MHB Integration using Beta and Gamma Functions

[Amad27]

Seems correct. Let us verify it by taking the limit as $x \to 1$.

What do you get ?

Does it agree with the value returned by wolfram.

The issue is with taking the limit.

$\displaystyle lim_{x\to1} \log(1-x)$ does not exist.

 

[alyafey22]

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x) $$

Take limits of both sides. Don't distribute the limit until you are sure the limit exists.

 

[Amad27]

Take limits of both sides. Don't distribute the limit until you are sure the limit exists.

$$\lim_{{x}\to{1}} -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x)$$
$$= \lim_{{x}\to{1}} -Li_2(x) +(\log(1-x))(1 - x) + 3x - \frac{x^2}{8} - Li_3(x)$$

The more interesting case is $\log(1-x)\cdot(1-x)$

$$y = \lim_{{x}\to{1}} (\log(1-x)\cdot(1-x))$$
Let $u = 1-x$ as $x\to1$ we have $u \to 0$

$$y = \lim_{{u}\to{0}} \log(u)\cdot(u) = \lim_{{u}\to{0}} \log(u\cdot e^u)$$

$$e^y = \lim_{{u}\to{0}} 0$$

But then we are back to where we started with $\log(0)$

 

[alyafey22]

$$\lim_{u \to 0} u \times \log(u) = \lim_{u \to 0} \frac{\log(u)}{\frac{1}{u}} $$

By L'Hospital rule the limit goes to 0.

 

[Amad27]

$$\lim_{u \to 0} u \times \log(u) = \lim_{u \to 0} \frac{\log(u)}{\frac{1}{u}} $$

By L'Hospital rule the limit goes to 0.

That is very nice, but the denominator is not indeterminate is it? $1/0$ is not indeterminate is it?

$$= \lim_{{x}\to{1}} -Li_2(x) +(\log(1-x))(1 - x) + 3x - \frac{x^2}{8} - Li_3(x)$$

$$= \lim_{{x}\to{1}} -Li_2(x) + \frac{23}{8} - Li_3(x)$$

$$= -\zeta(2) + \frac{23}{8} - \zeta(3)$$

Then

$$ \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^3} = -\frac{Li_2(x)}{x^2}+ \frac{1}{x} - \frac{\log(1-x)}{x} + \frac{\log(1-x)}{x^2} + \frac{2}{x} - \frac{1}{8} - \frac{Li_3(x)}{x^2}$$

Then we need to integrate:

$$\int_{0}^{1} -\frac{Li_2(x)}{(1-x)x^2}+ \frac{1}{(1-x)x} - \frac{\log(1-x)}{x(1-x)} + \frac{\log(1-x)}{(1-x)x^2} + \frac{2}{(1-x)x} - \frac{1}{8(1-x)} - \frac{Li_3(x)}{(1-x)x^2} \,dx$$

This is the most difficult part of all. Especially because if this is split up, a lot of it does not converge.

out of curiosity: could we use complex analysis to solve the sum we had? Also, can you also help me through some complex analysis, and contour integration? I am quite interested, but I don't have means to learn it except MHB. Help will be appreciated! =)

 

[alyafey22]

That is very nice, but the denominator is not indeterminate is it? $1/0$ is not indeterminate is it?

The limit is of the form $\frac{\infty}{\infty}$. which can be solved by L'Hospital rule.

Then we need to integrate:

$$\int_{0}^{1} -\frac{Li_2(x)}{(1-x)x^2}+ \frac{1}{(1-x)x} - \frac{\log(1-x)}{x(1-x)} + \frac{\log(1-x)}{(1-x)x^2} + \frac{2}{(1-x)x} - \frac{1}{8(1-x)} - \frac{Li_3(x)}{(1-x)x^2} \,dx$$

This is the most difficult part of all. Especially because if this is split up, a lot of it does not converge.

I guess this will be complicated , at least you tried :) . Why don't we stick to the solution using Harmonic numbers ?

out of curiosity: could we use complex analysis to solve the sum we had? Also, can you also help me through some complex analysis, and contour integration? I am quite interested, but I don't have means to learn it except MHB. Help will be appreciated! =)

Of course.I 'll answer this question in the other thread.

 

[Amad27]

The limit is of the form $\frac{\infty}{\infty}$. which can be solved by L'Hospital rule.
I guess this will be complicated , at least you tried :) . Why don't we stick to the solution using Harmonic numbers ?
Of course.I 'll answer this question in the other thread.

But how do we use that formula?

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Lets consider $n = \displaystyle \sqrt{k+1}$

Then the lower limit will be

$k = 0$

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Nevermind, I just realized the $H_n$ won't fit properly... Wow, this is difficult.

 

[alyafey22]

Ok , I will give you quick exercise

Find

$$\sum_{k\geq 0}\frac{H_k}{(k+1)^2}$$

 

[Amad27]

Ok , I will give you quick exercise

Find

$$\sum_{k\geq 0}\frac{H_k}{(k+1)^2}$$

Ok.

Let $n = j+1$

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2}= \left(1+\frac{2}{2} \right)\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(3-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2} =
\sum_{j=0}^\infty \frac{H_{j} + \frac{1}{j+1}}{(j+1)^2} = \sum_{j=0}^\infty \frac{H_{j}}{(j+1)^2}+ \sum_{j=0}^{\infty} + \frac{1}{(j+1)^3}$$

Originally, the formula states:

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

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

Then

$$\sum_{n=1}^\infty \frac{H_j}{(j+1)^2} = \sum_{n=0}^\infty \frac{H_j}{(j+1)^2} - 1 = (2)\zeta(3) - \frac{1}{2}(indeterminate)$$

I checked W|A, the sum involving the $q-2$ is indeterminate, which is bad news.

 

[alyafey22]

Ok.

Let $n = j+1$

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2}= \left(1+\frac{2}{2} \right)\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(3-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2} =
\sum_{j=0}^\infty \frac{H_{j} + \frac{1}{j+1}}{(j+1)^2} = \sum_{j=0}^\infty \frac{H_{j}}{(j+1)^2}+ \sum_{j=0}^{\infty} + \frac{1}{(j+1)^3}$$

Originally, the formula states:

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

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

Then

$$\sum_{n=1}^\infty \frac{H_j}{(j+1)^2} = \sum_{n=0}^\infty \frac{H_j}{(j+1)^2} - 1 = (2)\zeta(3) - \frac{1}{2}(indeterminate)$$

I checked W|A, the sum involving the $q-2$ is indeterminate, which is bad news.

By definition if the upper index of the sum is 0 then the sum is 0.

 

[Amad27]

By definition if the upper index of the sum is 0 then the sum is 0.

Then why does W|A return it as indeterminate?

 

[alyafey22]

Then why does W|A return it as indeterminate?

W|A can't solve it because the lower index is greater than the upper index.

So we have to add the following constrains $q>1$ and when $q=2$ the series is equal to 0. It is just a constrain that we defined on that particular case.

Any way, Let us assume that we can't find the summation when $q=2$ so we have to find

$$\sum_{k\geq 1} \frac{H_k}{k^2}$$

Can you think of a way to evaluate it without using that formula ?

 

[Amad27]

W|A can't solve it because the lower index is greater than the upper index.

So we have to add the following constrains $q>1$ and when $q=2$ the series is equal to 0. It is just a constrain that we defined on that particular case.

Any way, Let us assume that we can't find the summation when $q=2$ so we have to find

$$\sum_{k\geq 1} \frac{H_k}{k^2}$$

Can you think of a way to evaluate it without using that formula ?

Not really. I suppose the formula is the best way, and I suppose by definition the sum is $0$.

$$\sum_{n=1}^\infty \frac{H_j}{(j+1)^2} = \sum_{n=0}^\infty \frac{H_j}{(j+1)^2} - 1 = (2)\zeta(3) - 0 - 1 = 2\zeta(3) - 1$$

 

[alyafey22]

Hint :

use that

$$\int^1_0 \frac{1-x^n}{1-x}\,dx = H_n$$

 

[Amad27]

Hint :

use that

$$\int^1_0 \frac{1-x^n}{1-x}\,dx = H_n$$

That is what I did a while ago (with the other sum) and it became a mess as you saw.

$$\sum_{k=1}^{\infty} \frac{H_k}{k^2}$$

$$H_k = \int_{0}^{1} \frac{1-x^k}{1-x} \,dx$$

$$S = \int_{0}^{1} \frac{1}{1-x} \sum_{k=1}^{\infty} \frac{1-x^k}{k^2} \,dx$$

Assuming the parts of the sum are convergent we get:

$$S = \int_{0}^{1} \frac{1}{1-x} \left(\zeta(2) - \sum_{k=1}^{\infty} \frac{x^k}{k^2}\right) \,dx$$

$$S = \int_{0}^{1} \frac{1}{1-x} \left(\zeta(2) - Li_2(x)\right) \,dx$$

$$S = \lim_{{x}\to{1}} -\zeta(2)\log(1-x) - \int_{0}^{1} \frac{Li_2(x)}{1-x} \,dx$$

The Li_2(x)/(1-x) is the hard part. Integration by parts or some other strategy?

 

[alyafey22]

The limit doesn't exist so you cannot separate the two integrals because they are divergent as I said previously.

$$Li_2(x)+Li_2(1-x) = \zeta(2)-\log(x)\log(1-x)$$

$$\zeta(2)-Li_2(x) = Li_2(1-x)+\log(x)\log(1-x)$$

 

[Amad27]

The limit doesn't exist so you cannot separate the two integrals because they are divergent as I said previously.

$$Li_2(x)+Li_2(1-x) = \zeta(2)-\log(x)\log(1-x)$$

$$\zeta(2)-Li_2(x) = Li_2(1-x)+\log(x)\log(1-x)$$

$$S = \int_{0}^{1} \frac{Li_2(1-x) + \log(x)\log(1-x)}{1-x} \,dx$$

Assuming this integral is convergent, I suppose can split this.

$$S = -Li_3(1-x)]_{0}^{1} - \text{problem}$$

Okay, this is depressing.

And very hard.

 

[alyafey22]

Try integration by parts , don't quite so easily.

 

[Amad27]

Try integration by parts , don't quite so easily.

NEW IDEA!

$\log(1-x)\log(x) = -Li_2(1-x) - Li_2(x) + \zeta(2)$

$$S = \int_{0}^{1} \frac{-Li_2(x) + \zeta(2)}{1-x} \,dx$$

First the antiderivative:

$$S = (\zeta(2) - Li_2(x)) + \int \frac{\log^2(1-x)}{(1-x)} dx$$

Let $\log(1-x) = u \implies du = \frac{1}{1-x}$

$$S_{antiderivative} = (\zeta(2) - Li_2(x)) + \frac{\log^3(1-x)}{3}$$

$$S = \lim_{{x}\to{1}} -Li_2(x) + \frac{\log^3(1-x)}{3}$$

Wait, does this limit exist?

 

[alyafey22]

NEW IDEA!

$\log(1-x)\log(x) = -Li_2(1-x) - Li_2(x) + \zeta(2)$

$$S = \int_{0}^{1} \frac{-Li_2(x) + \zeta(2)}{1-x} \,dx$$

First the antiderivative:

$$S = (\zeta(2) - Li_2(x)) + \int \frac{\log^2(1-x)}{(1-x)} dx$$

I don't understand how you got that. Moreover , S shouldn't be used twice since the first is definite and the second is indefinite integral.

 

[Amad27]

I don't understand how you got that. Moreover , S shouldn't be used twice since the first is definite and the second is indefinite integral.

I will need some help. Can you start me off on the integral? I can't think anything that will help right now.

 

[alyafey22]

Here is the approach.

 

[Amad27]

Here is the approach.

Oh I see.

How is

$$\int_{0}^{1} \frac{\log(1-x)\log(x)}{x} \,dx = \int_{0}^{1} Li_2(x)/x \,dx$$

I believe $\log(1-x)\log(x) = -(Li_2(x) + Li_2(1-x)) + \zeta(2)$

??

 

[alyafey22]

Oh I see.

How is

$$\int_{0}^{1} \frac{\log(1-x)\log(x)}{x} \,dx = \int_{0}^{1} Li_2(x)/x \,dx$$

I believe $\log(1-x)\log(x) = -(Li_2(x) + Li_2(1-x)) + \zeta(2)$

??

Try integration by parts where I used that

$$Li_2(x) =- \int^x_0 \frac{\log(1-t)}{t}\,dt$$

 

[Amad27]

Try integration by parts where I used that

$$Li_2(x) =- \int^x_0 \frac{\log(1-t)}{t}\,dt$$

I see.

But how does this all relate to the original sum?

 

[alyafey22]

I see.

But how does this all relate to the original sum?

Which original sum ? Isn't the question how to solve

$$\sum_{n=1}^\infty \frac{H_n}{n^2}$$

without using the formula ?

 

[Amad27]

Which original sum ? Isn't the question how to solve

$$\sum_{n=1}^\infty \frac{H_n}{n^2}$$

without using the formula ?

No, the one for the integral.

 

[alyafey22]

No, the one for the integral.

It was just an exercise to know how to solve Euler sums without using the formula.

 

[Amad27]

It was just an exercise to know how to solve Euler sums without using the formula.

How how can you solve:

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

With the use of the formula?

Please, will you show me this example, and I can do the sum for the integral.

 

[alyafey22]

How how can you solve:

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

With the use of the formula?

Please, will you show me this example, and I can do the sum for the integral.

Look at this answer.