^{\frac{\pi}{2}}_0 \, \sqrt{\sin^{2}x} \,\, dx$$

Solution:

Rewrite as :

$$\int^{\frac{\pi}{2}}_0 \, \sin^{\frac{3}{2}}x \,\cos^{0}x\, dx$$

$$2x-1= \frac{3}{2}\,\, \Rightarrow \,\,\, x = \frac{5}{4}$$

$$2y-1= 0\,\, \Rightarrow \,\,\, y = \frac{1}{2}$$

$$\int^{\frac{\pi}{2}}_0 \, \sin^{\frac{3}{2}}x \, dx= \frac{B\left(\frac{5}{4}, \frac{1}{2}\right)}{2}=\frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{7}{4}\right)}$$

simplification is left to the reader as practice ...

Find the following integral :

$$\int^{\frac{\pi}{2}}_{0} (\sin x)^i \cdot (\cos x)^{-i}\, dx$$

Solution :

Don't be intimidated by the complex looking of the integral , we have the third beta integral : (Tmi)

$$2x-1= i \,\, \Rightarrow \,\,\, x = \frac{1+i}{2}$$

$$2y-1= -i \,\, \Rightarrow \,\,\, y = \frac{1-i}{2}$$

$$\,\frac{1}{2} \Gamma\left(\frac{1+i}{2}\right) \, \Gamma\left(\frac{1-i}{2}\right)$$

Now we see that we have to use ERF

$$\,\frac{1}{2} \Gamma\left(\frac{1+i}{2}\right) \, \Gamma\left(1-\frac{1+i}{2}\right)= \frac{\pi}{2\sin\left(\frac{\pi(1+i)}{2}\right)}= \frac{\pi}{2}\text{sech}\,\left( {\pi \over 2}\right)$$

Digamma function is explored in the next post

4.Integration using special functions (continued)

4.3.Digamma (psi) function :

General definition :

$$\psi(x) \, = \, \frac{\Gamma'(x)}{\Gamma(x)}$$

Assume that we have $$F(x)= \ln \left(\Gamma(x)\right)$$ then if we differentiate both sides we get $$f(x)=\frac{\Gamma'(x)}{\Gamma(x)}$$ then $f(x)= \psi(x)$ .

So clearly we see the strong relation between gamma and digamma through differentiation .

One of the most used results in solving integrals is the fact that can be obtained from the general definition $$\Gamma'(x)=\psi(x)\, \Gamma(x)$$.

Differentiating the gamma function is pretty unique since we can get the same function multiplied by digamma.

Assume we want to differentiate the following:

$$\frac{\Gamma(2x+1) }{\Gamma(x)}$$

We can use differentiating rule for quotients as the following :

$$\frac{2\Gamma'(2x+1)\Gamma(x)-\Gamma'(x)\Gamma(2x+1) }{\Gamma^2(x)}$$

Now we can use the result $$\Gamma'(x)=\psi(x)\, \Gamma(x)$$ :

$$\frac{2\Gamma(2x+1)\psi(2x+1)\Gamma(x)-\psi(x)\Gamma(x)\Gamma(2x+1) }{\Gamma^2(x)}$$

Now we can separate the numerator and simplify :

$$\frac{\Gamma(2x+1)}{\Gamma(x)} \left( 2\psi(2x+1)-\psi(x) \right)$$

Now that is interesting it is like we have just multilplied by the difference of psi if we neglect 2 .

A very interesting result is the following :

$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$

Poof :

we know by ERF that:

$$\Gamma(x)\Gamma(1-x)=\pi \csc(\pi x)$$

Now differentiate both sides :

$$\psi(x)\Gamma(x)\Gamma(1-x)-\psi(1-x)\Gamma(x)\Gamma(1-x)=-\pi^2 \csc(\pi x)\,\cot(\pi x)$$

Take $$-\Gamma(x)\Gamma(1-x)$$ as a common factor :

$$\Gamma(x)\Gamma(1-x)\left(\psi(1-x)-\psi(x)\right)=-\pi^2 \csc(\pi x)\,\cot(\pi x)$$

Now we see the ERF again so we can simplify to get :

$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$

That was basically a differentiating tutorial in the next post we will solve some integrals ...

[HW] $$\psi(1+x)-\psi(x)= \frac{1}{x}$$

4.Integration using special functions (continued)

4.3.Digamma function(continued) :

I gave this as HW :

$$\psi(1+x)-\psi(x) = \frac{1}{x}$$

Proof:

Let us start by the following :

$$\frac{\Gamma(1+x)}{\Gamma(x)}= x\text{ ---- (1) }$$

Now differentiate both sides :

$$\frac{\Gamma(1+x)}{\Gamma(x)}\left(\psi(1+x)-\psi(x)\right)= 1$$

$$\psi(1+x)-\psi(x)= \frac{\Gamma(x)}{\Gamma(1+x)}$$

The right hand side is the reciprocal of (1)

$$\psi(1+x)-\psi(x) = \frac{1}{x}$$

OK, Now that we have proved this identity we will use it in the next example.

Find the following integral :

$$\int^{\infty}_{0}\frac{\log(x) }{(1+x^2)^2}\, dx\text{ *}$$

Solution:

Let us try by finding the following :

$$\int^{\infty}_{0} \frac{x^a}{(1+x^2)^2}\, dx$$

Now use the following substitution : $$x^2= t$$

$$\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt$$

By the beta function this is equivalent to :

$$\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt = \frac{1}{2}B\left(\frac{a+1}{2},2-\frac{a+1}{2}\right)=\frac{1}{2}\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(2-\frac{a+1}{2}\right)$$

Now let the following :

$$F(a) =\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(2-\frac{a+1}{2}\right)$$

Differentiate with respect to a (Happy)

$$F'(a) =\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma\left(\frac{a+1}{2}\right)\Gamma \left(2-\frac{a+1}{2}\right)\left[\psi \left(\frac{a+1}{2}\right)-\psi \left(2-\frac{a+1}{2}\right) \right]$$

Now put a =0 so we have :

$$\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)\left[\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right)\right]$$

Now we use our identity :

$$\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right)=-\left(\psi \left(1+\frac{1}{2}\right)-\psi \left(\frac{1}{2}\right)\right)=-2$$

Also by some gamma manipulation we have :

$$\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)=\frac{1}{2}\,\Gamma^2 \left(\frac{1}{2}\right)=\frac{\pi}{2}$$

so the integral reduces to :

$$\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{-1}{2}}}{(1+t)^2}\, dt=-\frac{\pi}{4}$$

putting $$x^2= t$$ we have our result :

$$\int^{\infty}_{0}\frac{\log(x) }{(1+x^2)^2}\, dx=-\frac{\pi}{4}$$

To be continued ...

* In higher studies we usually donate the natural logarithm as log

4.Integration using special functions (continued)

4.3.Digamma function(continued) :

I introduced the
Weierstrass representation of gamma function :

$$\Gamma(x)=\frac{e^{-{\gamma}\,x}}{x}\,\prod^{\infty}_{n=1}\left(1+x/n \right)^{-1}e^{\frac{x}{n}}$$

Now to derive the digamma function :

First take the natural logarithm to both sides :

$$\log\left(\Gamma(x)\right)=-\gamma\, x\,-\,\log(x)\,+\sum^{\infty}_{n=1}-\ln\left(1+\frac{x}{n}\right)+{x\over n}$$

Now we shall differentiate with respect to x to get digamma :

$$\psi(x)=-\gamma\,-\frac{1}{x}\,+\sum^{\infty}_{n=1}\frac{\frac{-1}{n}}{1+\frac{x}{n}}+{1 \over n}$$

Further simplification will result in the following :

$$\psi(x)=-\gamma\,-\frac{1}{x}\,+\sum^{\infty}_{n=1}\frac{x}{n(n+x)}$$

Now let us evaluate some values :

1- $$\psi(1)=-\gamma\,-1 \,+\sum^{\infty}_{n=1}\frac{1}{n(n+1)}$$

$$\sum^{\infty}_{n=1}\frac{1}{n(n+1)}=1$$ (see why ? )

$$\fbox{[tex]\psi(1)\,=\,-\gamma$$}[/tex]

2- $$\psi\left(\frac{1}{2}\right)=-\gamma\,-2\,+\sum^{\infty}_{n=1}\frac{1}{n(2n+1)}$$

We need to find : $$\sum^{\infty}_{n=1}\frac{1}{n(2n+1)}$$

This requires the result $$\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n}=-\log(2)$$

So we can prove easily that :

$$\sum^{\infty}_{n=1}\frac{1}{n(2n+1)}=2-2\log(2)$$

hence , $$\fbox{ [tex]\psi \left( \frac{1}{2} \right)=-\gamma \,-\, 2\log (2)$$} [/tex]

Let us know have some examples on integrals :

Find the following integral :

$$\int^{\infty}_{0}e^{-a t}\, \log(t) \, dt$$

Let us start by noting that seems like a gamma function , but differentiated so :

$$F(b)= \int^{\infty}_{0}e^{-at}\, t^b\,dt$$

Now use the substitution $x=at$ we get

$$F(b)= \frac{1}{a}\int^{\infty}_{0}e^{-x}\, \left(\frac{x}{a}\right)^b\,dx$$

Now that looks exactly as the gamma definition so :

$$F(b)= \frac{1}{a}\int^{\infty}_{0}e^{-x}\, \left(\frac{x}{a}\right)^b\,dx=\frac{\Gamma(b+1)}{a^{b+1}}$$

Now differentiate with respect to b :

$$F'(b)= \frac{1}{a}\int^{\infty}_{0}e^{-x}\,\log\left(\frac{x}{a}\right) \left(\frac{x}{a}\right)^b\,dx=\frac{\Gamma(b+1) \psi (b+1)}{a^{b+1}}-\frac{\log(a)\Gamma(b+1)}{a}$$

Now put $b=0$ and $at=x$

$$\int^{\infty}_{0}e^{-at}\,\ln(t) \,dx=\frac{\psi(1)-\log(a)}{a}=-\frac{\gamma+\log(a)}{a}$$

To be continued ...

Integration lessons continued ...

4.Integration using special functions (continued)

4.3.Digamma (psi) function :

Prove the following identity :

$\displaystyle \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\,dz = \psi(a)$

Solution :

We begin with the double integral :

$$\displaystyle \int^{\infty}_0 \int^t_1 \,e^{-xz}\,dx\,dz=\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz$$

Using fubini theorem we also have :

$$\displaystyle \int^t_1 \int^{\infty}_0\,e^{-xz}\,dz \,dx = \int^t_1 \frac{1}{x}\,dx = \ln t$$

Hence we have the following :

$$\displaystyle \int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz = \ln(t) \text{ --------(1)}$$

We also know that :

$$\displaystyle \Gamma'(a) = \int^{\infty}_0 t^{a-1}e^{-t}\,\ln t \, dt \text{ --------(2)}$$

Substituting (1) in (2) we have :

$$\displaystyle \Gamma'(a) = \int^{\infty}_0 t^{a-1}e^{-t}\,\left(\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz \right)\, dt$$

$$\displaystyle \Gamma'(a) = \int^{\infty}_0 \int^{\infty}_{0}\frac{t^{a-1}e^{-t}e^{-z}-t^{a-1}e^{-t(z+1)}}{z}\, dz \, dt$$

Now we can use the fubini theorem :

$$\displaystyle \Gamma'(a) = \int^{\infty}_0 \int^{\infty}_{0}\frac{t^{a-1}e^{-t}e^{-z}-t^{a-1}e^{-t(z+1)}}{z}\, dt \, dz$$

$$\displaystyle \Gamma'(a) = \int^{\infty}_0 \frac{1}{z} \left( e^{-z}\int^{\infty}_{0}t^{a-1}e^{-t}\, dt-\int^{\infty}_0 t^{a-1}e^{-t(z+1)}\, dt \right)\, dz$$

But we can easily deduce using Laplace that :

$$\displaystyle \int^{\infty}_0 t^{a-1}e^{-t(z+1)}\,dt= \Gamma(a) \,(z+1)^{-a}$$

Aslo we have :

$$\displaystyle \int^{\infty}_{0}t^{a-1}e^{-t}\, dt=\Gamma(a)$$

Hence we can simplify our integral to the following :

$$\displaystyle \Gamma'(a) = \Gamma(a)\, \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z} dz$$

$$\displaystyle \frac{\Gamma'(a)}{\Gamma(a)} = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\, dz = \psi(a)$$

[HW] Prove that $$\displaystyle \int^{\infty}_{0}\, \frac{e^{-ax}-e^{-bx}}{x}\, dx=\ln\left({b\over a}\right)$$

you may use differentiation under the integral sign or double integration

[HW] Find $$\displaystyle \int^{\infty}_0 \left(e^{-x} -\frac{1}{x+1}\right) \frac{dx}{x}$$

To be continued ...

Integration lessons continued

4.Integration using special functions (continued)

4.3.Digamma function(continued) :

We have the following digamma property :

$\displaystyle \psi(s+1)\,=\, -\gamma \,+\, \int^{1}_{0}\frac{1-x^s}{1-x}$

Prove the following integral :

$$\displaystyle \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$

Solution :

As crazy as it looks , it becomes very easy to solve if we know how to start !

First note that since there is a log in the denominator that gives as an idea to differentiate ...

$$\displaystyle \text{Let : }F(c)=\int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$

Differentiate with respect to c :

$$\displaystyle F'(c)=\int_0^1 \frac{(1-x^a)(1-x^b)x^c}{(1-x)}dx$$

$$\displaystyle F'(c)=\int_0^1 \frac{(1-x^a-x^b+x^{a+b})x^c}{(1-x)}dx$$

$$\displaystyle F'(c)=\int_0^1 \frac{x^c\,-\,x^{a+c}\,-\,x^{b+c}\,+\,x^{a+b+c}}{(1-x)}dx$$

$$\displaystyle F'(c)=\int_0^1 \frac{(x^c\,-1)\,+\,(1-\,x^{a+c})\,+\,(1-\,x^{b+c})\,+\,(x^{a+b+c}-1)}{(1-x)}dx$$

$$\displaystyle F'(c)=-\int_0^1 \frac{1-x^c\,}{1-x}\,dx+\int_0^1\frac{1-x^{a+c}}{1-x}\,dx+\int_0^1\frac{1-x^{b+c}}{1-x}\,dx-\int_0^1\frac{1-x^{a+b+c}}{1-x}dx$$

Now use the following result :

$$\displaystyle \psi(s+1)=-\gamma+\int^{1}_0\frac{1-x^s}{1-x}\,dx$$

$$\displaystyle F'(c)=-\psi(c+1)-\gamma+\psi(a+c+1)+\gamma+\psi(b+c+1)+\gamma-\psi(a+b+c+1)-\gamma$$

$$\displaystyle F'(c)=-\psi(c+1)+\psi(a+c+1)+\psi(b+c+1)-\psi(a+b+c+1)$$

Integrate with respect to c we have :

$$\displaystyle F(c)=-\log\left[\Gamma(c+1)\right]+\log \left[\Gamma(a+c+1)\right] +\log\left[\Gamma(b+c+1)\right]-\log \left[\Gamma(a+b+c+1)\right] +C_1$$

Which reduces to :

$$\displaystyle \log\left[\frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma (c+1) \Gamma (a+b+c+1)} \right] +C_1$$

Now put c= 0 we have :

$$\displaystyle 0=\log \left[ \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\right]+C_1$$

$$\displaystyle C_1=-\log \left[ \frac{\Gamma(a+1)\Gamma(a+1)}{\Gamma(a+b+1)}\right]$$

So we have the following :

$$\displaystyle F(c)=\log\left[ \frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma(c+1) \Gamma (a+b+c+1)}\right] -\log\left[\frac{\Gamma(a+1)\Gamma(a+1)}{\Gamma(a+b+1)}\right]$$

Hence we have the result :

$$\displaystyle \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$

To be Continued ...

Integration lessons continued ...

4.Integration using special functions (continued)

4.3.Digamma function(continued) :

We continue with some more exercises :

Find the following integral :

$\displaystyle \int^{\infty}_0\left(e^{-bx}-\frac{1}{1+ax}\right)\,\frac{dx}{x}$​

Solution :

Let us first use the substitution t = ax so we get the following :

$$\displaystyle \int^{\infty}_0\left(e^{-\frac{bt}{a}}-\frac{1}{1+t}\right)\,\frac{dt}{t}$$

Now we must realize the result we get earlier in this series that :

$$\displaystyle \int^{\infty}_0 \frac{e^{-x}-(1+x)^{-a}}{x}\,dx = \psi(a)$$

But there we don't have $$\displaystyle e^{-x}$$ so let us add and subtract it :

$$\displaystyle \int^{\infty}_0\left(e^{-t}-e^{-t}+e^{-\frac{bt}{a}}-\frac{1}{1+t}\right)\,\frac{dt}{t}$$

$$\displaystyle \int^{\infty}_0 \left(e^{-t}-\frac{1}{1+t}\right)\,\frac{dt}{t}+\int^{\infty}_0 \frac{e^{-\frac{bt}{a}}-e^{-t}}{t}\, dt$$

$$\displaystyle \int^{\infty}_0 \left(e^{-t}-\frac{1}{1+t}\right)\frac{dt}{t}=-\gamma$$

We already also proved several posts ago that :

$$\displaystyle \int^{\infty}_0 \frac{e^{-\frac{bt}{a}}- e^{-t}}{t}\, dt= -\log\left(\frac{b}{a}\right) = \log\left(\frac{a}{b}\right)$$

Hence the result :

$$\displaystyle \int^{\infty}_0 \left(e^{-bx}-\frac{1}{1+ax} \right)\, \frac{dx}{x} = \log\left(\frac{a}{b} \right)-\gamma$$

Prove the following integral :

$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\text{coth}(x) \right) \, dx = \psi\left(\frac{a}{2}\right) - \ln\left(\frac{a}{2}\right) + \frac{1}{a}$​

Solution :

We know the following hyperoblic identity : (hopefully :D)

$$\displaystyle \text{coth}(x)= \frac{1+e^{-2x}}{1-e^{-2x}}$$

$$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\frac{1+e^{-2x}}{1-e^{-2x}} \right) \, dx$$

Now let 2x =t so we have :

$$\displaystyle \int^{\infty}_0 \, e^{-\left(\frac{at}{2}\right)} \left(\frac{1}{t}-\frac{1+e^{-t}}{2(1-e^{-t})} \right) \, dt$$

$$\displaystyle \int^{\infty}_0 \, \frac{e^{-\left(\frac{at}{2}\right)}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}+e^{\left(-\frac{at}{2}-t\right)}}{2(1-e^{-t})} \, dt$$

I will add and subtract some terms :

$$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}+e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t}\,-\,\frac{e^{-\left(\frac{at}{2}\right)}+e^{-\left(\frac{at}{2}\right)}-e^{-\frac{at}{2}}+e^{-\left(\frac{at}{2}\right)-t}}{2(1-e^{-t})} \, dt$$

$$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}}{1-e^{-t}}\,dt +\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}-e^{\left(-\frac{at}{2}-t\right)}} {2(1-e^{-t})}\, dt +\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t} \, dt$$

First : We use the identity :

$$\displaystyle \psi \left(x\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(x t\right)}}{1-e^{-t}}\, dt$$

$$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}}{1-e^{-t}}\, dt=\psi \left(\frac{a}{2}\right)$$

The second integral can easily be proven :

$$\displaystyle \int^{\infty}_0\frac{e^{-\left(\frac{at}{2}\right)}-e^{-\left(\frac{at}{2}-t\right)}}{2(1-e^{-t})}=\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}}{2}\,dt=\frac{1}{a}$$

We also proved earlier that :

$$\displaystyle \int^{\infty}_0\frac{e^{-t}\,-\,e^{-st}}{t} \, dt = \ln \left( s \right)$$

$$\displaystyle \int^{\infty}_0\frac{e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t} \, dt = -\ln \left(\frac{a}{2}\right)$$

Hence we have the full integral :

$$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\text{coth}(x) \right) \, dx = \psi\left(\frac{a}{2}\right) - \ln\left(\frac{a}{2}\right) + \frac{1}{a}$$

[HW] Prove : $$\displaystyle \psi \left(x\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(x t\right)}}{1-e^{-t}}\, dt$$

Zeta function ,,, is on the way Integration lessons (continued )

4.Integration using special functions (continued)

4.4.Riemann Zeta function :

You surely had heard about this interesting function, not only has it resulted in many interesting founding but also the yet to be solved Riemann Hypothesis makes it one of the most celebrated functions in history of mathematics. Its interesting aspect is the relation to prime numbers.

In this tutorial we will not go so deep in the proofs and analyticity of it, we will try to describe its properties and mainly focus on the integral representation and its relation to other functions.

Let us first start by defining the zeta function as the following :

$\displaystyle \zeta(s) \,=\, \sum_{n=1}^{\infty}\,\frac{1}{n^s}\,\,\, \,\, \text{Re}(s)>1$

There are more general representation of the zeta function but we will stick to the one we defined above. For the first glance it seems that zeta has no relations to other functions but it turned out that it has a strong relation to other functions such as gamma and digamma functions.

4.4.1 zeta and gamma representation :

$\displaystyle \Gamma(s) \, \zeta(s) \, = \int^{\infty}_{0}\, \frac{t^{s-1}}{e^t-1}\, dt$​

This relation turns out to be so much interesting since we can evaluate the right-hand integral using known results for both zeta and gamma.

This most famous value for zeta function is when $s=2$ which represents the infinite sum of reciprocals of squares :

$\displaystyle \zeta(2) =\sum^{\infty}_{n=1} \frac{1}{n^2}\,= \frac{\pi^2}{6}$

Actually we can derive all the values of $\displaystyle \zeta(2k) \,\,\,\, k\in \mathbb{Z}$ but unfortunately there is no known way to find the zeta for odd integers $\displaystyle \zeta(2k+1)\,$ .

Now let us use this result to find some integrals :

$\displaystyle \int^{\infty}_0 \frac{t}{e^t-1} \,dt$​

It follows directly from the gamma-zeta relation that taking $s=2$ we have :

$\displaystyle \Gamma(2) \, \zeta(2) \, = \int^{\infty}_{0}\, \frac{t}{e^t-1}\, dt$

$$\displaystyle \int^{\infty}_{0}\, \frac{t}{e^t-1}\, dt=\zeta(2)=\frac{\pi^2}{6}$$

Solve the following integral :

$\displaystyle \int_0^{\frac{\pi}{2}} \,\frac{\log \left(\sec(x) \right)}{\tan(x)} \,dx$​

use the substitution : $\displaystyle \sec(x)=e^t$

$$\displaystyle \int_0^{\infty} \, \frac{t}{e^{2t}-1}\,dt\, =\, \frac{1}{4} \cdot \int_0^{\infty}\,\frac{t}{e^t-1}\,dt=\frac{\pi^2}{24}$$

Another similar relation between zeta and gamma is the following identity :

$\displaystyle \Gamma(s)\, \zeta(s) \, (1-2^{1-s})\, = \int^{\infty}_0\, \frac{t^{s-1}}{e^t+1}\,$​

It follows from that equation that :

$$\displaystyle \int^{\infty}_0\, \frac{t}{e^t+1}\,dt = \frac{\pi^2}{12}$$

To be continued ...

Integration lessons continued ...

4.4.2.Zeta function and Bernoulli numbers

We will see in this section some values of the zeta function using an equation due to Euler , but let us first define the Bernoulli numbers .

It is usually defined as $$\displaystyle B_{k}$$ and the easiest way to derive them is find the coefficients of the power series of $$\displaystyle \frac{x}{e^x-1}$$

Definition

$$\displaystyle \frac{x}{e^x-1} = \sum_{k \geq 0} \frac{B_k}{k!}x^k$$​

Now let us derive some values for the Bernoulli numbers , rewrite the power series as

$$\displaystyle x = (e^x-1) \, \sum_{k\geq 0} \frac{B_k}{k!}x^k$$

$$\displaystyle x = \left( x+\frac{1}{2!}x^2 + \frac{1}{3!} x^3+ \frac{1}{4!} x^4+\cdots \right) \cdot \left( B_0 +B_1 \, x+\frac{B_2}{2!}\,x^2 +\frac{B_3}{3!} x^3+ \cdots \right)$$

$$\displaystyle x = B_0 x + \left(B_1+\frac{B_0}{2!} \right)x^2 + \left(\frac{B_0}{3!}+ \frac{B_2}{2!}+\frac{B_1}{2!}\right)x^3 +\left( \frac{B_0}{4!}+\frac{B_1}{3!}+\frac{B_2}{2!\, 2!}+\frac{B_3}{3!}\right) x^4+ \cdots$$

By comparing the terms we get the following values

$$\displaystyle {B_0 = 1 \, , \, B_1 =-\frac{1}{2} \, + \, B_2 = \frac{1}{6} \, , \, B_3=0 \, , \, B_4 = -\frac{1}{30} , \cdots }$$

Actually continuing with this we deduce

• $$\displaystyle B_{2k+1}=0 \,\, \, \, \, \,\,\, \, \, \, \, \forall \,\, \, \, \, \, k\in \mathbb{Z}^+$$
• Every Bernoulli number depends on all the numbers before it so it can be defined recursively .
• $$\displaystyle B_{2k} \, , \, k\geq 1$$ are alternating .

Now according to Euler we have the following interesting result :

$$\displaystyle \zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$​

PROOF

Let us start by the following definition due to Euler

$$\displaystyle \frac{\sin(z)}{z} = \prod_{n\geq1} \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$

Take the logarithm to both sides

$$\displaystyle \ln(\sin(z)) - \ln(z) = \sum_{n\geq1} \ln \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$

By differentiation with respect to z

$$\displaystyle \cot(z) -\frac{1}{z} = \sum_{n\geq1} \frac{-2 \frac{z}{n^2 \, \pi^2 }}{1-\frac{z^2}{n^2 \, \pi^2}}$$

By simple algaberic manipulation we have

$$\displaystyle z\cot(z) = 1+2\sum_{n\geq1} \frac{z^2}{z^2-n^2 \, \pi^2}$$

$$\displaystyle z\cot(z) = 1-2\sum_{n\geq1} \frac{z^2}{ n^2 \, \pi^2} \left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right)$$

Now using the power series expansion

$$\displaystyle \frac{1}{1-\frac{z^2}{\pi^2 \, n^2}}= \sum_{k\geq 0} \frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$$ converges for $$\displaystyle |z|< \pi \, n$$

$$\displaystyle \frac{z^2}{ n^2 \, \pi^2}\left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right) = \sum_{k\geq 1}\frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$$ notice the change in the index !

So now the sums becomes

$$\displaystyle z\cot(z) = 1-2\sum_{n\geq1} \sum_{k \geq 1} \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$$

Now if we invert the order of summation we have

$$\displaystyle z\cot(z) = 1-2\sum_{k\geq1} \sum_{n \geq 1} \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$$ (Wow do you notice !)

$$\displaystyle z\cot(z) = 1-2\sum_{k\geq1} \frac{\zeta(2k) }{\pi^{2k}}z^{2k} \text{ }\cdots(1)$$

Euler didn't stop here , he used power series estimation for $$\displaystyle z\cot(z)$$ using the Bernoulli numbers

staring by the equation $$\displaystyle \frac{x}{e^x-1} = \sum_{k\geq 0} \frac{B_k}{k!}x^k$$

by putting $$\displaystyle x=2iz$$ we have

$$\displaystyle \frac{2iz}{e^{2iz}-1} = \sum_{k\geq0} \frac{B_k}{k!}{(2iz)}^k$$

Which can be reduced directly to the following by noticing that $$\displaystyle B_{2k+1}=0, \,\,\, k>0$$

$$\displaystyle z \cot(z) = 1- \sum_{k\geq 1}(-1)^{k-1} B_{2k} \frac{2^{2k}}{(2k)!}z^{2k} \text{ } \cdots(2)$$

The result is immediate by comparing (1) and (2) $$\displaystyle \square$$

[HW] find $$\displaystyle \zeta(4) \, ,\, \zeta(6) \, , \, B_5 \, , \, B_6$$

To be continued ...

Re: Integration lessons continued ...

4.4.3. Hurwitz zeta and polygamma functions

Hurwitz zeta is a generalization of the zeta function by adding a parameter . This intimate relation between the two functions arises from multiple differentiations of the the digamma function .

Let us first start by defining the Hurwitz zeta function

Definition

$$\displaystyle \zeta(a,z) \, = \, \sum_{n\geq 0} \frac{1}{(n+z)^a}$$​

Note : according to this definition we have $$\displaystyle \zeta(a,1) = \zeta(a)$$

Let us define the polygamma function as the function produced by differentiating the digamma function and it is often denoted by $$\displaystyle \psi_{n}(z) \,\,\,\, \forall \,\, n\geq 0$$ . We define the digamma function by setting $n=0$ so it's denoted by $\psi_0(z)$ .

Other values can be found by the following recurrence relation $$\displaystyle \psi'_{n}(z) = \psi_{n+1}(z)$$ , so we have $$\displaystyle \psi_{1}(z) = \psi'_{0}(z)$$

We have the following relation between Hurwitz zeta and the polygamma function

DEFINITION

$$\displaystyle \psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z) \,\,\, \, \forall \,\, n\geq 1$$​

PROOF

We have already proved the following relation

$$\displaystyle \psi_{0}(z) = -\gamma -\frac{1}{z}+ \sum_{n\geq 1}\frac{z}{n(n+z)}$$

This can be written as the following

$$\displaystyle \psi_{0}(z) = -\gamma + \sum_{k\geq 0}\frac{1}{k+1}-\frac{1}{k+z}$$

By differentiating with respect to $z$

$$\displaystyle \psi_{1}(z) = \sum_{k\geq 0}\frac{1}{(k+z)^2}$$

$$\displaystyle \psi_{2}(z) = -2\sum_{k\geq 0}\frac{1}{(k+z)^3}$$

$$\displaystyle \psi_{3}(z) = 2 \cdot 3 \, \sum_{k\geq 0}\frac{1}{(k+z)^4}$$

$$\displaystyle \psi_{4}(z) = -2 \cdot 3 \cdot 4 \,\sum_{k\geq 0}\frac{1}{(k+z)^5}$$

$$\displaystyle \text{ }\cdot$$

$$\displaystyle \text{ }\cdot$$

$$\displaystyle \text{ }\cdot$$

$$\displaystyle \psi_{n}(z) = (-1)^{n+1}n!\,\sum_{k\geq 0}\frac{1}{(k+z)^{n+1}}$$

We realize the RHS is just the Hurwitz zeta function

$$\displaystyle \psi_{n}(z) = (-1)^{n+1}n!\,\zeta(n+1,z)$$

As required to prove $$\displaystyle \square$$.

By setting $z=1$ we have an equation in terms of the ordinary zeta function

$$\displaystyle \psi_{n}(1) = (-1)^{n+1}n!\,\zeta(n+1)$$

Now since we already proved in the preceding section that

$$\displaystyle \zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$

we can easily verify the following

$$\displaystyle \psi_{2k-1}(1)= (2k-1)!\, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$

$$\displaystyle \psi_{2k-1}(1)= (-1)^{k-1} B_{2k} \frac{2^{2k-2}}{k}{\pi}^{2k} \,\,\, \, , \,\, k\geq 1$$

This can be used to evaluate some values for the polygamma function

$$\displaystyle \psi_{1}(1)= \frac{\pi^2}{6} \,\, , \,\, \psi_{3}(1) = \frac{\pi^4}{15}$$

Other values can be evaluated in terms of the zeta function

$$\displaystyle \psi_{2}(1)= -2 \zeta(3) \,\, , \,\, \psi_{4}(1) = -24 \zeta(5)$$

To Be Continued ...

Re: Integration lessons continued ...

4.4.4. integrals involve zeta computations

Prove that

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,x \sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx = \frac{\pi}{16}-\frac{\pi^3}{192}$$

PROOF

Start by the change of variable, $$\displaystyle t=\frac{\pi}{2}-x$$

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,x \sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx\,=\frac{\pi}{4}\int^{\frac{\pi}{2}}_0 \,\sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx \text{ }\cdots (1)$$

We need to find :

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx$$

Let us start by the following :

$$\displaystyle F(a,b)=2\int^{\frac{\pi}{2}}_0 \,\sin^{2a-1}(x)\cos^{2b-1}(x) \, dx\,=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

Now let us differentiate with respect to $a$

$$\displaystyle \frac{\partial}{\partial a}(F(a,b))=4\int^{\frac{\pi}{2}}_0 \,\sin^{2a-1}(x)\cos^{2b-1}(x) \log(\sin x)\, dx\,=\frac{\Gamma(a)\Gamma(b) \left(\psi_0(a)-\psi_0(a+b)\right)}{\Gamma(a+b)}$$

Differentiate again but this time with respect to $b$

$$\displaystyle \frac{\partial}{\partial b}\left(F_a(a,b) \right)=8\int^{\frac{\pi}{2}}_0 \,\sin^{2a-1}(x)\cos^{2b-1}(x) \log(\sin x)\, \log( \cos x)dx\,$$

$$\displaystyle =\frac{\Gamma(a) \Gamma(b) \left( \psi_0^2(a+b) +\psi_0(a)\psi_0(b)-\psi_0(a)\psi_0(a+b)-\psi_0(b)\psi_0(a+b) +\psi_1(a+b)\right)}{\Gamma(a+b)}$$

putting $$\displaystyle a=b=1$$ we have the following

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin (x)\cos(x) \log(\sin x)\, \log( \cos x)dx\,= \frac{ \psi_0^2(2) +\psi_0^2(1)-\psi_0(1)\psi_0(2)-\psi_0(1)\psi(2) -\psi_1(2)}{8}$$

By simple algebra we arrive to

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin (x)\cos(x) \log(\sin x)\, \log( \cos x)dx\,= \frac{ (\psi_0(2) -\psi_0(1))^2-\psi_1(2)}{8}$$

We can easily see that

• $$\displaystyle \psi_0(1) = -\gamma$$
• $$\displaystyle \psi_0(2) =1 -\gamma$$

Now to evaluate $$\displaystyle \psi_1(2)$$ , we have to use the zeta function

we have already established the following relation :

$$\displaystyle \psi_1(z) = \sum_{k\geq 0} \frac{1}{(n+z)^2}$$

Now putting $$\displaystyle z =2$$ we have the following

$$\displaystyle \psi_1(2) = \sum_{k\geq 0} \frac{1}{(k+2)^2}$$

Let us write the first few terms in the expansion

$$\displaystyle \sum_{k\geq 0} \frac{1}{(k+2)^2} = \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+ \cdots$$

we see this is similar to $$\displaystyle \zeta(2)$$ but we are missing the first term

$$\displaystyle \psi_1(2) = \zeta(2)-1 = \frac{\pi^2}{6} -1$$

Collecting all these information together we have

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin (x)\cos(x) \log(\sin x)\, \log( \cos x)dx\,= \frac{1}{4}-\frac{\pi^2}{48}$$

substituting in (1) we get :

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,x\sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx =\frac{\pi}{16}-\frac{\pi^3}{192}$$

To Be Continued ...

Re: Integration lessons continued ...

4.4.5 Dirichlet eta function( an alternating form )

Dirichlet eta function is the alternating form of the zeta function

DEFINITION :

$$\displaystyle \eta(s) = \sum_{n\geq 1} \frac{(-1)^{n-1}}{n^s}$$​

The alternating form of the zeta function is easier to compute once we have established the main results of the zeta function because the alternating form is related to the zeta function through the relation

$$\displaystyle \eta(s) = \left( 1-2^{1-s} \right) \zeta(s)$$​

PROOF

We will start by the RHS

$$\displaystyle \left( 1-2^{1-s} \right) \zeta(s) = \zeta(s) - 2^{1-s} \zeta(s)$$

which can be written as sums of series

$$\displaystyle \sum_{n\geq 1} \frac{1}{n^s} - \frac{1}{2^{s-1}}\sum_{n\geq 1} \frac{1}{n^s}$$

$$\displaystyle \sum_{n\geq 1} \frac{1}{n^s} - 2\sum_{n\geq 1} \frac{1}{(2n)^s}$$

Clearly we can see that we are subtracting even terms twice , this is equivalent to

$$\displaystyle \sum_{n\geq 1} \frac{1}{(2n+1)^s} - \sum_{n\geq 1} \frac{1}{(2n)^s}$$

This looks easier to understand if we write the terms

$$\displaystyle \left(1+\frac{1}{3^s}+\frac{1}{5^s}+ \cdots\right) - \left( \frac{1}{2^s}+\frac{1}{4^s} + \frac{1}{6^s}+\cdots \right)$$

Rearranging the terms we establish the alternating form

$$\displaystyle 1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots = \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^s} =\eta(s) \, \square$$

We know feel tempted to evaluate some values

$$\displaystyle \eta(2) = \left(1-\frac{1}{2} \right) \zeta(2) = \frac{\pi^2}{12}$$

Actually there is a nice integration formula similar to that we had for zeta

$$\displaystyle \eta(s) \, \Gamma(s) = \int^{\infty}_{0} \frac{t^{s-1}}{e^t+1} \, dt$$​

PROOF :

Start by the RHS

$$\displaystyle \int^{\infty}_{0} \frac{t^{s-1}}{e^t+1} \, dt = \int^{\infty}_0 \frac{e^{-t} t^{s-1}}{1+e^{-t}} \, dt$$

Now using the power expansion we arrive to

$$\displaystyle \int^{\infty}_0 e^{-t} t^{s-1}dt \left(\sum_{n\geq 0} (-1)^n e^{-nt} \right)$$

$$\displaystyle \sum_{n\geq 0}(-1)^n\int^{\infty}_0 e^{-(n+1) t} t^{s-1}dt$$

Using Laplace transform we can solve the inner integral

$$\displaystyle \int^{\infty}_0 e^{-(n+1) t} t^{s-1}dt = \frac{\Gamma(s)}{(n+1)^s}$$

Hence we have the following

$$\displaystyle \Gamma(s) \sum_{n\geq 0}\frac{(-1)^n}{(n+1)^s}=\Gamma(s) \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^s}=\Gamma(s)\, \eta(s) \square$$

An easy result of the above integral

$$\displaystyle \int^{\infty}_{0}\frac{t}{e^t+1} \, dt = \Gamma(2)\, \eta(2) = \frac{\pi^2}{12}$$

We shall look at the polylogarithms in the next thread

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Re: Integration lessons continued ...

4.Integration using special functions (continued)

4.5. Polylogarithm

As the name suggests, this special function is closely related to the logarithms .

DEFINITION

$$\displaystyle \operatorname{Li}_{n}(z) = \sum_{k\geq 1} \frac{z^k}{k^n}$$​

Note : As we see , we use $$\displaystyle \operatorname{Li}_{n}(z)$$ to denote the polylogarithm . The name contains two parts , (poly) because we can choose different $n$ and produce many functions . (logarithm) because we can express it integrals of logarithms .

Through that representation we can see how closely it is related to the zeta function

$$\displaystyle \operatorname{Li}_{n}(1) = \sum_{k \geq 1} \frac{1}{k^n}= \zeta(n)$$

In particular we have for $n=2$

$$\displaystyle \operatorname{Li}_{2}(1) = \zeta(2) = \frac{\pi^2}{6}$$

Also we can relate it to the eta function though $z=-1$

$$\displaystyle \operatorname{Li}_{n}(-1) = \sum_{k\geq 1} \frac{(-1)^k}{k^n}=-\eta(n)$$

Also we shall see how to relate this function to the logarithm

putting $$\displaystyle n=1$$ we have the following

$$\displaystyle \operatorname{Li}_{\, 1}(z) = \sum_{n\geq 1} \frac{z^k}{k}$$

The power expansion on the left is quire famous

$$\displaystyle \sum_{n\geq 1} \frac{z^k}{k} = - \log(1-z)$$

There is an interesting recursive representation of this function

$$\displaystyle \operatorname{Li}_{\, n+1}(z) = \int^z_0 \frac{\operatorname{Li}_{\,n}(t) }{t}\, dt$$​

PROOF

Using the series representation we have

$$\displaystyle \int^z_0 \frac{1}{t} \left( \sum_{k\geq 1} \frac{t^k}{k^n}\, \right) dt$$

$$\displaystyle \sum_{k\geq 1}\frac{1}{k^n} \int^z_0 t^{k-1} \, dt$$

Integrating term by term we have

$$\displaystyle \sum_{k\geq 1}\frac{z^{k}}{k^{n+1}} \, = \operatorname{Li}_{\, n+1}(z) \,\, \square$$

We can differentiate the result to obtain

$$\displaystyle \frac{\partial}{\partial z}\operatorname{Li}_{\, n+1}(z) = \frac{1}{z} \operatorname{Li}_{\,n}(z)$$

Square formula

$$\displaystyle \operatorname{Li}_{\,n}(-z) + \operatorname{Li}_{\,n}(z) = 2^{1-n} \,\operatorname{Li}_{\,n}(z^2)$$​

PROOF

As usual we write the series representation of the LHS

$$\displaystyle \sum_{k\geq 1} \frac{z^k}{k^n}+\sum_{k\geq 1} \frac{(-z)^k}{k^n}$$

Listing the first few terms

$$\displaystyle z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\cdots +\left(-z+\frac{z^2}{2^n}-\frac{z^3}{3^n}+\cdots \right)$$

Clearly the odd terms will cancel so we are left with

$$\displaystyle 2\frac{z^2}{2^n}+2\frac{z^4}{4^n}+2\frac{z^6}{6^n}+\cdots$$

By simple manipulation

$$\displaystyle 2^{1-n} \left( z^{2}+\frac{(z^2)^2}{2^n}+\frac{(z^2)^3}{3^n}+ \cdots \right)$$

$$\displaystyle 2^{1-n} \sum_{k \geq 1} \frac{(z^2)^{k}}{k^n} = 2^{1-n} \, \operatorname{Li}_{\, n}(z^2) \, \square$$

[HW] prove that $$\displaystyle \operatorname{Li}_{\, 2}(z) = -\int^z_0 \frac{\log(1-t)}{t}\, dt$$ Dilogarithm

To Be Continued ...

Re: Integration lessons continued ...

4.5.1.Dilogarithms

Of all polylogarithms $$\displaystyle \operatorname{Li}_2(z)$$ is the most interesting one , in this section we will see why.

DEFINITION

$$\displaystyle \operatorname{Li}_2(z) = \sum_{k\geq 1} \frac{z^k}{k^2} = - \int^z_0 \frac{\log(1-t)}{t}\, dt$$​

The curious reader should try to prove the integral representation using the recursive definition we introduced in the previous section .

Some Functional equations

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) + \operatorname{Li}_2(-z) = - \frac{1}{2}\log^2(z)-\frac{\pi^2}{6}$$

PROOF

We will start by the following

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) = -\int^{\frac{-1}{z}}_0 \frac{\log(1-t)}{t}\, dt$$

Differentiate with respect to $z$

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2\left(\frac{-1}{z}\right) = \frac{1}{z^2} \left(-\frac{\log \left(1+\frac{1}{z} \right)}{\frac{-1}{z}} \right) = \frac{\log\left( 1+ \frac{1}{z} \right)}{z} = \frac{\log(1+z) - \log(z)}{z}$$

Now integrate with respect to $z$

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) = \int^{-z}_0 \frac{\log(1-t)}{t} \, dt - \frac{1}{2} \log^2(z) \,+ C$$

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) = -\operatorname{Li}_2 {-z} - \frac{1}{2} \log^2(z) \,+ C$$

To find the constant $C$ let $z = 1$

$$\displaystyle C= 2\operatorname{Li}_2\left(-1\right)$$

Now we must be aware that

$$\displaystyle \displaystyle \operatorname{Li}_{2}(-1) =-\eta(2) = \frac{-\pi^2}{12}$$

Hence we have

$$\displaystyle C= \frac{-\pi^2}{6}$$

which proves the result by simple rearrangement

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right)+\operatorname{Li}_2(-z) = -\frac{1}{2} \log^2(z) -\frac{\pi^2}{6} \, \square$$

We can let $$\displaystyle z=-1$$

$$\displaystyle 2\operatorname{Li}_2\left(1\right)=-\frac{1}{2} \log^2(-1) -\frac{\pi^2}{6} \,$$

knowing that $$\displaystyle \log(-1) = i \pi$$

$$\displaystyle 2\operatorname{Li}_2\left(1\right)=\frac{\pi^2}{2} -\frac{\pi^2}{6} \, = \frac{\pi^2}{3}$$

Hence we have $$\displaystyle \operatorname{Li}_2\left(1\right) = \frac{\pi^2}{6}$$ as expected .

Another function equation

$$\displaystyle \operatorname{Li}_2(z) + \operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,\,\, 0<z<1$$

PROOF

Start by the following

$$\displaystyle \operatorname{Li}_2\left(z\right) = -\int^{z}_0 \frac{\log(1-t)}{t} \, dt$$

Now integrate by parts to obtain

$$\displaystyle \operatorname{Li}_2\left(z\right)= -\int^z_0 \frac{\log(t)}{1-t} \, dt -\log(z) \log(1-z)$$

To solve $$\displaystyle \int^z_0 \frac{\log(t)}{1-t} \, dt \,$$ let $$\displaystyle t = 1-x$$

$$\displaystyle -\int^{1-z}_{1} \frac{\log(1-x)}{x} \, dx$$

For $$\displaystyle 0<z <1$$

$$\displaystyle \int^{1}_{1-z} \frac{\log(1-x)}{x} \, dx = \int^1_0 \frac{\log(1-x)}{x}\, dx - \int_0^{1-z} \frac{\log(1-x)}{x}\, dx$$

Now it is easy to see that

$$\displaystyle \int^{1}_{1-z} \frac{\log(1-x)}{x} \, dx =-\operatorname{Li}_2(1)+\operatorname{Li}_2(1-z)$$

$$\displaystyle \operatorname{Li}_2\left(z\right)= \operatorname{Li}_2(1)-\operatorname{Li}_2(1-z) -\log(z) \log(1-z)$$

$$\displaystyle \operatorname{Li}_2\left(z\right)+\operatorname{Li}_2(1-z)\, = \, \operatorname{Li}_2(1)-\log(z) \log(1-z)$$

Now since $$\displaystyle \operatorname{Li}_2(1) = \frac{\pi^2}{6}$$

$$\displaystyle \operatorname{Li}_2\left(z\right)+\operatorname{Li}_2(1-z)\, = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\, \square$$

We can easily deduce that for $$\displaystyle z=\frac{1}{2}$$

$$\displaystyle 2\operatorname{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{6}-\log^2\left(\frac{1}{2}\right) \,\,$$

$$\displaystyle \operatorname{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{12}-\frac{1}{2}\log^2 \left(\frac{1}{2}\right)$$

To be continued ...

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Re: Integration lessons continued ...

4.5.1.Dilogarithms (continued)

In this section we shall continue looking at some amazing results related to the dilogarithm.

Yet another functional equation

$$\displaystyle \operatorname{Li}_2(z) + \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = - \frac{1}{2} \log^2 (1-z) \,\,\,\, \, z<1$$​

PROOF

Start by the following

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = -\int^{\frac{z}{z-1}}_0 \frac{ \log(1-t)}{t}\, dt$$

Differentiate both sides with respect to $z$

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 \left(\frac{z}{z-1} \right) = \frac{1}{(z-1)^2}\left( \frac{ \log \left(1-\frac{z}{z-1}\right)}{\frac{z}{z-1}} \right)$$

Upon simplification we obtain

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 \left(\frac{z}{z-1} \right) = \frac{- \log(1-z)}{z(z-1)}$$

Using partial fractions decomposition

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 \left(\frac{z}{z-1} \right) = \frac{\log(1-z)}{1-z}+ \frac{\log(1-z)}{z}$$

Integrate both sides with respect to $z$

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = -\frac{1}{2} \log^2(1-z) - \operatorname{Li}_2(z) +C$$

put $z=-1$ to find the constant

$$\displaystyle \operatorname{Li}_2 \left(\frac{1}{2} \right) = -\frac{1}{2} \log^2(2) - \operatorname{Li}_2(-1) +C$$

Remember that

• $$\displaystyle \operatorname{Li}_2 \left(\frac{1}{2} \right) = \frac{\pi^2}{12}-\frac{1}{2} \log^2\left(\frac{1}{2} \right)$$
• $$\displaystyle \operatorname{Li}_2(-1) = -\frac{\pi^2}{12}$$
• $$\displaystyle \log^2 \left( \frac{1}{2} \right) = \log^2 \left(2 \right)$$

Hence we deduce that $C=0$ , so

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = -\frac{1}{2} \log^2(1-z) - \operatorname{Li}_2(z)$$

which can be written as

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right)+\operatorname{Li}_2(z) = -\frac{1}{2} \log^2(1-z) \, \,\square$$

Well, that doesn't end here , prove

$$\displaystyle \frac{1}{2} \operatorname{Li}_2 (z^2) = \operatorname{Li}_2 (z)+\operatorname{Li}_2 (-z)$$​

PROOF

We will continue with the same manner

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 (z^2) = -\int^{z^2}_0 \frac{\log(1-t)}{t}\, dt$$

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 (z^2) = -2 \frac{\log(1-z^2)}{z}$$

$$\displaystyle \frac{1}{2} \frac{d}{dz} \operatorname{Li}_2 (z^2) = -\frac{\log(1-z)}{z}- \frac{\log(1+z)}{z}$$

$$\displaystyle \frac{1}{2}\operatorname{Li}_2 (z^2) = \operatorname{Li}_2(z)+\operatorname{Li}_2(-z) \,+ C$$

putting $$\displaystyle z=1$$ we get $$\displaystyle C=0$$ , hence the result

$$\displaystyle \frac{1}{2}\operatorname{Li}_2 (z^2) = \operatorname{Li}_2(z)+\operatorname{Li}_2(-z) \, \square$$

Prove

$$\displaystyle \operatorname{Li}_2\left( \frac{\sqrt{5}-1}{2} \right) = \frac{\pi^2}{10} - \log^2 \left( \frac{\sqrt{5}-1}{2}\right)$$​

PROOF

First we add the two functional equations of this section to obtain

$$\displaystyle \operatorname{Li}_2 \left( \frac{z}{z-1} \right) + \frac{1}{2} \operatorname{Li}_2 (z^2) - \operatorname{Li}_2(-z) = -\frac{1}{2} \log^2 (1-z)$$

Now let $$\displaystyle z = \frac{1-\sqrt{5}}{2}$$

• $$\displaystyle z^2 = \frac{1-2\sqrt{5}+5}{4}= \frac{3-\sqrt{5}}{2}$$

• $$\displaystyle \frac{z}{z-1} = \frac{\sqrt{5}-1}{1+\sqrt{5}} = \frac{3-\sqrt{5}}{2}$$

$$\displaystyle \frac{3}{2} \operatorname{Li}_2 \left( \frac{3-\sqrt{5}}{2} \right) - \operatorname{Li}_2 \left( \frac{\sqrt{5}-1}{2} \right) = -\frac{1}{2} \log^2 \left( \frac{\sqrt{5}+1}{2}\right)$$ ----(1)

We already established the following functional equation

$$\displaystyle \operatorname{Li}_2 (z) + \operatorname{Li}_2 (1-z) = \frac{\pi^2}{6} - \log(z) \log(1-z)$$

putting $$\displaystyle z = \frac{3-\sqrt{5}}{2}$$

$$\displaystyle \operatorname{Li}_2 \left( \frac{3-\sqrt{5}}{2}\right) + \operatorname{Li}_2 \left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{6} - \log\left(\frac{3-\sqrt{5}}{2} \right) \log \left(\frac{\sqrt{5}-1}{2}\right)$$ -----(2)

Solving (1) , (2) for $$\displaystyle \operatorname{Li}_2 \left(\frac{\sqrt{5}-1}{2}\right)$$ we get our result $$\displaystyle \square$$

[HW] prove that $$\displaystyle \operatorname{Li}_2 \left( \frac{3-\sqrt{5}}{2} \right) = \frac{\pi^2}{15} - \frac{1}{4}\log^2 \left( \frac{3-\sqrt{5}}{2}\right)$$ .

To Be continued ...

Integration lessons continued ...

4.5.2.Exercises

In this section I will give some exercises that involve Polylogarithms

Find the following integral

$$\displaystyle I=\int^1_0 \, \frac{\log(1-x) \log(x) }{x} \, dx$$​

where $$\displaystyle \log$$ represents the natural logarithm

Integrate by parts by Integrating

$$\displaystyle \frac{\log(1-x)}{x} = - \text{Li}_2(x)$$

Differentiating

$$\displaystyle \log(x) = \frac{1}{x}$$

So we have

$$\displaystyle I=-\log(x) \text{Li}_2(x) |^1_0+\int^1_0 \frac{\text{Li}_2(t)}{t}\, dt$$

$$\displaystyle I=\text{Li}_3(1) = \sum_{k\geq 1} \frac{1}{k^3} = \zeta(3)$$

Evaluate the following integral

$$\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt \,\,\,\,\, 0<x<1$$​

Integrating by parts we get the following

$$\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) -\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt$$

Now we are left with the following integral

$$\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt = \int^{1}_{1-x} \frac{\text{Li}_2 (1-t)}{t} \, dt$$

Using a result we obtained earlier

$$\displaystyle \int^{1}_{1-x} \frac{\frac{\pi^2}{6} -\text{Li}_2(t) - \log(1-t) \log(t)}{t} \, dt$$

$$\displaystyle -\frac{\pi^2}{6}\log(1-x)- \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt -\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} \, dt$$

The first integral

• $$\displaystyle \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt =\text{Li}_3(1)-\text{Li}_3(1-x)$$

The second integral is the same as the first exercise

• $$\displaystyle \int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} =\text{Li}_3(1) +\log(1-x)\text{Li}_2(1-x)-\text{Li}_3(1-x)$$

Collecting the results together we obtain

$$\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3)$$

$$\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3)$$

This section concludes the PolyLogarithm discussion , I will start the Hypergeometric function in the next thread .

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Integration lessons continued ...

4.Integration using special functions (continued)

4.6. Ordinary Hypergeometric function

Ordinary or Gauss hypergoemtric function is a nice generalization of power expansion or series representations of many functions . Before we start with the definition we will explain some notations .

Defined the raising factorial as follows

$$(z)_n = \begin{cases} 1 & n = 0 \\ {} \\ \frac{\Gamma(z+n)}{\Gamma(z)}=z(z+1) \cdots (z+n-1) & n > 0. \end{cases}$$

We can easily prove that $(1)_n = 1\cdot 2\cdot 3 \cdots n = n!$ and $(2)_n = 2\cdot 3\cdot 4 \cdots (n+1)=(n+1)!$

Using this definition we defnie the Gauss hypergoemtric function as follows

$${}_2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}.$$

4.6.1 Famous functions using Hypergemtric representations

We can represent the famous functions using the hypergeomtric function

$$\displaystyle z{}_2F_1 (1,1;2;-z)= z\sum_{n\geq 0}\frac{(1)_n (1)_n}{(2)_n}\frac{(-z)^n}{n!}=\sum_{n\geq 0}\, (-1)^n\frac{ \, n!}{(n+1)!}z^{n+1}=\sum_{n\geq 0}(-1)^{n}\frac{z^{n+1}}{n+1}=\log(1+z)$$

$$\displaystyle _2F_1(a,1;1;z)= \sum_{n\geq 0}\frac{(a)_n \, (1)_n}{(1)_n}\frac{z^n}{n!}=\sum_{n\geq 0}\frac{(a)_n }{n!}z^n=(1-z)^{-a}$$

$$\displaystyle z \, _2F_1\left(\tfrac{1}{2}, \tfrac{1}{2}; \tfrac{3}{2};z^2\right)=\sum_{n\geq 0}\frac{\left( \frac{1}{2}\right)_n\, \left( \frac{1}{2}\right)_n}{\left( \frac{3}{2}\right)_n}\frac{z^{2n+1}}{n!}= \sum_{n \geq 0}\frac{ \left( \frac{1}{2}\right)_n}{2n+1}\frac{z^{2n+1}}{n!}= \sum_{n \geq 0}\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^n \, n!\, (2n+1)}z^{2n+1}$$

Which can be written as

$$\displaystyle z \, _2F_1\left(\tfrac{1}{2}, \tfrac{1}{2}; \tfrac{3}{2};z^2\right)= \sum_{n \geq 0}\frac{(2n)!}{4^n \, (n!)^2\, (2n+1)}z^{2n+1}=\sum_{n \geq 0}\frac{{2n \choose n}}{4^n \, (2n+1)}z^{2n+1}= \arcsin(z)$$

To be continued ...

Re: Integration lessons continued ...

4.Integration using special functions (continued)

4.6.1 Famous functions using Hypergemtric representations(continued)

In the previous section we discussed how to convert a hypergeometric representation into an elementary function or so called (Taylor expansion ) . Now we consider converting the Taylor expansion into the equivalent hypergeomtric representation .

Suppose the following

$$\displaystyle _2 F_1 ( a, b; c; z ) = \sum_{k \geq 0} t_k \,\,\, , \,\,\, t_0 =1 \,\,\,\,\,\,\,\,\, (1)$$​

Now consider the ratio

$$\displaystyle \frac{t_{k+1}}{t_k} = \frac{(k+a) (k+b) }{(k+c) (k+1)} z \,\,\,\,\,\,\,\,\, (2)$$​

Using this definition if we have a power series equivalent to (1) , we can easily find the terms $a,b,c$ and $z$.

Examples

Find the hypergeometric representation of the following

• $$\displaystyle f(z) = e^ z$$

We can easily find the Taylor expansion as

$$\displaystyle f(z) = e^ z= \sum_{k\geq 0} \frac{z^k}{k!}$$

Hence we have

$$\displaystyle \frac{t_{k+1} }{t_k} = \frac{z}{k+1}$$

Comparing to (2) we conclude

$$\displaystyle e^z ={} _2 F_1 (-,-;-;z)$$

• $$\displaystyle f(z) = \cos(z) = \sum_{k\geq 0} \frac{(-1)^k z^{2k}}{(2k)!}$$

By the same approach

$$\displaystyle \frac{t_{k+1}}{t_k} = -\frac{1}{(2k+2)(2k+1)}z= \frac{1}{(k+1)\left(k+\frac{1}{2} \right)}\frac{-z^2}{4}$$

Hence we have by comparing to (2)

$$\displaystyle \cos(z)={}_2 F_1 \left(-,- ; \frac{1}{2}; \frac{-z^2}{4} \right)$$

• $$\displaystyle f(z) = (1-z)^{-a}= \sum_{k\geq 0}\frac{(a)_k}{k!}z^k$$

$$\displaystyle \frac{t_{k+1}}{t_k} =\frac{(k+a)}{(k+1)}z= \frac{(k+a)(k+1)}{(k+1)(k+1)}z$$

Hence we have

$$\displaystyle (1-z)^{-a} ={} _2F_1(a,1;1;z)$$

[HW] Find the hypergeoemtric representations of the following functions

• $$\displaystyle \arcsin(z)$$
• $$\displaystyle \sin(z)$$

To be continued ...

integration lessons continued ...

4.Integration using special functions (continued)

4.6.2 Integral representation

In this section we give some transformations that will be useful to evaluate some integrals . We start this section by proving the integral representation .

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$​

Proof

Start by the RHS

$$\displaystyle \int_0^1 t^{b-1}(1-t)^{c-b-1} \, (1-tz)^{-a}\, dt$$

Using the expansion of $$\displaystyle (1-tz)^{-a}$$ we have

$$\displaystyle \int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{k\geq 0}\frac{(a)_k}{k!}\, (tz)^k$$

Interchanging the integral with the series

$$\displaystyle \sum_{k\geq 0}\frac{(a)_k}{k!}\, z^k \, \int_0^1 t^{k+b-1}(1-t)^{c-b-1}\, dt$$

Recalling the beta function we have

$$\displaystyle \sum_{n\geq 0}\frac{(a)_k \Gamma(k+b) \Gamma(c-b)}{ \Gamma(k+c)}\, \frac{z^k}{k!}$$

Using the identity that

$$\displaystyle \beta(c-b,b) = \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}$$

$$\displaystyle \beta(c-b,b)\sum_{k\geq 0}\frac{(a)_k \Gamma(k+b) \Gamma(c)}{\Gamma(b) \Gamma(k+c)}\, \frac{z^k}{k!}= \beta(c-b, b) \int^1_0 \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!}$$

As required $$\displaystyle \square$$

We can consider the case $$\displaystyle z=1$$ so we have

$$\displaystyle _2F_1(a,b;c;1)= \frac{1}{\beta(c-b,b) \,}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-t)^a}\, dt$$

$$\displaystyle _2F_1(a,b;c;1)=\frac{1}{\beta(c-b,b) }\int_0^1 t^{b-1}(1-t)^{c-b-a-1}\, dt =\frac{\beta(b,c-b-a)}{\beta(c-b,b) }=\frac{\Gamma(c-b-a)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$$

Some transformations

Start by transforming $$\displaystyle t \to 1-t$$

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 t^{c-b-1}(1-t)^{b-1}(1-(1-t)z)^{-a}\, dt =\int_0^1 t^{c-b-1}(1-t)^{b-1}(1-z+tz)^{-a}\, dt$$

Taking $$\displaystyle 1-z$$ as a common factor

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\frac{1}{(1-z)^a}\int_0^1 t^{c-b-1}(1-t)^{b-1}\left( 1-\frac{z}{z-1} \, t\right)^{-a}\, dt$$

Which can be written as

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\frac{\beta(c-b,b) {}_2F_1(a,c-b;c;z)}{(1-z)^a}$$

Evidently we have the interesting form

$$\displaystyle _2F_1(a,b;c;z)=(1-z)^{-a}{} _2F_1(a,c-b;c;z)$$

we will use this transformation later to find other special cases for $$\displaystyle z$$.

To be continued ...

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Integration lessons (continued)

4.Integration using special functions (continued)

4.7 Error Function

The error function is an interesting function that has many applications in probability, statistics and physics.

Definition

$$\displaystyle \text{erf}(x)=\frac{2}{\sqrt{\pi}}\int^x_0 e^{-t^2}\,dt$$

Complementary error function

$$\displaystyle \text{erfc}(x)=1-\text{erf}(x)$$

Imaginary error function

$$\displaystyle \text{erfi}(x) = -i \text{erf}(ix)$$

Properties :

The error function is odd

$$\text{erf}(-x) =\frac{2}{\sqrt{\pi}}\int^{-x}_0 e^{-t^2}\,dt=-\frac{2}{\sqrt{\pi}}\int^{x}_0 e^{-t^2}\,dt=-\text{erf}(x)$$

Real part and imaginary parts

$$\Re \,\text{erf}(z)=\frac{\text{erf}(z)+\text{erf}(\bar{z})}{2}$$

$$\Im \,\text{erf}(z)=\frac{\text{erf}(z)-\text{erf}(\bar{z})}{2i}$$

Using complex variables it can be done using $\text{erf}(\bar{z})=\overline{\text{erf}(z)}\,\square$

Exercises

$$I = \int^x_0 e^{t^2}\,dt$$

The function has no elementary anti-derivative so we represent it using the error function.

So we need to find a function $f$ such that $\frac{d}{dz}f=e^{z^2}$

Consider the imaginary error function

$$\text{erfi}(x)=-i\text{erf}(ix) = -i\frac{2}{\sqrt{\pi}}\int^{ix}_0e^{-t^2}\,dt$$

By differentiating both sides we have

$$\frac{d}{dx}\text{erfi}(x) = \frac{2}{\sqrt{\pi}}e^{x^2}$$

Hence we have

$$e^{x^2}=\frac{d}{dx}\frac{\sqrt{\pi}}{2}\text{erfi}(x)$$

By integrating both sides we have

$$\int e^{x^2}\,dx=\frac{\sqrt{\pi}}{2}\text{erfi}(x)+C$$

4.7.1 Relation to other functions

Hypergeomtric function

$$\text{erf}(x) = \frac{2x}{\sqrt{\pi}} {}_1F_1\left( \frac{1}{2},\frac{3}{2},-x^2\right)$$

By expanding the hypergeometirc function

$$\frac{2x}{\sqrt{\pi}} {}_1F_1\left( \frac{1}{2},\frac{3}{2},-x^2\right)=\frac{2x}{\sqrt{\pi}}\sum_{k\geq 0}\frac{\left(\frac{1}{2} \right)_k}{\left(\frac{3}{2} \right)_k}\frac{(-x^2)^k}{k!}$$

$$\frac{2x}{\sqrt{\pi}} {}_1F_1\left( \frac{1}{2},\frac{3}{2},-x^2\right)=\frac{x}{\sqrt{\pi}} \sum_{k\geq 0}\frac{(-x^2)^k}{(\frac{1}{2}+k)k!}$$

Notice that this is actually the expanded error function

$$\frac{2}{\sqrt{\pi}}\int^x_0 e^{-t^2}\,dt =\frac{2}{\sqrt{\pi}}\int^x_0\sum_{k\geq 0}\frac{(-x^2)^k}{k!}\,dx = \frac{2}{\sqrt{\pi}}\sum_{k\geq 0}\frac{(-x)^{2k+1}}{(2k+1)k!}$$

The function can be written using the incomplete Gamma function

Incomplete Gamma function

$$\text{erf}(x)=1-\frac{\Gamma\left(\frac{1}{2},x^2 \right)}{\sqrt{\pi}}$$

Exercises

$$\int^\infty_0 \text{erfc}(x)\,dx=\frac{1}{\sqrt{\pi}}$$

Using the complementary error function

$$\int^\infty_0 (1-\text{erf}(x))\,dx$$

Integration by parts we have

$$I= x(1-\text{erf}(x))\left. \right]^\infty_0+\frac{2}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\,dx$$

Now we compute $\text{erf}(\infty)$

$$\text{erf}(\infty) =\frac{2}{\sqrt{\pi}}\int^\infty_0 e^{-t^2}\,dt = \frac{2}{\sqrt{\pi}}\times\frac{\sqrt{\pi}}{2}=1$$

Hence we have the following

$$I=\frac{2}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\,dx = \frac{1}{\sqrt{\pi}}$$