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?