MHB Integration using Beta and Gamma Functions

[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

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

 

[Amad27]

The iteration variable here is $k$ not $n$.
Use the following property

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

From your hints, I believe I am to do the following:

$$\int n{z}^{(n-1)} \,dz = z^n$$

But in any case, $z$ is not the iteration variable, so we can't integrate with respect to $z$ now can we?

 

[alyafey22]

From your hints, I believe I am to do the following:

$$\int n{z}^{(n-1)} \,dz = z^n$$

But in any case, $z$ is not the iteration variable, so we can't integrate with respect to $z$ now can we?

I mean use the series

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

And integrate with respect to $z$.

 

[Amad27]

I mean use the series

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

And integrate with respect to $z$.

mmmmm...

Isnt $n$ the iteration variable? $z$ could merely be considered a constant (like in a partial derivative).

Then how can you integrate with respect to $z$ and not $n$?

How do you know, which is an iteration variable?

Thanks!

 

[alyafey22]

The iteration variable is the one that changes in the summation. It might be easier for you if I expanded the summation as

$$\sum_{n\geq 1} z^{n-1} = 1+z+z^2+z^3 +\cdots $$

As we see the iteration variable is $n$ and is the power of $z$. Can you see how we can integrate with respect to $z$. Integrating with respect to $n$ doesn't make sense because it is changing as we iterate.

 

[Amad27]

The iteration variable is the one that changes in the summation. It might be easier for you if I expanded the summation as

$$\sum_{n\geq 1} z^{n-1} = 1+z+z^2+z^3 +\cdots $$

As we see the iteration variable is $n$ and is the power of $z$. Can you see how we can integrate with respect to $z$. Integrating with respect to $n$ doesn't make sense because it is changing as we iterate.

Okay, I'll ask and "do" at the same exact time.

(Q1) So, the iteration variable is sort of like a constant, since in the expansion it is actually a number, which is why we don't integrate with respect to it?

(A1) I'll do the integration as a well;

$$\int z^{n-1} \,dz = \frac{z^n}{n}$$

So we have the $n$ in the denominator,

We need now $\frac{1}{(2n-1)}$ left to work with. In

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

 

[alyafey22]

Okay, I'll ask and "do" at the same exact time.

(Q1) So, the iteration variable is sort of like a constant, since in the expansion it is actually a number, which is why we don't integrate with respect to it?

For the time being thing of the iteration variable as an index that helps us find the terms of the summation or sequence. Most of the time it is not of that interest for us. Integrating or differentiating with respect to it doesn't make sense because it is discrete . It only represents a subset of the integers. What we are interested in manipulating is the variable that the iteration is performed on. In that sense we have

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

Now we need to get $(2n-1)$ in the denominator. Can you think of a way to do this ? Always think about integration.

 

[Amad27]

For the time being thing of the iteration variable as an index that helps us find the terms of the summation or sequence. Most of the time it is not of that interest for us. Integrating or differentiating with respect to it doesn't make sense because it is discrete . It only represents a subset of the integers. What we are interested in manipulating is the variable that the iteration is performed on. In that sense we have

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

Now we need to get $(2n-1)$ in the denominator. Can you think of a way to do this ? Always think about integration.

Hello, what does discrete mean? Secondly, We can't integrate anything to get $(2n - 1)$ in the denominator because it is supposedly a constant isn't it?

Thanks! A lot!

 

[alyafey22]

Hello, what does discrete mean? Secondly,

Not continuous.

We can't integrate anything to get $(2n - 1)$ in the denominator because it is supposedly a constant isn't it?

Thanks! A lot!

What do you mean by constant ? We should integrate with respect to $z$ , right ?

 

[Amad27]

Not continuous.
What do you mean by constant ? We should integrate with respect to $z$ , right ?

First then, how is an iteration variable non continuous? Because it change like $n = 1,2,3,4,5...$?? Perhaps?

Then secondly, I didnt see that $log$ sorry!

We had

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

We have a fraction successfully, all we need now is the $n$.

We can multiply and divide by $n$

$$u = -\frac{2n}{2n - n} = -\frac{2}{1}$$

Never mind, this won't work.

 

[alyafey22]

First then, how is an iteration variable non continuous? Because it change like $n = 1,2,3,4,5...$?? Perhaps?

Yes , as I said it only helps us locate a certain element or find the sum up to that element.

Then secondly, I didnt see that $log$ sorry!

We had

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

We have a fraction successfully, all we need now is the $n$.

We can multiply and divide by $n$

$$u = -\frac{2n}{2n - n} = -\frac{2}{1}$$

Never mind, this won't work.

I don't understand the step you made!

Let us focus on generating the $2n-1$ term on the denominator. You can do a trick. For example you can substitute $z^2$ to get

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

Now if we intgrate we get $2n+1$ in the denominator but we want $2n-1$ ! How to get around this ?

 

[Amad27]

Yes , as I said it only helps us locate a certain element or find the sum up to that element.
I don't understand the step you made!

Let us focus on generating the $2n-1$ term on the denominator. You can do a trick. For example you can substitute $z^2$ to get

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

Now if we intgrate we get $2n+1$ in the denominator but we want $2n-1$ ! How to get around this ?

Well, if we differentiate we will get,

$$\frac{-1}{-2z}$$

We can't really integrate

$$-\log(1 -z^2)$$

I mean you can, but it wont help since we will keep on rolling with logs then.Thanks!

 

[alyafey22]

Well, if we differentiate we will get,

$$\frac{-1}{-2z}$$

We can't really integrate

$$-\log(1 -z^2)$$

I mean you can, but it wont help since we will keep on rolling with logs then.Thanks!

You have done the differentiation wrongly. Don't think of the right hand side yet. Let us focus on getting the $2n-1$ term , first.

 

[Amad27]

You have done the differentiation wrongly. Don't think of the right hand side yet. Let us focus on getting the $2n-1$ term , first.

Hi,

We aren't getting anywhere, can we try something new? I mean this:

We are trying to evaluate $\psi(1/2)$

Can you show me a step-by-step evaluation for $\psi(1)$ so that I can see what you are doing and I can then apply that, rather than shooting arrows in the sky.

I honestly feel that is a better strategy once i get an experienced FEEL at it.

What do you say? I think it is better; but ultimately the choice depends on you =)

 

[alyafey22]

Can you show me a step-by-step evaluation for $\psi(1)$

There is no thing to evaluate since we have

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

Putting $z=0$ we get $\psi(1) = -\gamma$.

I honestly feel that is a better strategy once i get an experienced FEEL at it.

What do you say? I think it is better; but ultimately the choice depends on you =)

No problem. Ok, we arrived at the following

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

Divide by $z^2$

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

Integrate with respect to $z$

$$\sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$$

Put $z=1$

$$\sum_{n=1}^\infty \frac{1}{n(2n-1)} = -\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$Can you evaluate the integral on the right ?

 

[Amad27]

There is no thing to evaluate since we have

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

Putting $z=0$ we get $\psi(1) = -\gamma$.
No problem. Ok, we arrived at the following

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

Divide by $z^2$

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

Integrate with respect to $z$

$$\sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$$

Put $z=1$

$$\sum_{n=1}^\infty \frac{1}{n(2n-1)} = -\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$Can you evaluate the integral on the right ?

Thank you very much @ZaidAlyafey, kudos to you, and I really appreciate this; the method is much clearer this way as I can understand better, and as I read carefully, all at once. So thank you ZaidAlyafey, I can't thank you enough! I owe you =)

The INTEGRAL:
----------------------
$$-\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$

$$B(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)} \,dt$$

Let's differentiate with respect to $b$

$$B_b(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)}\ln(1-t) \,dt$$

Let $t = x^2 \implies dt = 2x dx$

$$B_b(a,b) = (2) \int_{0}^{1} {x}^{2a-1}{(1-x^2)}^{(b-1)}\ln(1-x^2) \,dx$$

We need $b - 1 = 0 \implies b = 1$
We need $2a - 1 = -2 \implies a = -1/2$

$$B_b((-1/2),1) = (2) \int_{0}^{1} {x}^{-2}\ln(1-x^2) \,dx$$

Darn... We still can't do anything =(

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

Let $\psi(1/2) = u$

$$I = 1/2B_b(-1/2, 1) = 1/2B(-1/2, 1)(-\gamma - u)$$

The sum is $u = \psi(1/2)$ so

$$u = (-1/2)B(-1/2, 1)\gamma - (1/2)B(-1/2, 1)(u)$$

It is very surprising.

I couldn't compute it but how in the world is $B(-1/2, 1) = -2$? I computed it on WolframAlpha, but I am stumped.

$$u = \gamma + u$$

$$0 = \gamma$$

Some error happened unfortunately.

 

[alyafey22]

Thank you very much @ZaidAlyafey, kudos to you, and I really appreciate this; the method is much clearer this way as I can understand better, and as I read carefully, all at once. So thank you ZaidAlyafey, I can't thank you enough! I owe you =)

The INTEGRAL:
----------------------
$$-\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$

$$B(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)} \,dt$$

Let's differentiate with respect to $b$

$$B_b(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)}\ln(1-t) \,dt$$

Let $t = x^2 \implies dt = 2x dx$

$$B_b(a,b) = (2) \int_{0}^{1} {x}^{2a-1}{(1-x^2)}^{(b-1)}\ln(1-x^2) \,dx$$

We need $b - 1 = 0 \implies b = 1$
We need $2a - 1 = -2 \implies a = -1/2$

$$B_b((-1/2),1) = (2) \int_{0}^{1} {x}^{-2}\ln(1-x^2) \,dx$$

Darn... We still can't do anything =(

You will run into a cycle. We want to find the integral without using the digamma function. Fortunately , there is an anti-derivative for the integral. Please , give it more thoughts.

The sum is $u = \psi(1/2)$ so

$$u = (-1/2)B(-1/2, 1)\gamma - (1/2)B(-1/2, 1)(u)$$

It is very surprising.

I couldn't compute it but how in the world is $B(-1/2, 1) = -2$? I computed it on WolframAlpha, but I am stumped.

$$u = \gamma + u$$

$$0 = \gamma$$

Some error happened unfortunately.

Actually $I \neq \psi(1/2)$. Remember that we have

$$\psi(1/2) = -\gamma -\sum \frac{1}{n(2n-1)}$$

$$I = -\int^1_0 \frac{\log(1-x^2)}{x^2}\,dx=-\sum \frac{1}{n(2n-1)}= u + \gamma$$

, hence $u+\gamma = u +\gamma$ which holds true. As I said you run into a loop by doing this method.

As how to evaluate $B(-1/2,1)$ by definition we have

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

It is known that $\Gamma(z+1) = z\Gamma(z) $ so $\Gamma(1/2) = (-1/2)\Gamma(-1/2)$.

 

[Amad27]

You will run into a cycle. We want to find the integral without using the digamma function. Fortunately , there is an anti-derivative for the integral. Please , give it more thoughts.
Actually $I \neq \psi(1/2)$. Remember that we have

$$\psi(1/2) = -\gamma -\sum \frac{1}{n(2n-1)}$$

$$I = -\int^1_0 \frac{\log(1-x^2)}{x^2}\,dx=-\sum \frac{1}{n(2n-1)}= u + \gamma$$

, hence $u+\gamma = u +\gamma$ which holds true. As I said you run into a loop by doing this method.

As how to evaluate $B(-1/2,1)$ by definition we have

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

It is known that $\Gamma(z+1) = z\Gamma(z) $ so $\Gamma(1/2) = (-1/2)\Gamma(-1/2)$.

Okay.

Do you mind giving a small hint?

So will the integral require gamma or beta?

So I can start it knowing a fact =)

Thanks!

 

[alyafey22]

Okay.

Do you mind giving a small hint?

So will the integral require gamma or beta?

So I can start it knowing a fact =)

Thanks!

It can be solved using elemetnary functions. Moreover , you can find an antiderivative of

$$\int \frac{\log(1-x^2)}{x^2}\,dx$$

Hint: start by $t=1/x^2$.

 

[Amad27]

It can be solved using elemetnary functions. Moreover , you can find an antiderivative of

$$\int \frac{\log(1-x^2)}{x^2}\,dx$$

Hint: start by $t=1/x^2$.

How did you find out that specific substition? It is not obvious to start with $t = 1/x^2$ so how did you figure out that we should make that substitution?

It may seem that I am being pedantic, but knowing the reason behind every step is important.

Sorry if this is too unusual, tell me so. =)

 

[alyafey22]

How did you find out that specific substition? It is not obvious to start with $t = 1/x^2$ so how did you figure out that we should make that substitution?

It may seem that I am being pedantic, but knowing the reason behind every step is important.

Sorry if this is too unusual, tell me so. =)

I was going to sleep then I remembered that I made a type. The substitution was supposed to be $t=1/x$. I am really sorry .

As for the reason for the sub , we have $1/x^2$ which is the derivative multiplied by a constant.

 

[alyafey22]

Ok , here is the how to do it

$$\int \frac{\log(1-x^2)}{x^2}\,dx = -\int \log(1-1/t^2) = -\int\log(t^2-1)-2\log(t)\,dt$$

This can be written as

$$-\int\log(t-1)\,dt -\int \log(t+1)\,dt+2\int\log(t)\,dt $$

All the functions above has an ant-derivative.

 

[Amad27]

Ok , here is the how to do it

$$\int \frac{\log(1-x^2)}{x^2}\,dx = -\int \log(1-1/t^2) = -\int\log(t^2-1)-2\log(t)\,dt$$

This can be written as

$$-\int\log(t-1)\,dt -\int \log(t+1)\,dt+2\int\log(t)\,dt $$

All the functions above has an ant-derivative.

Hello,

Okay let's get

$$-\int \log(t-1)\,dt = -(t-1)\ln(t-1) + (t-1)$$
$$-\int \ln(1 + t) dt = -(1+t)\ln(t+1) + (t+1)$$
$$2\int \ln(t) dt = 2t\ln(t) - 2t $$

$$ \implies -(t-1)\ln(t-1) + (t-1) -(1+t)\ln(t+1) + (t+1) + 2t\ln(t) - 2t$$

How can we change the lower bound in terms of $t$

$$ t = 1/x$$
$$t = 1/0$$

Thanks!

 

[alyafey22]

How can we change the lower bound in terms of $t$

$$ t = 1/x$$
$$t = 1/0$$

Thanks!

We just put $t=1/x$. Try then to find the integral on the interval [0,1].

 

[Amad27]

We just put $t=1/x$. Try then to find the integral on the interval [0,1].

But we CANNOT change the lower bound.

$t = 1/0$ because $x = 0$ was the lower bound.

 

[alyafey22]

But we CANNOT change the lower bound.

$t = 1/0$ because $x = 0$ was the lower bound.

The lower and upper limits before the substitution were $0$ and $1$ consecutively. If we make the substitution they are $\infty$ and $1$. You have the choice between $x$ and $t$. Remember if we have an improper integral we take the limit

$$\int^\infty_1 f(x) \,dx = \lim_{t \to \infty }\int^t_1 f(x) \,dx$$