Integration using Beta and Gamma Functions

[Amad27]

Interestingly, I seem to have an integral I have posted before, but I want to take a different approach to it.

$\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \,dx$

The beta function states,

$B(x,y) = \int_{0}^{1} {t}^{x-1}({1-t}^{y-1}) \,dx$

So, I was just thinking if there a possible way to compute this integral using the beta function also knowing

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

EDIT

For some reason the exponent in {1-t}^{y-1} does not work properly, but I think you know what I mean!

 

[alyafey22]

I am not aware of solving that integral using beta function. Just wanted to note that there is another version of the beta function that can be helpful in solving other integrals

\(\displaystyle \int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}} \, = B(x,y)\)

Maybe you can represent your integral using the hypegeometric function

\(\displaystyle B(b , c-b) F(a,b,c;z) = \int^1_0 x^{b-1}(1-x)^{c-b-1} (1-zx)^{-a} \, dx\)

 

[Amad27]

I am not aware of solving that integral using beta function. Just wanted to note that there is another version of the beta function that can be helpful in solving other integrals

\(\displaystyle \int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}} \, = B(x,y)\)

Maybe you can represent your integral using the hypegeometric function

\(\displaystyle B(b , c-b) F(a,b,c;z) = \int^1_0 x^{b-1}(1-x)^{c-b-1} (1-zx)^{-a} \, dx\)

I see, is it because of the natural log that I can't use the beta?

 

[alyafey22]

I see, is it because of the natural log that I can't use the beta?

Not necessarily. We can get the log by differentiation w.r.t to a variable. For example

\(\displaystyle -\int^\infty_0 \frac {t^{x-1} \log(1+t)}{(1+t)^{x+y}}\,dx = \frac{\partial}{\partial y}B(x,y)\)

This can be used to solve many problems , like

\(\displaystyle \int^\infty_0 \frac{\log(1+x^2)}{1+x^2}\,dx\)

 

[Amad27]

Not necessarily. We can get the log by differentiation w.r.t to a variable. For example

\(\displaystyle -\int^\infty_0 \frac {t^{x-1} \log(1+t)}{(1+t)^{x+y}}\,dx = \frac{\partial}{\partial y}B(x,y)\)

This can be used to solve many problems , like

\(\displaystyle \int^\infty_0 \frac{\log(1+x^2)}{1+x^2}\,dx\)

Polya said it is better to try an easier problem first. So let's try,

\(\displaystyle \int^\infty_0 \frac{\log(1+x^2)}{1+x^2}\,dx\)

I have two curiosities, let's try this one,

and also, can you possibly show me a way to do this integral (that you suggested using series?? Please? please...? I'll really appreciate it. So

Lets consider the Beta function here.

$$B(a, b) = \int_{0}^{\infty} \frac{{x}^{a-1}}{{(1+x)}^{(a)}\cdot{(1+x)}^{b}} \,dx$$

Lets consider $b$

$$B'(a,b) = (-)\int_{0}^{\infty}\frac{{x}^{(a-1)}\ln(1+x)}{{(1+x)}^{(a+b)}} \,dx$$

Basically the $x$ in the numerator must have the same meaning as the $x$ in the denominator, so we can use $x^2$ instead of just $x$.

Is this how we can solve the problem? Any leads?

 

[Amad27]

Not necessarily. We can get the log by differentiation w.r.t to a variable. For example

\(\displaystyle -\int^\infty_0 \frac {t^{x-1} \log(1+t)}{(1+t)^{x+y}}\,dx = \frac{\partial}{\partial y}B(x,y)\)

This can be used to solve many problems , like

\(\displaystyle \int^\infty_0 \frac{\log(1+x^2)}{1+x^2}\,dx\)

Ok, I tried hard, but I get stuck at $\Gamma(1/2)$ Here is the working (to your problem).

\(\displaystyle B'(a, b) = -\int^\infty_0 \frac {t^{x-1} \log(1+t)}{(1+t)^{x+y}}\,dx = \frac{\partial}{\partial y}B(x,y)\)

The issue is we have $t = x^2$ not $t = x$, so we must compute $dt$

$t = x^2$
$dt = 2x dx$

This makes

$$\int_{0}^{\infty} \frac{{x}^{(2a - 1)}\ln(1+x^2)}{{(1+x^2)}^{(a+b)}} \,dx$$

We must have $2a - 1 = 0$ which is only possible if $a = 1/2$

We must have $(a + b) = (1/2 + b) = 1$which is only possible if $b = 1/2$

Now we have

$$\frac{\partial}{\partial y}B(1/2,1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)}$$

The numerator is an issue.

$$\Gamma(1/2) = (-1/2)!$$

Impossible...

 

[alyafey22]

Can you take another look at the problem ? You have some mistakes.

 

[Amad27]

Can you take another look at the problem ? You have some mistakes.

I suppose its the formatting.

I missed out the sign, so the gammas will be negative.

Also, I was just evaluating $B(a,b)$ I had NOT taken the partial derivative.

How do you go about this problem? The derivative if we took after computing the gammas would be 0...

 

[alyafey22]

Ok , you said that \(\displaystyle \Gamma(1/2)\) is impossible but this is not correct. Actually we know that it is equal to $\sqrt{\pi}$. The gamma function agrees with the factorial only at positive integers so there is no meaning of saying $\Gamma(x) = (x-1)!$ where $x$ is not a positive integer.

To differentiate the gamma function we need the digamma function which is the logarithmic derivative of the gamma function defined as

\(\displaystyle \Gamma'(x) = \Gamma(x) \psi(x)\)

 

[Amad27]

Ok , you said that \(\displaystyle \Gamma(1/2)\) is impossible but this is not correct. Actually we know that it is equal to $\sqrt{\pi}$. The gamma function agrees with the factorial only at positive integers so there is no meaning of saying $\Gamma(x) = (x-1)!$ where $x$ is not a positive integer.

To differentiate the gamma function we need the digamma function which is the logarithmic derivative of the gamma function defined as

\(\displaystyle \Gamma'(x) = \Gamma(x) \psi(x)\)

Oh, ok, so let me start this partially over. We have,

$$\pd{B(a,b)}{b} = \pd{\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}}{b}$$

So $\Gamma(a)$ is just a constant. Let $u = a + b \implies du/db = 1$

$$ = (-)\cdot\Gamma(a) \cdot (\frac{\Gamma(b)\psi(b) - \Gamma(a+b)\psi(a+b)}{(\Gamma(a+b))^2})$$

So we substitute $(a,b) = (1/2, 1/2)$

So that would become,

$$=(-)\cdot \sqrt{\pi}\cdot(\frac{\sqrt{\pi}\psi(1/2) - 1\psi(1)}{1}$$

So, now we are left with the $\psi$ digamma to figure out...

$\psi(n) = \frac{\Gamma'(n)}{\Gamma(n)}$

But the derivative of the gamma relates back to the digamma...

How can we compute digammas?

Thanks!

 

[alyafey22]

You are making a small mistake when taking the derivative

$$ d_b (B(a,b)) = \frac{\Gamma(a+b)\Gamma(a)\Gamma(b)\psi(b) - \Gamma(a)\Gamma(b)\psi(a+b)\Gamma(a+b)}{\Gamma^2(a+b)}$$

This can be simplified to
$$d_b (B(a,b)) = \frac{\Gamma(a)\Gamma(b) (\psi(b)-\psi(a+b))}{\Gamma(a+b)} = B(a,b) (\psi(b)-\psi(a+b))$$

Can you see where your mistake is ?

 

[Amad27]

You are making a small mistake when taking the derivative

$$ d_b (\Gamma(a,b)) = \frac{\Gamma(a+b)\Gamma(a)\Gamma(b)\psi(b) - \Gamma(a)\Gamma(b)\psi(a+b)\Gamma(a+b)}{\Gamma^2(a+b)}$$

This can be simplified to
$$d_b (\Gamma(a,b)) = \frac{\Gamma(a)\Gamma(b) (\psi(b)-\psi(a+b))}{\Gamma(a+b)} = B(a,b) (\psi(b)-\psi(a+b))$$

Can you see where your mistake is ?

But, you did not define $\Gamma(a,b)$ What does that function mean, Gamma takes place in one variable; Beta takes place in two like,

$B(a,b)$, what is the definition of $\Gamma(a,b)?$

Thanks a lot!

 

[alyafey22]

But, you did not define $\Gamma(a,b)$ What does that function mean, Gamma takes place in one variable; Beta takes place in two like,

$B(a,b)$, what is the definition of $\Gamma(a,b)?$

Thanks a lot!

Oh , I am really sorry . That was supposed to be the beta function. I hope I didn't confuse you !

 

[Amad27]

Oh , I am really sorry . That was supposed to be the beta function. I hope I didn't confuse you !

I read the part where I took the derivative of the Beta and compared it to yours.

:sigh: I see, I forgot the $\Gamma(a + b)$ in the numerator... foolish mistake.

Then you can get

$$B'(a,b) =\Gamma(a)\cdot\frac{\Gamma(b)\psi(b)-\Gamma(b)\psi(a+b)}{\Gamma(a+b)}$$

$$B_b(a,b) = \frac{(\Gamma(a)\Gamma(b))(\psi(b)-\psi(a+b)}{\Gamma(a+b)}$$

$$B_b(a,b) = B(a,b)(\psi(b) - \psi(a+b))$$

Okay, so we have this part, and

$b = 1/2$
$a = 1/2$

$$B_b(1/2, 1/2) = B(1/2, 1/2)(\psi(1/2) - \psi(1))$$

I'm still confused, how can you find values of $\psi$ the digamma?

From wikipedia,

$$\psi(x) = \int_{0}^{\infty} \frac{{e}^{-t}}{t} - \frac{{e}^{-xt}}{1-{e}^{-t}} \,dt$$

$$\psi(1) = \int_{0}^{\infty} \frac{{e}^{-t}}{t} - \frac{{e}^{-t}}{1-{e}^{-t}} \,dt$$

$$\psi(1) = \int_{0}^{\infty} \frac{{e}^{-t}}{t} \,dt - \int_{0}^{\infty} \frac{{e}^{-t}}{1-{e}^{-t}} \,dt$$

Lets compute the first integral.

$$I_1 = \int_{0}^{\infty} \frac{{e}^{-t}}{t} \,dt$$ First, the antiderivative.

$$I_1 = \int \frac{{e}^{-t}}{t} dt$$

Let $u = 1/t$ and let $dv = {e}^{-t} dt$
$du = -1/t^2$ and $v = -{e}^{-t}$

$$I_1 = \frac{-{e}^{-t}}{t} - \int \frac{{e}^{-t}}{t^2} dt$$

mmmmmmmmm... yeah, that is quite hard to compute, $I_1$ antiderivative is the problem. But I know there is a specified value already. But would you mind helping me out with the antiderivative of $I_1$ as well?

from the data-tables, $\psi(1) = -\gamma$ and $\psi(1/2) = -(2\ln(2) + \gamma)$

$$B_b(1/2, 1/2) = B(1/2, 1/2)(\psi(1/2) - \psi(1))$$
$$B_b(1/2, 1/2) = \frac{\Gamma(1/2)\Gamma(1/2)(\psi(1/2) - \psi(1/2))}{\Gamma(1)}$$
$\Gamma(1/2) = \sqrt{\pi}$

$$B_b(1/2, 1/2) = \frac{\pi(\psi(1/2) - \psi(1))}{1}$$

$$B_b(1/2, 1/2) = (\pi)(-(2\ln(2) + \gamma) + \gamma)$$

$$B_b(1/2, 1/2) = -2\ln(2)\pi$$

What do you think? (And also please take consideration of the integral definition of $\psi(1)$ and tell me how to find the antiderivative...)

By the way, WolframAlpha can't compute this integral, how can I know the final answer to it?

Thanks!

 

[alyafey22]

Lets compute the first integral.

$$I_1 = \int_{0}^{\infty} \frac{{e}^{-t}}{t} \,dt$$ First, the antiderivative.

Unfortunately , the integral in hand is divergent. So you can not separate the two integrals as you did.

$$I_1 = \int \frac{{e}^{-t}}{t} dt$$

The antiderivative can not be represented using elementary functions. Though, we can use the Exponential integral to represent it.

where we define

\(\displaystyle E_1(z) = \int^\infty_z \frac{e^{-t}}{t}\,dt\)

As we see from wikipedia the quote

the definition becomes ambiguous due to branch points at 0

For the time point you can think of branch points as 0 for the logarithm.

---

Generally to find the values of the digamma we use the series representation

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

You see that the value for $\psi(1)$ is straight forward.

Please try to find the value for $\psi(1/2)$ using the series representation.

 

[Amad27]

Unfortunately , the integral in hand is divergent. So you can not separate the two integrals as you did.

The antiderivative can not be represented using elementary functions. Though, we can use the Exponential integral to represent it.

where we define

\(\displaystyle E_1(z) = \int^\infty_z \frac{e^{-t}}{t}\,dt\)

As we see from wikipedia the quote
For the time point you can think of branch points as 0 for the logarithm.

---

Generally to find the values of the digamma we use the series representation

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

You see that the value for $\psi(1)$ is straight forward.

Please try to find the value for $\psi(1/2)$ using the series representation.

Darn it.....

How did you figure out that,

$\int_{0}^{\infty} \frac{{e}^{-t}}{t}\,dt$

Is divergent without evaluating it??

I would love to know the way of knowing the divergence of an integral, so how did you find out? Secondly,

I am not experienced with series at all. Like nothing, I haven't done series yet but when I get to series I'll surely try the method.

For now I relied on wikipedia that,

$\psi(1) = -\gamma$

But my answer to (your) integral,

$=-2\ln(2)\pi$

Is it correct? WolframAlpha can't compute the integral.

 

[alyafey22]

Darn it.....

How did you figure out that,

$\int_{0}^{\infty} \frac{{e}^{-t}}{t}\,dt$

Is divergent without evaluating it??

I would love to know the way of knowing the divergence of an integral, so how did you find out?

$$\int^x_0 \frac{{e}^{t}}{t} > \int^x_0 \frac{1}{t} $$

The integral on the right is divergent because of the branch point at $0$.

Secondly,

I am not experienced with series at all. Like nothing, I haven't done series yet but when I get to series I'll surely try the method.

Are you familiar with Maclurain series. If so , I think you can evaluate the integral. Generally , series that evaluate to logs are easy to handle. Please , try it. It is not that difficult.

For now I relied on wikipedia that,

$\psi(1) = -\gamma$

But my answer to (your) integral,

$=-2\ln(2)\pi$

Is it correct? WolframAlpha can't compute the integral.

I will have to see the full evaluations. Can you post it ?

 

[Amad27]

$$\int^x_0 \frac{{e}^{t}}{t} > \int^x_0 \frac{1}{t} $$

The integral on the right is divergent because of the branch point at $0$.
Are you familiar with Maclurain series. If so , I think you can evaluate the integral. Generally , series that evaluate to logs are easy to handle. Please , try it. It is not that difficult.
I will have to see the full evaluations. Can you post it ?

I'm not quite sure what a Maclurain series is unfortunately, so I'll have to research on that. Meanwhile, I did post my full workings before though. I'll repost it here:

$$B'(a,b) =\Gamma(a)\cdot\frac{\Gamma(b)\psi(b)-\Gamma(b)\psi(a+b)}{\Gamma(a+b)}$$

$$B_b(a,b) = \frac{(\Gamma(a)\Gamma(b))(\psi(b)-\psi(a+b)}{\Gamma(a+b)}$$

$$B_b(a,b) = B(a,b)(\psi(b) - \psi(a+b))$$

Okay, so we have this part, and

$b = 1/2$
$a = 1/2$

$$B_b(1/2, 1/2) = B(1/2, 1/2)(\psi(1/2) - \psi(1))$$

from the data-tables, $\psi(1) = -\gamma$ and $\psi(1/2) = -(2\ln(2) + \gamma)$

$$B_b(1/2, 1/2) = B(1/2, 1/2)(\psi(1/2) - \psi(1))$$
$$B_b(1/2, 1/2) = \frac{\Gamma(1/2)\Gamma(1/2)(\psi(1/2) - \psi(1/2))}{\Gamma(1)}$$
$\Gamma(1/2) = \sqrt{\pi}$

$$B_b(1/2, 1/2) = \frac{\pi(\psi(1/2) - \psi(1))}{1}$$

$$B_b(1/2, 1/2) = (\pi)(-(2\ln(2) + \gamma) + \gamma)$$

$$B_b(1/2, 1/2) = -2\ln(2)\pi$$

Also, I wrote that WolframAlpha can't compute the integral. How can I check the actual answer?

 

[alyafey22]

$$\int^\infty_0 \frac{\log(1+x^2) }{1+x^2} = \pi \log(2)$$

I used the numerical value which can be found in wolfram by putting N[integral] then this can be seen to be very close to $\pi \log(2)$.

 

[Amad27]

$$\int^\infty_0 \frac{\log(1+x^2) }{1+x^2} = \pi \log(2)$$

I used the numerical value which can be found in wolfram by putting N[integral] then this can be seen to be very close to $\pi \log(2)$.

Hey,

Would you mind checking what I did wrong? I got $2ln(2)\pi$

Why the 2 times in mine??

Thanks!

 

[alyafey22]

Ok, I tried hard, but I get stuck at $\Gamma(1/2)$ Here is the working (to your problem).

\(\displaystyle B'(a, b) = -\int^\infty_0 \frac {t^{x-1} \log(1+t)}{(1+t)^{x+y}}\,dx = \frac{\partial}{\partial y}B(x,y)\)

The issue is we have $t = x^2$ not $t = x$, so we must compute $dt$

$t = x^2$
$dt = 2x dx$

This makes

$$\int_{0}^{\infty} \frac{{x}^{(2a - 1)}\ln(1+x^2)}{{(1+x^2)}^{(a+b)}} \,dx$$

Maybe you forgot 2 in that substitution ?

 

[Amad27]

Maybe you forgot 2 in that substitution ?

Okay, I have to check something...

Which 2 in the substitution, and which substitution overall? I'll restart. Evaluate...

Step 1: The Basic Evaluations
$$\int_{0}^{\infty} \frac{\ln(1+x^2)}{(1+x^2)}\,dx$$
$$\implies B(a, b) = \int_{0}^{\infty} \frac{{t}^{a-1}}{{(1+t)}^{(a+b)}} \,dt$$
$$ \implies \pd{B(a,b)}{b} = (-)\cdot\int_{0}^{\infty} \frac{\ln(1+t){t}^{a-1}}{{(1+t)}^{(a+b)}} \,dt$$
$$ t = x^2 \implies dt = 2x dx$$
$$\implies \pd{B(a,b)}{b} = (-)\cdot\int_{0}^{\infty} \frac{(2x)\ln(1+x^2){x^2}^{a-1}}{{(1+x^2)}^{(a+b)}} \,dx$$
$$\implies \pd{B(a,b)}{b} = (-2)\cdot\int_{0}^{\infty} \frac{\ln(1+x^2){x}^{2a-1}}{{(1+x^2)}^{(a+b)}} \,dx$$
$$ \implies 2a - 1 = 0 \implies a = \frac{1}{2}$$
$$\implies a + b = 1 \implies \frac{1}{2} + b = 1 \implies b = \frac{1}{2}$$Step 1.5: Remember the integral, its not completely the beta derivative.
$$\pd{B(a,b)}{b} = (-2)\cdot\int_{0}^{\infty} \frac{\ln(1+x^2){x}^{2a-1}}{{(1+x^2)}^{(a+b)}} \,dx$$
$$ \int_{0}^{\infty} \frac{\ln(1+x^2){x}^{2a-1}}{{(1+x^2)}^{(a+b)}} \,dx$$ $$= \frac{\pd{B(a,b)}{b}}{(-2)}$$Step 2: The partial derivative of the Beta Function
$$\pd{B(a,b)}{b} = B(a,b)(\psi(b) - \psi(a+b))$$
$$ \implies \pd{B(\frac{1}{2},\frac{1}{2})}{b} = B(\frac{1}{2}, \frac{1}{2})(\psi(\frac{1}{2}) - \psi(1))$$
$$ \implies B(\frac{1}{2}, \frac{1}{2}) = \frac{\pi}{1}$$
$$\implies \psi(\frac{1}{2}) = (-)\cdot(2\ln(2) + \gamma)$$
$$\implies \psi(1) = -\gamma$$

Step 3: Evaluate the Beta derivative at 1/2, 1/2
$$ \pd{B(\frac{1}{2},\frac{1}{2})}{b} = B(\frac{1}{2}, \frac{1}{2})(\psi(\frac{1}{2}) - \psi(1))$$
$$ \implies \pd{B(\frac{1}{2},\frac{1}{2})}{b} = (\frac{\pi}{1})(-2\ln(2)\pi - \gamma + \gamma$$
$$ = \frac{-2\ln(2)\pi}{1}$$

Step 4: Solve for the actual integral.

$$I = \frac{\pd{B(a,b)}{b}}{(-2)}$$
$$ \implies I = \frac{-2\ln(2)\pi}{-2} = \pi\ln(2)$$

Which is the required result.

Thanks @Zaid for showing me an awesome method! ! !.

Do you perhaps have any other interesting integrals like this, which involve Beta and Gamma? Perhaps another improper one like this? I want to try another one!

Thanks Zaid, you are awesome!

 

[alyafey22]

First of all we have to prove the value of $\psi(1/2)$. Since you are not familiar with maclurain series I will make things easier for you.

First of all start with the definition

$$\psi(z+1) = -\gamma +\sum \frac{z}{n(n+z)}$$

So we need to find the value of

$$\tag {1} \sum_{n\geq 1} \frac{1}{n(2n-1)}$$

You can start by the geometric series

$$ \tag{2} \sum_{n\geq 1}z^n = \frac{1}{1-z}$$

Can you reach (1) from (2) ? You can integrate, differentiate and plug some value for $z$.

 

[Amad27]

First of all we have to prove the value of $\psi(1/2)$. Since you are not familiar with maclurain series I will make things easier for you.

First of all start with the definition

$$\psi(z+1) = -\gamma +\sum \frac{z}{n(n+z)}$$

So we need to find the value of

$$\tag {1} \sum_{n\geq 1} \frac{1}{n(2n-1)}$$

You can start by the geometric series

$$ \tag{2} \sum_{n\geq 1}z^n = \frac{1}{1-z}$$

Can you reach (1) from (2) ? You can integrate, differentiate and plug some value for $z$.

So $z = -1/2$ so that we get $\psi(1/2)$ We need therefore,

$$\sum_{1}^{\infty} \frac{-1}{2n(n - \frac{1}{2})}$$
$$ = (-)\cdot\sum_{1}^{\infty} \frac{1}{n(2n - 1)}$$

I'm not sure what you did, you do not have the $(-)$ on yours. Why?

$$\frac{1}{n(2n - 1)} = \frac{A}{n} + \frac{B}{2n-1}$$
$$\implies 1 = A(2n - 1) + Bn$$

$$ \implies 1 = B/2 \implies B = 2$$
$$\implies 1 = -A \implies A = 1$$

$$\frac{1}{n(2n - 1)} = \frac{1}{n} + \frac{2}{2n-1}$$

$$= (-)\cdot\sum_{1}^{\infty} \frac{1}{n(2n - 1)}$$
$$ = (-)\cdot\sum_{1}^{\infty} \frac{1}{n} + \frac{2}{2n-1}}$$

The series diverges because of the harmonic series,

$$\sum_{1}^{\infty} \frac{1}{n}$$

So, In the end it is,

$-\gamma$ + divergence

 

[alyafey22]

So $z = -1/2$ so that we get $\psi(1/2)$ We need therefore,

$$\sum_{1}^{\infty} \frac{-1}{2n(n - \frac{1}{2})}$$
$$ = (-)\cdot\sum_{1}^{\infty} \frac{1}{n(2n - 1)}$$

I'm not sure what you did, you do not have the $(-)$ on yours. Why?

I just ignored the sign because it is not necessary for the evaluation.

$$\frac{1}{n(2n - 1)} = \frac{A}{n} + \frac{B}{2n-1}$$
$$\implies 1 = A(2n - 1) + Bn$$

$$ \implies 1 = B/2 \implies B = 2$$
$$\implies 1 = -A \implies A = 1$$

$$\frac{1}{n(2n - 1)} = \frac{1}{n} + \frac{2}{2n-1}$$

$$= (-)\cdot\sum_{1}^{\infty} \frac{1}{n(2n - 1)}$$
$$ = (-)\cdot\sum_{1}^{\infty} \frac{1}{n} + \frac{2}{2n-1}}$$

The series diverges because of the harmonic series,

$$\sum_{1}^{\infty} \frac{1}{n}$$

So, In the end it is,

$-\gamma$ + divergence

Always avoid to separate integrals or series until you are sure the sums are convergent.
Start by the geometric series I provided.

 

[Amad27]

I just ignored the sign because it is not necessary for the evaluation.

Always avoid to separate integrals or series until you are sure the sums are convergent.
Start by the geometric series I provided.

I am seriously confused here unforunately.

What did you mean "integrate" or differentiate a series?

It is not possible to get $(1)$ from $(2)$ because $(1)$ does not contain a $z$ term.

 

[alyafey22]

I am seriously confused here unforunately.

No problem . I assure you that this is going to be a very good exercise.

What did you mean "integrate" or differentiate a series?

Can you take a look at this http://www.math24.net/differentiation-and-integration-of-power-series.html. Please ask, If you don't understand something.

It is not possible to get $(1)$ from $(2)$ because $(1)$ does not contain a $z$ term.

You can put a value for $z$ to arrive to (1).

 

[Amad27]

No problem . I assure you that this is going to be a very good exercise.

Can you take a look at this http://www.math24.net/differentiation-and-integration-of-power-series.html. Please ask, If you don't understand something.
You can put a value for $z$ to arrive to (1).

Thank you very much @ZaidAlyafey, I appreciate (very very very much) the pains you are taking here for me! Thank you! I read the tutorial, and it seems to be easier to understand. Let's start with the ratio sum you gave and try to differentiate it first.$$\sum_{n=1}^{\infty} \frac{1}{1-z} = z^n$$

So we must differentiate with respect to $z$ to get

$$n{z}^{(n-1)} = \sum_{n=1}^{\infty} n{z}^{(n-1)}$$

Ok. I admit, this is tough, I am not sure if I am headed correct here. Ideas?

Thanks @ZaidAlyafey!

 

[alyafey22]

Thank you very much @ZaidAlyafey, I appreciate (very very very much) the pains you are taking here for me! Thank you! I read the tutorial, and it seems to be easier to understand. Let's start with the ratio sum you gave and try to differentiate it first.$$\sum_{n=1}^{\infty} \frac{1}{1-z} = z^n$$

The sum I provided was ,

$$\sum_{n=0}^{\infty}z^n= \frac{1}{1-z}$$

I made a mistake that should start from 0 not 1.

Now, to make things easier for you , expand it

$$1+z+z^2+\cdots= \frac{1}{1-z}$$

Differentiate term by term both sides. Then write it in the form of a series.

 

[Amad27]

The sum I provided was ,

$$\sum_{n=0}^{\infty}z^n= \frac{1}{1-z}$$

I made a mistake that should start from 0 not 1.

Now, to make things easier for you , expand it

$$1+z+z^2+\cdots= \frac{1}{1-z}$$

Differentiate term by term both sides. Then write it in the form of a series.

I'll assume the LHS continues on forever.

$$1 + 2z + 3z^2 + 4z^3 + 5z^4 ... (n){z}^{(n-1)} = LHS'$$
$$\implies LHS' = \sum_{n=1}^{\infty}n{z}^{(n-1)}$$

$$RHS = \frac{1}{1-z}$$
$$RHS' = \ln(1-z)$$

$$ \implies \ln(1-z) = \sum_{n=1}^{\infty}n{z}^{(n-1)}$$

Any progress?

 

[alyafey22]

$$RHS' = \ln(1-z)$$

Nice but you integrated one side and differentiated the other. I guess you got the concept , though.

 

[Amad27]

Nice but you integrated one side and differentiated the other. I guess you got the concept , though.

Oh god, how much worse can I get?

I'll retry, maybe you can guide me a little further please?

$$RHS' = \frac{1}{{(z-1)}^{2}} = (n)\sum_{n=1}^{\infty}{z}^{n-1} = LHS'$$

But how does this come back to the original problem? Which was evaluating,

$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)}$$
?

Thanks!

 

[alyafey22]

Oh god, how much worse can I get?

No problem, I know that the concept is new to you.

I'll retry, maybe you can guide me a little further please?

$$RHS' = \frac{1}{{(z-1)}^{2}} = (n)\sum_{n=1}^{\infty}{z}^{n-1} = LHS'$$

You can not take $n$ out side the sum because it is the iteration variable.

But how does this come back to the original problem? Which was evaluating,

$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)}$$
?

Thanks!

Ok, our purpose is have $n$ in the denominator. How can we do that ?

Hint : use integration.

 

[Amad27]

No problem, I know that the concept is new to you.
You can not take $n$ out side the sum because it is the iteration variable.
Ok, our purpose is have $n$ in the denominator. How can we do that ?

Hint : use integration.

Really? $n$ is an iteration variable? Then in for example,

$$\sum_{k=1}^{n} 4nx$$

How are you able to take the $n$ out?

Okay, I am supposed to integrate something.
Would you mind telling what I should integrate please? I am sort of confused in this.

I know the sum is supposed to converge to 0, so I know the answer, but not how to get to it.

Thanks @ZaidAlyafey!

 

[alyafey22]

Really? $n$ is an iteration variable? Then in for example,

$$\sum_{k=1}^{n} 4nx$$

How are you able to take the $n$ out?

The iteration variable here is $k$ not $n$.

Okay, I am supposed to integrate something.
Would you mind telling what I should integrate please? I am sort of confused in this.

I know the sum is supposed to converge to 0, so I know the answer, but not how to get to it.

Thanks @ZaidAlyafey!

Use the following property

$$\int^x_0 t^{n-1}\,dt = \frac{x^n}{n}$$