MHB Integration using Beta and Gamma Functions

[Amad27]

Look at this answer.

In the answer,

how is

$$\frac{H_n}{(n+1)^2} = \frac{H_{n+1}}{(n+1)^2} - \frac{1}{(n+1)^3}$$

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

$$= \frac{1}{(n+1)^2}\left(H_{n+1} - \frac{1}{(n+1)}\right)$$

??

 

[alyafey22]

The replier used that

$$H_{n+1} = H_n +\frac{1}{(n+1)}$$

Hence

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

I think he made a mistake because the index should change when making the substitution in the second line.

 

[Amad27]

The replier used that

$$H_{n+1} = H_n +\frac{1}{(n+1)}$$

Hence

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

I think he made a mistake because the index should change when making the substitution in the second line.

Wait, why would the index change?

$$\sum_{k=1}^{n} \frac{1}{k} = H_n$$

$$\sum_{k=1}^{n+1} \frac{1}{k} = \sum_{k=1}^{n} \frac{1}{k} + \frac{1}{n+1} = H_n + \frac{1}{n+1}$$

?

 

[alyafey22]

The replier claimed that

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

But this is not correct , can you see why ?

 

[alyafey22]

By the way , I got a question you might like. Find

$$\int^1_0 \mathrm{Li}_k(x)\,dx$$

 

[Amad27]

By the way , I got a question you might like. Find

$$\int^1_0 \mathrm{Li}_k(x)\,dx$$

For the other post, yes I see why it is not correct.

If you let u= n+1, then when n=1, $u=2$, which should be the index.

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

$$\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} \int_{0}^{1} \frac{x^n}{n^k} \,dx$$

From Fubini's theorem, I suppose we were allowed to interchange.

$$\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} [\frac{x^{n+1}}{(n+1)n^k}]_{0}^{1}$$

$$\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}$$

Let $S = \displaystyle \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}$

$$\frac{1}{(n+1)n^k}$$

I think we should decompose this.

$$ = \frac{A}{(n+1)}$$

But we cannot because of $k$. I am quite stumped.

 

[alyafey22]

Ok, I will give you the starting point

$$\sum \frac{1}{(n+1)n^k} = \sum \frac{1+n-n}{(n+1)n^k} =\sum \frac{1}{n^k}-\frac{1}{(1+n)n^{k-1}}$$

Hence this can be written as

$$S_k = \zeta(k) -S_{k-1} \, \,\,\, k\geq 2$$

 

[Amad27]

Ok, I will give you the starting point

$$\sum \frac{1}{(n+1)n^k} = \sum \frac{1+n-n}{(n+1)n^k} =\sum \frac{1}{n^k}-\frac{1}{(1+n)n^{k-1}}$$

Hence this can be written as

$$S_k = \zeta(k) -S_{k-1} \, \,\,\, k\geq 2$$

So,

This goes in a continuous cycle.

$$L = \zeta(k) - L$$

$L = \zeta(k)/2$

What can we do?

 

[alyafey22]

So,

This goes in a continuous cycle.

$$L = \zeta(k) - L$$

How is that true ?

 

[Amad27]

How is that true ?

Nevermind:

$S(k) = \zeta(k) - S(k-1)$

then

$S(k-1) = \zeta(k-1) - S(k-2)$

Then

$S(k-2) = \zeta(k-2) - S(k-3)$

So,

$S(k) = \zeta(k) - (\zeta(k-1) - S(k-2))$

$S(k) = \zeta(k) - \zeta(k-1) + S(k-2)$

$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) - S(k-3)$

I really don't understand how to sum the series.

 

[alyafey22]

Nevermind:

$S(k) = \zeta(k) - S(k-1)$

then

$S(k-1) = \zeta(k-1) - S(k-2)$

Then

$S(k-2) = \zeta(k-2) - S(k-3)$

So,

$S(k) = \zeta(k) - (\zeta(k-1) - S(k-2))$

$S(k) = \zeta(k) - \zeta(k-1) + S(k-2)$

$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) - S(k-3)$

I really don't understand how to sum the series.

You are on the correct path but you have to know the stopping criteria. That means the base case.

we know that

$$S(1) = \sum_{n\geq 1}\frac{1}{n(n+1)} = 1$$

So what is the general form for $S(k)$ ?

 

[Amad27]

You are on the correct path but you have to know the stopping criteria. That means the base case.

we know that

$$S(1) = \sum_{n\geq 1}\frac{1}{n(n+1)} = 1$$

So what is the general form for $S(k)$ ?

As you said earlier,

$$S(k) = \zeta(k) - S(k-1)$$

$$S(2) = \zeta(2) - S(1)$$

$$S(3) = \zeta(3) - S(2)$$

$$S(4) = \zeta(4) - S(3)$$

I am not sure what you mean here?

 

[alyafey22]

$$S(2) = \zeta(2) - S(1)$$

$$S(3) = \zeta(3) - S(2)$$

$$S(4) = \zeta(4) - S(3)$$

If you sum up these three formulas what do you get ? Can you generalize ?

 

[Amad27]

If you sum up these three formulas what do you get ? Can you generalize ?

$$S(k) = \zeta(k) + \int_{0}^{1} Li_{k-1}(x) \,dx$$

 

[alyafey22]

You already did it correctly , what is the problem , exactly ?

Each time you are subtracting a value from $k$ so eventually we will reach $S(1)$.

Nevermind:

$S(k) = \zeta(k) - S(k-1)$

then

$S(k-1) = \zeta(k-1) - S(k-2)$

Then

$S(k-2) = \zeta(k-2) - S(k-3)$

So,

$S(k) = \zeta(k) - (\zeta(k-1) - S(k-2))$

$S(k) = \zeta(k) - \zeta(k-1) + S(k-2)$

$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) - S(k-3)$

I really don't understand how to sum the series.

This also can be written as

$$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) -\cdots -(-1)^n\zeta(k-(n-1))+(-1)^nS(k-n)$$

 

[Amad27]

You already did it correctly , what is the problem , exactly ?

Each time you are subtracting a value from $k$ so eventually we will reach $S(1)$.
This also can be written as

$$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) -\cdots -(-1)^n\zeta(k-(n-1))+(-1)^nS(k-n)$$

Would you mind giving it away? I don't think I am getting anywhere.

 

[alyafey22]

We have

$$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) -\cdots -(-1)^n\zeta(k-(n-1))+(-1)^nS(k-n)$$

Let $n=(k-1)$

$$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) -\cdots -(-1)^{k-1}\zeta(2)+(-1)^{k-1}S(1)$$

 

[Amad27]

We have

$$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) -\cdots -(-1)^n\zeta(k-(n-1))+(-1)^nS(k-n)$$

Let $n=(k-1)$

$$S(k) = \zeta(k) - \zeta(k-1) + \zeta(k-2) -\cdots -(-1)^{k-1}\zeta(2)+(-1)^{k-1}S(1)$$

$$S(k) = \sum_{k=1}^{n} (-1)^{k+1} \zeta(n-k)$$

 

[alyafey22]

$$S(k) = \sum_{k=1}^{n} (-1)^{k+1} \zeta(n-k)$$

No , that is not correct. Try writing the partial sums of your series. Is it the same ?

 

[Amad27]

No , that is not correct. Try writing the partial sums of your series. Is it the same ?

$$S(k) = \sum_{z=0}^{k-1} \zeta(k - z)$$

 

[alyafey22]

$$S(k) = \sum_{z=0}^{k-1} \zeta(k - z)$$

1- The sums should alternate , right ?

2- The last term is $\zeta(k-(k-1))=\zeta(1)$ which is divergent.

3- Where is the last term $(-1)^{k-1}S(1)$ ?

 

[Amad27]

1- The sums should alternate , right ?

2- The last term is $\zeta(k-(k-1))=\zeta(1)$ which is divergent.

3- Where is the last term $(-1)^{k-1}S(1)$ ?

Lets see.We had:

$$ S_k = \zeta(k) - S_{k-1} $$

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

$$S_k = \zeta(k) - Li_{k-1}(x)$$

But I can't seem to get how to have a series representation...

 

[alyafey22]

That should be the final answer

\begin{align}
\int_0^1\mathrm{Li}_p(x)\,\mathrm{d}x
&=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}\sum_{k=1}^\infty\frac1{k(k+1)}\\
&=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}
\end{align}

You can use induction if you wish.

 

[Amad27]

That should be the final answer

\begin{align}
\int_0^1\mathrm{Li}_p(x)\,\mathrm{d}x
&=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}\sum_{k=1}^\infty\frac1{k(k+1)}\\
&=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}
\end{align}

You can use induction if you wish.

How did you possibly derive that?

By the way, I looked at this page:

http://mathhelpboards.com/calculus-10/integration-lessons-continued-12240.html#post63930

The one you posted on 30 November. Did you think of those proofs such as the one involving $H_k^(p)$ yourself? Firstly, you are extremely smart, secondly, how do you think of these ideas?

Thanks!

 

[alyafey22]

How did you possibly derive that?

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.

By the way, I looked at this page:

http://mathhelpboards.com/calculus-10/integration-lessons-continued-12240.html#post63930

The one you posted on 30 November. Did you think of those proofs such as the one involving $H_k^{(p)}$ yourself? Firstly, you are extremely smart, secondly, how do you think of these ideas?

Thanks!

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.

 

[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!