Integration using Beta and Gamma Functions

[Amad27]

You start by noticing a pattern, then you finish by induction.
For example try to evaluate

$$\int^1_0 Li_3(x)\,dx$$

Do you see the pattern ? Try other values.
Basically , once you work with a problem for a long time , it becomes easier to establish some proofs. Usually when you read about these topics you get more ideas and techniques to solve similar problems.

Very cool! I will try to prove the dilog duplication formula.

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

Meanwhile, have you ever solved:

$$\int_{0}^{1} \log^3(1-x)\log^3(x) \,dx$$

??

 

[alyafey22]

Meanwhile, have you ever solved:

$$\int_{0}^{1} \log^3(1-x)\log^3(x) \,dx$$

??

No. Nevertheless , it is worth trying.

Another problem is the following

$$\int_{0}^{1} \frac{\log^3(1-x)\log^3(x)}{x} \,dx$$

 

[Amad27]

No. Nevertheless , it is worth trying.

Another problem is the following

$$\int_{0}^{1} \frac{\log^3(1-x)\log^3(x)}{x} \,dx$$

That is a great problem, but difficult. We could (to begin with) try integration by parts, but that would create a mess.

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

The only way I see is to multiply by $\displaystyle \log^3(x)\log(1-x)$ But it will make it weird.

$$\sum_{n=1}^{\infty} \frac{\log^3(x)\log(1-x)\cdot x^{n+1}(H_n)}{n+1} = \frac{-\log^{3}(1-x)\log^3(x)}{2}$$

$$(-2)\cdot \sum_{n=1}^{\infty} \frac{\log^3(x)\log(1-x)\cdot x^{n}(H_n)}{n+1} = \frac{\log^{3}(1-x)\log^3(x)}{x}$$

$$(-2)\cdot \sum_{n=1}^{\infty} \frac{(H_n)}{n+1} \int_{0}^{1} \frac{\log^3(x)\log(1-x)\cdot x^{n}}{1} \,dx = \int_{0}^{1} \frac{\log^{3}(1-x)\log^3(x)}{x} \,dx$$

Will you give a Small hint? Nothing too large to giveaway. =)

 

[alyafey22]

That is a great problem, but difficult. We could (to begin with) try integration by parts, but that would create a mess.

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

There shouldn't be a minus sign!

The only way I see is to multiply by $\displaystyle \log^3(x)\log(1-x)$ But it will make it weird.

$$\sum_{n=1}^{\infty} \frac{\log^3(x)\log(1-x)\cdot x^{n+1}(H_n)}{n+1} = \frac{-\log^{3}(1-x)\log^3(x)}{2}$$

$$(-2)\cdot \sum_{n=1}^{\infty} \frac{\log^3(x)\log(1-x)\cdot x^{n}(H_n)}{n+1} = \frac{\log^{3}(1-x)\log^3(x)}{x}$$

$$(-2)\cdot \sum_{n=1}^{\infty} \frac{(H_n)}{n+1} \int_{0}^{1} \frac{\log^3(x)\log(1-x)\cdot x^{n}}{1} \,dx = \int_{0}^{1} \frac{\log^{3}(1-x)\log^3(x)}{x} \,dx$$

Will you give a Small hint? Nothing too large to giveaway. =)

Good attempt but I don't think this will help. The idea is to get rid of that $\log(1-x)$ because it makes the problem harder to tackle.

- - - Updated - - -

Ok , I think we can use the following generating function , instead

$$\sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

It looks promising.

 

[Amad27]

There shouldn't be a minus sign!
Good attempt but I don't think this will help. The idea is to get rid of that $\log(1-x)$ because it makes the problem harder to tackle.

- - - Updated - - -

Ok , I think we can use the following generating function , instead

$$\sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

It looks promising.

I am assuming you don't derive these generation functions yourself?

So, where do you find these generating functions??

$$\sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

$$\sum_{n\geq 1} \frac{H^2_n x^{n+1}}{n+1} = \frac{-\log^3(1-x)}{3}+\int\frac{\operatorname{Li}_2(x)}{1-x} dx$$

The polylogarithmic integral of the RHS is extremely messy, what can be done?

 

[alyafey22]

I am assuming you don't derive these generation functions yourself?

So, where do you find these generating functions??

I saw it somewhere on the internet.

$$\sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

$$\sum_{n\geq 1} \frac{H^2_n x^{n+1}}{n+1} = \frac{-\log^3(1-x)}{3}+\int\frac{\operatorname{Li}_2(x)}{1-x} dx$$

The polylogarithmic integral of the RHS is extremely messy, what can be done?

No , it is not that difficult.

Note: It would be much better if you don't have that n+1 in the sum. To avoid it you first divide by x then integrate in the first step.

 

[Amad27]

I saw it somewhere on the internet.
No , it is not that difficult.

Note: It would be much better if you don't have that n+1 in the sum. To avoid it you first divide by x then integrate in the first step.

Hello,

It would be extremely hard to integrate.

$$\sum_{n=1}^{\infty}H_{n}^{2}x^{n} = \frac{\log^2(1-x) + Li_2(x)}{(1-x)}$$

$$\sum_{n=1}^{\infty}H_{n}^{2}x^{n-1} = \frac{\log^2(1-x) + Li_2(x)}{x(1-x)}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}^{2}x^{n}}{n} = \int\frac{\log^2(1-x)}{x(1-x)} + \int \frac{Li_2(x)}{x(1-x)} $$

$$\int \frac{\log^2(1-x)}{x(1-x)} dx$$
Let $dv = \frac{\log^2(1-x)}{(1-x)} dx$
Let $u = \frac{1}{x} dx$

$$\int \frac{\log^2(1-x)}{x(1-x)} dx = \frac{-\log^3(1-x)}{3x} - \int \frac{\log^3(1-x)}{3x^2} dx$$

Let $t = 1/x \implies x = 1/t \implies dt = -1/x^2 \implies dx = -1/t^2 dt$

$$ = \log^3(t - 1) - \log^3(t)$$

$$ I = \int \frac{\log^2(1-x)}{x(1-x)} dx = \frac{-\log^3(1-x)}{3x} + \frac{1}{3}\int \frac{\log^3(-(t - t)) - \log^3(t)}{t^2} dt$$

Im lost trying to integrate..

 

[alyafey22]

Hello,

It would be extremely hard to integrate.

$$\sum_{n=1}^{\infty}H_{n}^{2}x^{n} = \frac{\log^2(1-x) + Li_2(x)}{(1-x)}$$

$$\sum_{n=1}^{\infty}H_{n}^{2}x^{n-1} = \frac{\log^2(1-x) + Li_2(x)}{x(1-x)}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}^{2}x^{n}}{n} = \int\frac{\log^2(1-x)}{x(1-x)} + \int \frac{Li_2(x)}{x(1-x)} $$

No no no. Try partial fractions not integration by parts.

 

[Amad27]

No no no. Try partial fractions not integration by parts.

I was tired yesterday, so I'll say that was my excuse for not realizing partial fractions...

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

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

The integral is extremely difficult and lengthy (W|A) states.

So, what can be done?

 

[alyafey22]

I was tired yesterday, so I'll say that was my excuse for not realizing partial fractions...

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

That should be $\log^2(1-x)$ not $\log^2(x)$. I think we can have an antiderivative for $$\int \frac{\mathrm{Li}_2(x)}{1-x}$$

 

[Amad27]

That should be $\log^2(1-x)$ not $\log^2(x)$. I think we can have an antiderivative for $$\int \frac{\mathrm{Li}_2(x)}{1-x}$$

I really am tired...

$$\frac{\log^2(1-x) + Li_2(x)}{x} - \frac{\log^2(1-x) + Li_2(x)}{(1-x)}$$

The integral gives:

$$\int \frac{\log^2(1-x) + Li_2(x)}{x} - \frac{\log^2(1-x) + Li_2(x)}{(1-x)}dx = Li_3(x) + \int\frac{\log^2(1-x)}{x} dx +\frac{\log^3(1-x)}{3} - \int \frac{Li_2(x)}{1-x} dx$$

I checked WolframAlpha, we do have an antiderivative for the dilog but it is complicated, so I tried this:

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

We get:

$$\int \frac{Li_2(x)}{1-x} dx = (-)\int \frac{\log(x)\log(1-x)}{1-x}dx - Li_3(1-x) -\zeta(2)\log(1-x)$$

It is still complicated to integrate, but I think integration by parts is what we would do next.

 

[alyafey22]

$$\frac{\log^2(1-x) + Li_2(x)}{x} - \frac{\log^2(1-x) + Li_2(x)}{(1-x)}$$

That should be plus sign not minus between the two terms.

 

[Amad27]

That should be plus sign not minus between the two terms.

But it still doesn't help the integration process?

 

[alyafey22]

But it still doesn't help the integration process?

Sometimes a wrong sign can make your life miserable.

 

[Amad27]

Sometimes a wrong sign can make your life miserable.

I like your humour.

By the way, I was wondering. Are you familiar with real analysis proofs? Because as I was cruising around, a lot of what you do here is based off of real analysis isn't it?

Can you perhaps link me to a page with all these generating functions? I mean, I tried searching online but got no results.

I did a lot of work on this. And this is what I have to now:

$$\sum_{n=1}^{\infty} \frac{H_{n}^{2}x^n}{n} = \log^2(1-x)\log(x) + 2Li_3(1-x) +2\zeta(2)\log(1-x) - \int \frac{Li_2(x)}{1-x} dx$$

The left over integral is very hard, but I am initially able to do this:

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

I remember you showed me a series a while ago, another gen. function.

I got:

$$2\sum_{n=1}^{\infty} \frac{x^{n+1}H_n}{(n+1)^2} = \int \frac{\log^2(1-x)}{x} dx$$

But the issue: How can I do the sum? The $x^{n+1}$ will cause a very big issue.

 

[alyafey22]

By the way, I was wondering. Are you familiar with real analysis proofs? Because as I was cruising around, a lot of what you do here is based off of real analysis isn't it?

I took a course in real analysis a year ago. To justify all the steps we need a firm knowledge of real analysis which I think I don't have.

Can you perhaps link me to a page with all these generating functions? I mean, I tried searching online but got no results.

Well , it is difficult to find them. They are not collected in one paper but scattered in many places.

I did a lot of work on this. And this is what I have to now:

$$\sum_{n=1}^{\infty} \frac{H_{n}^{2}x^n}{n} = \log^2(1-x)\log(x) + 2Li_3(1-x) +2\zeta(2)\log(1-x) - \int \frac{Li_2(x)}{1-x} dx$$

I am not sure how you got that ? Maybe you are messing something ?

 

[Amad27]

I took a course in real analysis a year ago. To justify all the steps we need a firm knowledge of real analysis which I think I don't have.
Well , it is difficult to find them. They are not collected in one paper but scattered in many places.
I am not sure how you got that ? Maybe you are messing something ?

I have a way of approaching this Ill post tomorrow, it is quite late here in Turkmenistan.

Anyway, have you actually solved this problem yourself?

 

[alyafey22]

Anyway, have you actually solved this problem yourself?

No , we are solving it together. I just thought of it when you posted that problem.

 

[Amad27]

No , we are solving it together. I just thought of it when you posted that problem.

Thats good. Collaboration for success.

For about two days I do not have time to work on this, since half my 11th grade semester is about to be over, and I am in a time crunch. Is it fine if we start working on this intensively in 2-3 days?

Thanks for cooperating!By the way:

http://mathhelpboards.com/calculus-10/integration-using-beta-gamma-functions-12437-9.htmlWe were using the wrong sum for the log^2 integral!

$$\sum_{n=1}^{\infty} \frac{x^k}{k} = -\log(1-x)$$

It doest equal $\frac{\log(1-x)}{x-1}$

??

 

[alyafey22]

Thats good. Collaboration for success.

For about two days I do not have time to work on this, since half my 11th grade semester is about to be over, and I am in a time crunch. Is it fine if we start working on this intensively in 2-3 days?

Thanks for cooperating!

Ok , no problem.

By the way:

http://mathhelpboards.com/calculus-10/integration-using-beta-gamma-functions-12437-9.htmlWe were using the wrong sum for the log^2 integral!

$$\sum_{n=1}^{\infty} \frac{x^k}{k} = -\log(1-x)$$

It doest equal $\frac{\log(1-x)}{x-1}$

??

I don't see what is wrong !

 

[Amad27]

Ok , no problem.
I don't see what is wrong !

OK,

I am ready to begin the problem again.

Just a question.

By any chance, do you have any other generating functions?

I don't see a way to use that one.

 

[alyafey22]

Just a question.

By any chance, do you have any other generating functions?

I found an interesting thread .

I don't see a way to use that one.

$$\sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

$$\sum_{n\geq 1} H^2_n x^{n-1} = \frac{\log^2(1-x)}{x}+\frac{\log^2(1-x)}{1-x}+\frac{\operatorname{Li}_2(x)}{1-x}+\frac{\operatorname{Li}_2(x)}{x}$$$$\sum_{n\geq 1} \frac{H^2_n}{n} x^{n} = \int^x_0\frac{\log^2(1-t)}{t}\,dt +\int^x_0\frac{\operatorname{Li}_2(t)}{1-t}\,dt-\frac{1}{3}\log^3(1-x)+\operatorname{Li}_3(x)$$

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

Hence we get

$$\sum_{n\geq 1} \frac{H^2_n}{n} x^{n} = -\log(1-t)\operatorname{Li}_2(t)-\frac{1}{3}\log^3(1-x)+\operatorname{Li}_3(x)$$

$$\sum_{n\geq 1} \frac{H^2_n}{n} x^{n-1} = -\frac{\log(1-x)\operatorname{Li}_2(x)}{x}-\frac{1}{3}\frac{\log^3(1-x)}{x}+\frac{\operatorname{Li}_3(x)}{x}$$

$$\sum_{n\geq 1} \frac{H^2_n}{n^2} x^{n} =\frac{ \operatorname{Li}^2_2(x)}{2}-\frac{1}{3}\int^x_0\frac{\log^3(1-t)}{t}\,dt+\operatorname{Li}_4(x)$$

$$\frac{1}{3}\int^x_0\frac{\log^3(1-t)}{t}\,dt =\operatorname{Li}_4(x)+\frac{ \operatorname{Li}^2_2(x)}{2}-\sum_{n\geq 1} \frac{H^2_n}{n^2} x^{n} $$

 

[Amad27]

I found an interesting thread .
$$\sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

$$\sum_{n\geq 1} H^2_n x^{n-1} = \frac{\log^2(1-x)}{x}+\frac{\log^2(1-x)}{1-x}+\frac{\operatorname{Li}_2(x)}{1-x}+\frac{\operatorname{Li}_2(x)}{x}$$$$\sum_{n\geq 1} \frac{H^2_n}{n} x^{n} = \int^x_0\frac{\log^2(1-t)}{t}\,dt +\int^x_0\frac{\operatorname{Li}_2(t)}{1-t}\,dt-\frac{1}{3}\log^3(1-x)+\operatorname{Li}_3(x)$$

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

Hence we get

$$\sum_{n\geq 1} \frac{H^2_n}{n} x^{n} = -\log(1-t)\operatorname{Li}_2(t)-\frac{1}{3}\log^3(1-x)+\operatorname{Li}_3(x)$$

$$\sum_{n\geq 1} \frac{H^2_n}{n} x^{n-1} = -\frac{\log(1-x)\operatorname{Li}_2(x)}{x}-\frac{1}{3}\frac{\log^3(1-x)}{x}+\frac{\operatorname{Li}_3(x)}{x}$$

$$\sum_{n\geq 1} \frac{H^2_n}{n^2} x^{n} =\frac{ \operatorname{Li}^2_2(x)}{2}-\frac{1}{3}\int^x_0\frac{\log^3(1-t)}{t}\,dt+\operatorname{Li}_4(x)$$

$$\frac{1}{3}\int^x_0\frac{\log^3(1-t)}{t}\,dt =\operatorname{Li}_4(x)+\frac{ \operatorname{Li}^2_2(x)}{2}-\sum_{n\geq 1} \frac{H^2_n}{n^2} x^{n} $$

This is good, but we still cannot do our integral. Here we don't have $\log^3(t)$ as a factor.

 

[alyafey22]

$$\int^x_0\frac{\log^3(1-t)}{t}\,dt =3\operatorname{Li}_4(x)+\frac{3}{2}\operatorname{Li}^2_2(x)-3\sum_{n\geq 1} \frac{H^2_n}{n^2} x^{n} $$

$$\int^1_0 \frac{\log^3(1-x)\log^3(x)}{x}\,dx$$

Integration by parts

$$\int^1_0 \frac{\log^3(1-x)\log^3(x)}{x}\,dx = 9\int^1_0 \frac{\operatorname{Li}_4(x)\log^2(x)}{x}\,dx+\frac{9}{2}\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx-9\sum_{n\geq 1} \frac{H^2_n}{n^2} \int^1_0x^{n-1}\log^2(x)\,dx $$

$$\int^1_0 \frac{\log^3(1-x)\log^3(x)}{x}\,dx = 9\int^1_0 \frac{\operatorname{Li}_4(x)\log^2(x)}{x}\,dx+\frac{9}{2}\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx-18\sum_{n\geq 1} \frac{H^2_n}{n^5} $$

The first integral

$$\int^1_0 \frac{\operatorname{Li}_4(x)\log^2(x)}{x}\,dx = 2\zeta(7)$$

I still have to think of a way to solve

$$\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx$$

 

[Amad27]

$$\int^x_0\frac{\log^3(1-t)}{t}\,dt =3\operatorname{Li}_4(x)+\frac{3}{2}\operatorname{Li}^2_2(x)-3\sum_{n\geq 1} \frac{H^2_n}{n^2} x^{n} $$

$$\int^1_0 \frac{\log^3(1-x)\log^3(x)}{x}\,dx$$

Integration by parts

$$\int^1_0 \frac{\log^3(1-x)\log^3(x)}{x}\,dx = 9\int^1_0 \frac{\operatorname{Li}_4(x)\log^2(x)}{x}\,dx+\frac{9}{2}\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx-9\sum_{n\geq 1} \frac{H^2_n}{n^2} \int^1_0x^{n-1}\log^2(x)\,dx $$

$$\int^1_0 \frac{\log^3(1-x)\log^3(x)}{x}\,dx = 9\int^1_0 \frac{\operatorname{Li}_4(x)\log^2(x)}{x}\,dx+\frac{9}{2}\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx-18\sum_{n\geq 1} \frac{H^2_n}{n^5} $$

The first integral

$$\int^1_0 \frac{\operatorname{Li}_4(x)\log^2(x)}{x}\,dx = 2\zeta(7)$$

I still have to think of a way to solve

$$\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx$$

Lets actually pass on this integral, I have a different contour integration question, which may or may not help with this later on.

Anyway, I hope Ill catch you there!

 

[alyafey22]

Here is the evaluation of the last integral. There might be a mistake in a constant

$$\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx = \sum_{n,k\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\log^2(x)\,dx= 2\sum_{n,k\geq 1}\frac{1}{(nk)^2(n+k)^3}$$

$$\sum_{n,k\geq 1}\frac{1}{n^2k^2(n+k)^3} = \sum_{n,k\geq 1}\frac{n+k-k}{n^3k^2(n+k)^3} = \sum_{n,k\geq 1}\frac{1}{n^3k^2(n+k)^2}-\sum_{n,k\geq 1}\frac{1}{n^3k(n+k)^3}$$

First sum

$$\begin{align}\sum_{n,k\geq 1}\frac{1}{n^3k^2(n+k)^2} &= \sum_{n,k\geq 1}\frac{1}{n^4k^2(n+k)}-\sum_{n,k\geq 1}\frac{1}{n^4k(n+k)^2}\\
&=\sum_{n,k\geq 1}\frac{1}{n^5k^2}-\sum_{n,k\geq 1}\frac{1}{n^5k(n+k)}-\sum_{n,k\geq 1}\frac{1}{n^5k(n+k)}+\sum_{n,k\geq 1}\frac{1}{n^5(n+k)^2}\\
&=\zeta(5)\zeta(2)-2\sum_{n \geq 1}\frac{H_n}{n^5}+\sum_{n \geq 1}\frac{H^{(2)}_n}{n^5}
\end{align}$$

$$\begin{align}\sum_{n,k\geq 1}\frac{1}{n^3k(n+k)^3} &= \sum_{n,k\geq 1}\frac{1}{n^4k(n+k)^2}-\sum_{n \geq 1}\frac{H_n^{(3)}}{n^4}\\
&= \sum_{n,k\geq 1}\frac{H^{(2)}_n}{n^5}-\sum_{n,k\geq 1}\frac{H_n}{n^5}-\sum_{n \geq 1}\frac{H_n^{(3)}}{n^4}
\end{align}$$