Integration using Beta and Gamma Functions

[Amad27]

This is where a confusion happens. The inner $z$ is independent of the outer $z$ so to avoid that we always avoid such symbols. The inner $z$ is a dummy variable so we can choose any thing we want $x$ , $y$ or even $z$. So it is always preferable to use a symbol different than the boundary of integration for example $x$

$$\int^z_0 x^2\log(1-x)\,dx$$

Why not just avoid all confusion, and directly integrate form 0 to 1?anyway,

$\psi(5/4), z = 1/4$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)$

Let $x = z^4$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(4n+1)} = -\int_{0}^{1} \ln(1-z^4) \,dz$

So we must figure out $\displaystyle -\int_{0}^{1} \ln(1-z^4) \,dz$ then add $\gamma$ to it. The integral is tough though.

The antiderivative is too intense to even look at. How would you evaluate the integral?

Out of curiosity, can you get a closed-form by using series. Maybe that could help. I was looking at the challenges section of the forum.

One of them is your logarithmic integral #3, which remains unsolved..

I just had a thought... if you don't mind, maybe we can try the integral, since no one has done it in two months or is that not allowed? Thanks.

 

[alyafey22]

Why not just avoid all confusion, and directly integrate form 0 to 1?

Sure , that might be better :)

anyway,

$\psi(5/4), z = 1/4$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)$

Let $x = z^4$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(4n+1)} = -\int_{0}^{1} \ln(1-z^4) \,dz$

So we must figure out $\displaystyle -\int_{0}^{1} \ln(1-z^4) \,dz$ then add $\gamma$ to it. The integral is tough though.

The antiderivative is too intense to even look at. How would you evaluate the integral?

It might not be as tought as you think , use the property

$$(1-z^4) = (1-z)(1+z)(1+z^2)$$

Out of curiosity, can you get a closed-form by using series. Maybe that could help. I was looking at the challenges section of the forum.

One of them is your logarithmic integral #3, which remains unsolved..

I just had a tough... if you don't mind, maybe we can try the integral, since no one has done it in two months or is that not allowed? Thanks.

Can you give me the link ?

 

[Amad27]

Sure , that might be better :)
It might not be as tought as you think , use the property

$$(1-z^4) = (1-z)(1+z)(1+z^2)$$
Can you give me the link ?

Hey there!

$\displaystyle \log((1-z)(1+z)(1+z^2)) = \log(1-z) + \log(1+z) + \log(1+z^2)$

$\displaystyle \int_{0}^{1} \log((1-z)(1+z)(1+z^2)) \,dz = \int_{0}^{1} \log(1-z) + \log(1+z) + \log(1+z^2) \,dz = \lim_{z\to 1} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + \int_{0}^{1} \log(1 + z^2) \,dz$

Currently, I am running out of sleep, so I'll continue the integration tomorrow. Please do not give hints, just let me know if the setup is correct. Thanks.

NEXT:
----------
The integral was: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html

I spent hours and hours trying this, but later find out it requires special functions...

Question #1:
Do you know where I can get a quick course on special functions and more IMPORTANTLY, how to implement them into integration?

Functions such as $\zeta(s), L_1, L_2$ I am not sure what the ones with the $L$ are, supposed logarithmic integrals for $\frac{1}{\ln(x)}$ But perhaps you could guide me to somewhere they talk about this and how to implement this in depth.

Question #2:
You posted a hint there as well. I assume you want me to integrate the series, LHS, RHS to get it into some sort of form?

Question #3:
What is that series anyway? I am not familiar with the $H_kx^k$ idea or what It is. I tried putting it online, but no result?

Question #4:
Can you perhaps show me a page where they talk about all these laws, such as the one with dilogs etc shown here: http://mathhelpboards.com/calculus-10/higher-order-reflected-logarithms-13087.html

Thank you @ZaidAlyafey, I appreciate your guidance here in this forum, it is highly valuable to me! Thanks a bunch =)

 

[alyafey22]

Hey there!

$\displaystyle \log((1-z)(1+z)(1+z^2)) = \log(1-z) + \log(1+z) + \log(1+z^2)$

$\displaystyle \int_{0}^{1} \log((1-z)(1+z)(1+z^2)) \,dz = \int_{0}^{1} \log(1-z) + \log(1+z) + \log(1+z^2) \,dz = \lim_{z\to 1} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + \int_{0}^{1} \log(1 + z^2) \,dz$

Currently, I am running out of sleep, so I'll continue the integration tomorrow. Please do not give hints, just let me know if the setup is correct. Thanks.

Let me see the full solution.

NEXT:
----------
The integral was: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html

I spent hours and hours trying this, but later find out it requires special functions...

There are many ways to solve that integral. You can solve it using the beta function. It is a good practice , try it .

Question #1:
Do you know where I can get a quick course on special functions and more IMPORTANTLY, how to implement them into integration?

Functions such as $\zeta(s), L_1, L_2$ I am not sure what the ones with the $L$ are, supposed logarithmic integrals for $\frac{1}{\ln(x)}$ But perhaps you could guide me to somewhere they talk about this and how to implement this in depth.

These are called dilogarithms. I discussed lots of special functions on this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post14488.html#post14488.

Question #2:
You posted a hint there as well. I assume you want me to integrate the series, LHS, RHS to get it into some sort of form?

Exactly.

Question #3:
What is that series anyway? I am not familiar with the $H_kx^k$ idea or what It is. I tried putting it online, but no result?

$H_k$ is called the harmonic number. The series containg them are called Euler sums.

Question #4:
Can you perhaps show me a page where they talk about all these laws, such as the one with dilogs etc shown here: http://mathhelpboards.com/calculus-10/higher-order-reflected-logarithms-13087.html

You can refer to the link above on Advanced integration techniques. In this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post23071.html#post23071I discuss polylogarithms.

 

[Amad27]

Let me see the full solution.
There are many ways to solve that integral. You can solve it using the beta function. It is a good practice , try it .
These are called dilogarithms. I discussed lots of special functions on this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post14488.html#post14488.
Exactly.
$H_k$ is called the harmonic number. The series containg them are called Euler sums.
You can refer to the link above on Advanced integration techniques. In this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post23071.html#post23071I discuss polylogarithms.

The improper integrals must be divergent. The limit cannot exist because of $\ln(1-z)$ the limit as $x \to 1$ does not exist.

Question #1:
Also where did you learn special functions? What is a possible textbook, since i can access that offline. Your page was amazing but my Internet can't load all LaTex making it immensely hard to read. But I tried.

I read the page, I was able to prove the integral.
$\displaystyle Li_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2} = -\int_{0}^{z} \frac{log(1-x)}{x}\,dx$

$\displaystyle \sum_{n=1}^{\infty} {x}^{n-1} = \frac{1}{1-x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln\left({1-x}\right)$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = \frac{-\ln(1-x)}{x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{z^n}{n^2} = \int_{0}^{z} \frac{-\ln(1-x)}{x} \,dx$

Let me know.

Question #2:

How can you integrate $Li_2$? I tried from $0 \to 1$

$\displaystyle \int_{0}^{1} Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$

$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$

$$(An + B)(n+1) + D(n^2) = 1$$

Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$

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

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

The $1/n$ is the problem, it is the harmonic series, which diverges.

Thanks (BTW: Where did you find the dilog reflection law?)

Thanks =)

 

[alyafey22]

The improper integrals must be divergent. The limit cannot exist because of $\ln(1-z)$ the limit as $x \to 1$ does not exist.

I must see your full solution. If you are talking about $\int^1_0\log(1-z)dz$ then this integral converges and it is value is $-1$.

Question #1:
Also where did you learn special functions? What is a possible textbook, since i can access that offline. Your page was amazing but my Internet can't load all LaTex making it immensely hard to read. But I tried.

Oh , that is so bad. Unfortunately , I learned each special function separately by searching the internet about its properties. I don't reckon seeing a book that explains all special functions.

I read the page, I was able to prove the integral.
$\displaystyle Li_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2} = -\int_{0}^{z} \frac{log(1-x)}{x}\,dx$

$\displaystyle \sum_{n=1}^{\infty} {x}^{n-1} = \frac{1}{1-x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln\left({1-x}\right)$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = \frac{-\ln(1-x)}{x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{z^n}{n^2} = \int_{0}^{z} \frac{-\ln(1-x)}{x} \,dx$

Let me know.

That is good , you are making a big progress.

Question #2:

How can you integrate $Li_2$? I tried from $0 \to 1$

$\displaystyle \int_{0}^{1} Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$

$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$

$$(An + B)(n+1) + D(n^2) = 1$$

Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$

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

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

The $1/n$ is the problem, it is the harmonic series, which diverges.

Very nice. Now separate the sum into

\(\displaystyle \sum \frac{1}{n^2 }+\sum \frac{1}{n+1}-\frac{1}{n}\)

The right is a telescoping series , isn't it ?

Thanks (BTW: Where did you find the dilog reflection law?)

Thanks =)

http://mathhelpboards.com/challenge-questions-puzzles-28/dilogarithmic-integration-parts-10876.html?highlight=reflectionare two proofs.

 

[Amad27]

I must see your full solution. If you are talking about $\int^1_0\log(1-z)dz$ then this integral converges and it is value is $-1$.
Oh , that is so bad. Unfortunately , I learned each special function separately by searching the internet about its properties. I don't reckon seeing a book that explains all special functions.
That is good , you are making a big progress.
Very nice. Now separate the sum into

\(\displaystyle \sum \frac{1}{n^2 }+\sum \frac{1}{n+1}-\frac{1}{n}\)

The right is a telescoping series , isn't it ?
http://mathhelpboards.com/challenge-questions-puzzles-28/dilogarithmic-integration-parts-10876.html?highlight=reflectionare two proofs.

Thank you ZaidAlyafey, you are a very encouraging person, which I appreciate. I appreciate your help here as well, I have learned a lot from you, especially about series.

I went to the proof page. I read your proof, it was very nice :D But it stated,

take $x = 1$ we get C = $Li_2$, but how? What is $f(x)$ if $x = 1$? I won't be defined.

The integral first.

I need $\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz$ But the question is how...

$\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz = ?$

I am thinking of a strategy for this. I checked WolframAlpha. The integral is quite intense, I think by parts is the best idea.

It was intense, I did it on paper. I ran into a loop, typing it would be very hard, I'll do the last few steps here.

$I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx) + 0.5I]$

$-I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

$I = \displaystyle -x\ln(1 + x^2) + 2x^2\arctan(x) - 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

Finally,

$I_{Final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [(-1\ln(2)) + 2\arctan(1) - 4[\arctan(1) - 0.5\log(2) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - 4[\pi/4 - \log(\sqrt{2}) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - \pi + \log(4) - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [\log(2) - \frac{\pi}{2} - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$$2\int x\arctan(x) \,dx = x^2\arctan(x) - (x/2)\log(1+x^2) + 0.5\int log(1 + x^2)$$

Ahhh this is crazy. So much going around, now I am back at the log(1+x^2) integral... Can this be done otherwise?

Lets try the: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html integral.

$\displaystyle \int_{0}^{1} \log^2(x)\log^2(1-x) \,dx$

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

$$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

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

If it has to do with integration of $H_k$ then I have no clue. Well, I sort of do.

$\displaystyle H_k = \sum_{n=1}^{k} \frac{1}{n}$

$\displaystyle (H_k)(x^k) = \sum_{n=1}^{k} \frac{x^k}{n}$

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

$\displaystyle \sum_{k=1}^{\infty} \int_{0}^{z}H_k(x^k) \,dx = \int_{0}^{z} \frac{-\log(1-x)}{1-x} \,dx$

I have just one issue.

$H_k = \displaystyle \sum_{n=1}^{k} \frac{1}{n}$
$x^k = x^k$

$H_k (x^k) = \displaystyle \sum_{n=1}^{k} \frac{x^k}{n)$

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n}$ is what we must figure out, which seems ultimately difficult.

using Fubini's and Tonelli's theorems,

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n} = \sum_{n=1}^{k} \frac{1}{n} \sum_{k=1}^{\infty} \frac{x^k}{1}$

$\displaystyle \sum_{k=1}^{\infty} x^{k} = \frac{1}{1-x} - 1 = \frac{x}{1-x}$

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

Okay, I am confused.

How can you integrate $H_k$? It becomes extremely difficult. Any suggestions? Thanks

 

[alyafey22]

I went to the proof page. I read your proof, it was very nice :D But it stated,

take $x = 1$ we get C = $Li_2$, but how? What is $f(x)$ if $x = 1$? I won't be defined.

I started by setting

$$f(x) = \mathrm{Li}_2(x) + \mathrm{Li}_2(1-x)$$

Hence setting $x\to 1 $ we get

$$c = \lim_{x\to 1 } f(x) +\log(x)\log(1-x) = \mathrm{Li}_2(1) + \mathrm{Li}_2(0)+ 0 = \mathrm{Li}_2(1) =\sum \frac{1}{n^2} = \zeta(2)$$

The integral first.

I need $\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz$ But the question is how...

$\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz = ?$

I am thinking of a strategy for this. I checked WolframAlpha. The integral is quite intense, I think by parts is the best idea.

It was intense, I did it on paper. I ran into a loop, typing it would be very hard, I'll do the last few steps here.

$I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx) + 0.5I]$

$-I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

$I = \displaystyle -x\ln(1 + x^2) + 2x^2\arctan(x) - 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

Finally,

$I_{Final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [(-1\ln(2)) + 2\arctan(1) - 4[\arctan(1) - 0.5\log(2) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - 4[\pi/4 - \log(\sqrt{2}) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - \pi + \log(4) - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [\log(2) - \frac{\pi}{2} - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$$2\int x\arctan(x) \,dx = x^2\arctan(x) - (x/2)\log(1+x^2) + 0.5\int log(1 + x^2)$$

Ahhh this is crazy. So much going around, now I am back at the log(1+x^2) integral... Can this be done otherwise?

Maybe you are doing it the wrong way ? Integrate $dx$ and differentiate $\log(1+x^2)$

Lets try the: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html integral.

$\displaystyle \int_{0}^{1} \log^2(x)\log^2(1-x) \,dx$

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

$$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

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

If it has to do with integration of $H_k$ then I have no clue. Well, I sort of do.

$\displaystyle H_k = \sum_{n=1}^{k} \frac{1}{n}$

$\displaystyle (H_k)(x^k) = \sum_{n=1}^{k} \frac{x^k}{n}$

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

$\displaystyle \sum_{k=1}^{\infty} \int_{0}^{z}H_k(x^k) \,dx = \int_{0}^{z} \frac{-\log(1-x)}{1-x} \,dx$

I have just one issue.

$H_k = \displaystyle \sum_{n=1}^{k} \frac{1}{n}$
$x^k = x^k$

$H_k (x^k) = \displaystyle \sum_{n=1}^{k} \frac{x^k}{n)$

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n}$ is what we must figure out, which seems ultimately difficult.

using Fubini's and Tonelli's theorems,

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n} = \sum_{n=1}^{k} \frac{1}{n} \sum_{k=1}^{\infty} \frac{x^k}{1}$

$\displaystyle \sum_{k=1}^{\infty} x^{k} = \frac{1}{1-x} - 1 = \frac{x}{1-x}$

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

Okay, I am confused.

How can you integrate $H_k$? It becomes extremely difficult. Any suggestions? Thanks

I am confused too :) . It will be difficult for you trying to solve many different problems using different approaches. Euler sums require a descent knowledge of polylogarithms. Let us stick with finding the integral by differentiating Beta ?

Maybe we can start out by and easier one

$$\int^1_0 \log(x)\log(1-x) \, dx$$

PS: Nice that you are trying to post your attempts

 

[Amad27]

I started by setting

$$f(x) = \mathrm{Li}_2(x) + \mathrm{Li}_2(1-x)$$

Hence setting $x\to 1 $ we get

$$c = \lim_{x\to 1 } f(x) +\log(x)\log(1-x) = \mathrm{Li}_2(1) + \mathrm{Li}_2(0)+ 0 = \mathrm{Li}_2(1) =\sum \frac{1}{n^2} = \zeta(2)$$
Maybe you are doing it the wrong way ? Integrate $dx$ and differentiate $\log(1+x^2)$

I am confused too :) . It will be difficult for you trying to solve many different problems using different approaches. Euler sums require a descent knowledge of polylogarithms. Let us stick with finding the integral by differentiating Beta ?

Maybe we can start out by and easier one

$$\int^1_0 \log(x)\log(1-x) \, dx$$

PS: Nice that you are trying to post your attempts

Thank you Zaid. I will post my attempts on the first integral (the one with $\ln(1-z^4)$) tomorrow, I want to try this second one now.

$$\int^1_0 \log(x)\log(1-x) \, dx$$

I actually asked $Li_2$ because of this, we notice that.

$\log(x)\log(1-x) = -Li_2(x) - Li_2(1-x) + \pi^2/6$

$$\int^1_0 \log(x)\log(1-x) \, dx = \int_{0}^{1} -Li_2(x) - Li_2(1-x) + \pi^2/6 \,d$$

$Li_2(x) = \displaystyle \sum_{k=1}^{\infty} \frac{x^k}{k^2}$

$\displaystyle \int_{0}^{1} Li_2(x) \,dx = \displaystyle \sum_{k=1}^{\infty} \int_{0}^{1} \frac{x^k}{k^2} \,dx = \sum_{k=1}^{\infty} \frac{1}{k^2(k+1)} $

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2(k+1)} = \sum_{k=1}^{\infty}\frac{1}{k^2} - \frac{1}{k} + \frac{1}{(k+1)}$

$\displaystyle = \zeta(2) + \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k}$

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k}$

$S_1 = \frac{1}{2} - 1 $
$S_2 = \frac{1}{2} - 1 + \frac{1}{3} - \frac{1}{2}$
$S_3 = \frac{1}{2} - 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{4} - \frac{1}{3}$

$S_n = -1 + \frac{1}{n+1}$
$\displaystyle \lim_{n\to\infty} S_n = -1$

$Sum_{final} = \zeta(2) - 1$

Next We go to $\Li_2(1-x)$

$\Li_2(1-x) = \displaystyle \sum_{k=1}^{\infty} \frac{(1-x)^k}{k^2}$

$\int_{0}^{1} \Li_2(1-x) \,dx = \displaystyle \sum_{k=1}^{\infty}\int_{0}^{1} \frac{(1-x)^k}{k^2} \,dx$

$$ = \sum_{k=1}^{\infty} 0 = 0$$

because $Li_2(x)$ works for $0 < x < 1$,
$Li_2(1-x)$ should have domain $ 1 > x > 0$

So we take the limit as $x \to 1$ which is $-infty$ This is the hard part.

If I could get some help with $Li_2(1-x)$ I should be ready to roll!

Thanks. I would like to try the integral with Euler sums rather than differentiating Beta. Mostly because the Euler method is cool. And second, I have never learned multivariable calculus formally, just from some web reading, so when it comes to differentials (multivariable, will be required) I won't be able to do anything. So I would like to try euler sums!

 

[alyafey22]

Use \mathrm{Li}_p to display polylogarithms in a better way.

 

[Amad27]

Could you rewrite that? It doesn't display well on my machine.

Its okay. I got it,

$\displaystyle \lim_{x\to 1} (1-x) = 0$

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

$\displaystyle = \int_{0}^{1} -(\zeta(2) - 1) + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} -2\zeta(2) + 2 + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} 2 - \zeta(2) \,dx$

$= 2 - \zeta(2)$

Can we try the second power integral with Euler Sums ?? Please?

 

[alyafey22]

Its okay. I got it,

$\displaystyle \lim_{x\to 1} (1-x) = 0$

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

$\displaystyle = \int_{0}^{1} -(\zeta(2) - 1) + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} -2\zeta(2) + 2 + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} 2 - \zeta(2) \,dx$

$= 2 - \zeta(2)$

Some how you got the idea but it should be like this

$$\int^1_0 \mathrm{Li}_2(1-t) \,dt = \int^1_0 \mathrm{Li}_2(x)\,dx $$

using the change of the variable $t = 1-x$.

Can we try the second power integral with Euler Sums ?? Please?

Ok, but you have to revise your calculations and focus more. Let me see your second attempt.

 

[Amad27]

Some how you got the idea but it should be like this

$$\int^1_0 \mathrm{Li}_2(1-t) \,dt = \int^1_0 \mathrm{Li}_2(x)\,dx $$

using the change of the variable $t = 1-x$.
Ok, but you have to revise your calculations and focus more. Let me see your second attempt.

Okay, I'll do a full version here:

$$\int_{0}^{1} \log(x)\log(1-x) \,dx$$
$$\log(x)\log(1-x) = -(L_2(x) + L_2(1-x)) + \zeta(2)$$
$$= \int_{0}^{1} -(L_2(x) + L_2(1-x)) \,dx + \zeta(2)$$

Evaluation of $\int_{0}^{1} L_2(x) \,dx$:
$$L_2(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}$$
$$\int_{0}^{1} L_2(x) \,dx = \sum_{k=1}^{\infty} \frac{1}{k^2(k+1)}$$
$$= \zeta(2) + \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k^2}$$
$$S_1 = \frac{1}{2} - \frac{1}{1}$$
$$S_n = \frac{1}{n+1} - 1$$
$$\lim_{n\to\infty} S_n = -1$
Integral of $L_2(x) = \zeta(2) - 1$

Evaluation of $L_2(1-x)$:

The sum is exactly like the one as $L_2(x)$ so integral of $L_2(1-x) = \zeta(2) - 1$

Evaluation of all combined:
$$= \int_{0}^{1} -(L_2(x) + L_2(1-x)) \,dx + \zeta(2)$$
$$= \int_{0}^{1} -2(\zeta(2) - 1) \,dx + \zeta(2)$$
$$ = -2\zeta(2) + 2 + \zeta(2)$$
$$I_{final} = 2 - \zeta(2)$$

I began working on the power integral.

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx$$

$$\sum_{k=1}^{\infty} \frac{x^k}{k} = -\frac{\log(1-x)}{1-x}$$
$$\int\sum_{k=1}^{\infty} \frac{x^k}{k} \,dx = \int -\frac{\log(1-x)}{1-x} \,dx$$
Let $u = \log(1-x) \implies du = \frac{-1}{1-x} dx$
$$\int\sum_{k=1}^{\infty} \frac{x^k}{k} \,dx = \frac{\log^2(1-x)}{2} $$
$$\log^2(1-x) = 2\sum_{k=1}^{\infty} \frac{x^{k+1}}{k(k+1)}$$

We see that:

$$\sum_{k=1}^{\infty} \frac{z^k}{k} = -\frac{\log(1-z)}{1-z}$$
Let $z = 1 - x$, it guarantees $|z| < 1$
$$(-)\sum_{k=1}^{\infty} \frac{(x+1)^k}{k} = \frac{\log(x)}{x}$$
$$(-2)\sum_{k=1}^{\infty} \frac{(x+1)^{k+1}}{k(k+1)} = \log^2(x)$$

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx = (-4)\cdot\int_{0}^{1} \sum_{k=1}^{\infty} \frac{(x+1)^{k+1}}{k(k+1)})(\sum_{k=1}^{\infty} \frac{x^{k+1}}{k(k+1)})\,dx$$

So far so good?

$$= (-4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{(x+1)^{k+1}}{k(k+1)}) \frac{x^{k+1}}{k(k+1)}) \,dx = (-4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} (\frac{(2)^{k+1}}{k(k+1)}) - \frac{(1)}{k(k+1)}))(\frac{1}{k(k+1)}) = (-4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \frac{(2^{k+1} - 1)}{k^2(k+1)^2}$$

I must've done something wrong somewhwre.

Ah-Ha I found the mistake. I got the wrong square log.

It still won't be correct. I just can't get the squared log.

 

[alyafey22]

Evaluation of all combined:
$$= \int_{0}^{1} -(L_2(x) + L_2(1-x)) \,dx + \zeta(2)$$
$$= \int_{0}^{1} -2(\zeta(2) - 1) \,dx + \zeta(2)$$
$$ = -2\zeta(2) + 2 + \zeta(2)$$
$$I_{final} = 2 - \zeta(2)$$

You should not write it like that. It should be

$$ -\int^1_0 L_2(x)dx- \int^1_0 L_2(1-x) dx = -2\int^1_0 L_2(x)dx = -2(\zeta(2)-1) $$

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

How did you get that ? are you missing the harmonic number ?

 

[Amad27]

You should not write it like that. It should be

$$ -\int^1_0 L_2(x)dx- \int^1_0 L_2(1-x) dx = -2\int^1_0 L_2(x)dx = -2(\zeta(2)-1) $$
How did you get that ? are you missing the harmonic number ?

mmm... Why can't I write the integral like that? I just factored out a $(-1)$ Is that not correct?

Secondly.

I did miss out the harmonic number unforunately. But anyhow.

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

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

Actually,

$$\sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} = \frac{-\log^2(1-x)}{2} \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $$

Since $\displaystyle \sum_{k=1}^{\infty} (H_k)(x^k) = -\frac{\log(1-x)}{1-x}$ this implies the fact that $\displaystyle \sum_{k=1}^{\infty} (H_k)((1-x)^k) = -\frac{\log(x)}{x}$

$$ (-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1} = \log^2(x)$$

$$\implies \log^2(x)\log^2(1-x) = \left((-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left((-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right)$$

$$\implies \int_{0}^{1}\log^2(x)\log^2(1-x) \,dx = \int_{0}^{1} \left((-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left((-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot\int_{0}^{1} \left( \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left( \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot \left( \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{(H_k)(1-x)^{k+1}}{k+1} \cdot \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot ( \sum_{k=1}^{\infty} \sum_{k=1}^{\infty})\frac{(H_k)^2}{(k+1)^2} \cdot \int_{0}^{1} (1-x)^{k+1}\cdot(x)^{k+1}\,dx$$

Realize, $\displaystyle \int_{0}^{1} (1-x)^{(k+2) - 1}\cdot(x)^{(k+2) - 1} \,dx = B(k+2, k+2) = \frac{\Gamma^2(k+2)}{\Gamma(2(k+2)}$

Finally, $\displaystyle = (4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty}\frac{(H_k)^2}{(k+1)^2} \cdot \frac{\Gamma^2(k+2)}{\Gamma(2(k+2)}$

Let's see if we can simplify $\displaystyle \frac{(H_k)^2}{(k+1)^2} \cdot \frac{\Gamma^2(k+2)}{\Gamma(2(k+2)}$

$$\Gamma(k+2) = (k+1)!$$
$$\Gamma(2(k+2)) = (2k + 1)!$$

$$= \frac{(H_k)^2}{(k+1)^2} \cdot \frac{(k+1)!}{((2k+1)!}$$

Finally, $\displaystyle = (4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty}\frac{(H_k)^2}{(k+1)^2} \cdot \frac{(k+1)!}{(2k+1)!} $

I have no clue how to evaluate this sum...

 

[alyafey22]

mmm... Why can't I write the integral like that? I just factored out a $(-1)$ Is that not correct?

Ok , you should revise it because you made a small mistake.

\(\displaystyle = \int_{0}^{1} -2(\zeta(2) - 1) \,dx + \zeta(2)\)

There shouldn't be an integral there because you already integrated, right ?

$$= (4)\cdot\int_{0}^{1} \left( \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left( \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot \left( \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{(H_k)(1-x)^{k+1}}{k+1} \cdot \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

Are you sure you can do this ? I mean

$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

Let us try

$$\sum_{n \geq 0}r^n = \frac{1}{1-r} $$

Is it the case that ?

$$\sum_{n \geq 0}r^n \cdot \sum_{n \geq 0}r^n = \sum_{n \geq 0} \sum_{n \geq 0}r^{2n}$$

 

[Amad27]

Ok , you should revise it because you made a small mistake.
There shouldn't be an integral there because you already integrated, right ?
Are you sure you can do this ? I mean

$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

Let us try

$$\sum_{n \geq 0}r^n = \frac{1}{1-r} $$

Is it the case that ?

$$\sum_{n \geq 0}r^n \cdot \sum_{n \geq 0}r^n = \sum_{n \geq 0} \sum_{n \geq 0}r^{2n}$$

Question #1:

So, the answer I had for $\log(x)\log(1-x)$ is incorrect? ??

Question #2:
What are the requirements for:
$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

??

Also, mmmm. Can you give me a hint? I have the parts.

I had $\log^(1-x)$ and $\log^2(x)$ written as separate sums, but I can't take the sum because the harmonic number is there.

$\displaystyle \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $

$\displaystyle (-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1} = \log^2(x) $

Lets try the sum of $\log^2(1-x)$ first.

$\displaystyle \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$
$$\sum_{k=1}^{\infty} x^k = \frac{x}{1-x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = \int \frac{x}{1-x} \,dx$$
$$(-2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{x^{k+1}}{k+1} = (-2H_k)\cdot\int \frac{x}{1-x} \, dx = \log^2(1-x) = (2H_k)\cdot(x + \log(1-x))$$

$$\therefore \log^2(1-x) = (2H_k)\cdot(x + \log(1-x))$$

Now, let's work on $\log^2(x)$

$\displaystyle \implies \log^2(x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)((1-x)^{k+1})}{{k+1}} $

$$= \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{1} = \frac{1}{x}$$
$$= \sum_{k=1}^{\infty} \frac{(1-x)^{k}}{1} = \frac{(1-x)}{x}$$
$$= \sum_{k=1}^{\infty} (-)\cdot \frac{(1-x)^{k+1}}{k+1} = \int \frac{(1-x)}{x}$$
By part-integration, $\displaystyle (-)\cdot \sum{k=1}^{\infty} \frac{(1-x)^{k+1}}{k+1} = \log(x) - x$

Recall, $\log^(x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)((1-x)^{k+1})}{{k+1}} $

So we must multiply the previous sum by $(2H_k)$ to get:

$$\log^2(x) = (-2H_k)\cdot \sum{k=1}^{\infty} \frac{(1-x)^{k+1}}{k+1} = 2H_k(\log(x) - x)$$

Finally,

$$\log^2(x)\log^2(1-x) = \left(2H_k(\log(x) - x)\right)\cdot\left((2H_k)\cdot(x + \log(1-x))\right) = (2H_k)\left(\log(x) + \log(1-x)\right)$$

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx = (2H_k) \int_{0}^{1} \log(x) + \log(1-x) \,dx$$

Considering that $\displaystyle \int \log(u) \,du = u(\log(u) - 1)$

$$= (2H_k)\cdot(-2) = -4(H_k)$$

I don't get it. I got

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx = (2H_k) \int_{0}^{1} \log(x) + \log(1-x) \,dx = -4(H_k)$$

But there is no way to compute $H_k$. What can be done?

 

[alyafey22]

Question #1:

So, the answer I had for $\log(x)\log(1-x)$ is incorrect? ??

It is correct but you should have not written it like this

$$\int^1_0(-2)(\zeta(2)-1)\,dx $$

You already integrated , right ? and you got the value to be $
\zeta(2)-1$ . Why integrating again ?

Question #2:
What are the requirements for:
$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

??

I must say , I don't know.

Also, mmmm. Can you give me a hint? I have the parts.

$$(-2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{x^{k+1}}{k+1} = (-2H_k)\cdot\int \frac{x}{1-x} \, dx = \log^2(1-x) = (2H_k)\cdot(x + \log(1-x))$$

No , you can't multiply both sides by $H_k$ because the value for $k$ depends on the summation , it is like you are doing that

$$\int x \,dx = \frac{x^2}{2}+c$$

Then

$$\int x^2 \,dx = \frac{x^3}{2}+c$$

which is not correct since the $x$ cannot be inserted inside the integration. The same for $H_k$ you cannot insert it inside the summation. Also when you muliply by $x$ the outer $x$ is different than the inner $x$.

I will give the hint once we agree on the above notes.

 

[Amad27]

It is correct but you should have not written it like this

$$\int^1_0(-2)(\zeta(2)-1)\,dx $$

You already integrated , right ? and you got the value to be $
\zeta(2)-1$ . Why integrating again ?
I must say , I don't know.
No , you can't multiply both sides by $H_k$ because the value for $k$ depends on the summation , it is like you are doing that

$$\int x \,dx = \frac{x^2}{2}+c$$

Then

$$\int x^2 \,dx = \frac{x^3}{2}+c$$

which is not correct since the $x$ cannot be inserted inside the integration. The same for $H_k$ you cannot insert it inside the summation. Also when you muliply by $x$ the outer $x$ is different than the inner $x$.

I will give the hint once we agree on the above notes.

Okay.

For the first integral of $\log(x)\log(1-x)$ I agree. I just realized it was a typo. I didn't pay enough attention, sorry, but I hope you know what I meant. Anyhow,

For the other harder integral, I understand. Because (H_k) is related to $k$ so we can't simply multiply it after.

I was just thinking that:

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

So what we had:

$\displaystyle \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $

$\displaystyle (-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1} = \log^2(x) $

First: $\log^2(1-x)$

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

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

But the integral in the numerator does not converge.

I must say, you have given an extremely tough challenge.

 

[alyafey22]

Okay , here is the hint

Only expand $\log^2(1-x)$ as series. Leave $\log^2(x)$ as it is.

You will see later , why!

 

[Amad27]

Okay , here is the hint

Only expand $\log^2(1-x)$ as series. Leave $\log^2(x)$ as it is.

You will see later , why!

Hello ZaidAlyafey,

Lets see what we got here:

$$\log^2(1-x) = (-2)\cdot\sum_{k=1}^{\infty}\frac{(H_k)(x)^{(k+1)}}{(k+1)}$$

$$\log^2(1-x)\log^2(x) = (2)\cdot\sum_{k=1}^{\infty} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)}$$

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

$$\int_{0}^{1} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)} \,dx = \frac{H_k}{k+1}\cdot \int_{0}^{1} \log^2(x)\cdot (x)^{k+1} \,dx$$

The integral is the tough part.

It is very well displayed by the second partial derivative of the beta function but I yet have no clue how to compute trigammas etc.

Through integration by parts I was able to do it.

$$\int_{0}^{1} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)} \,dx = \frac{H_k}{k+1}\cdot \int_{0}^{1} \log^2(x)\cdot (x)^{k+1} \,dx = \frac{H_k}{k+1} \cdot \frac{(2)}{(k+2)^3}$$

So we have,

$$I = \sum_{k=1}^{\infty}\frac{2H_k}{(k+1)(k+2)^3}$$

What is left is, is to find this difficult sum. It is tough because of the harmonic number...

I'm not sure...

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

This is quite difficult =)

 

[alyafey22]

$$I = \sum_{k=1}^{\infty}\frac{2H_k}{(k+1)(k+2)^3}$$

Very good , nice work! Now we should use partial fractions , right ?

 

[Amad27]

Very good , nice work! Now we should use partial fractions , right ?

That was one of my attempts, but it doesn't work, I'll try again here:

The harmonic number is not a constant. Partial fraction requires the numerator to be a constant! Ah - now I see why I was failing miserably.

 

[alyafey22]

And you may use the following

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

If looking for a proof .

 

[Amad27]

And you may use the following

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

If looking for a proof .

Thanks for the hint, but I don't think it will apply in this case.

The denominator is not one variables, it is a product of two binomials, which the theorem you stated does not apply for.

 

[alyafey22]

Thanks for the hint, but I don't think it will apply in this case.

The denominator is not one variables, it is a product of two binomials, which the theorem you stated does not apply for.

That's why you have to use partial fractions.

 

[Amad27]

That's why you have to use partial fractions.

We had the following,

$$\frac{2H_k}{(k+1)(k+2)^3}$$

$$\frac{A}{k+1}+\frac{B}{k+2}+\frac{C}{(k+2)^2}+\frac{D}{(k+2)^3} = \frac{2H_k}{(k+1)(k+2)^3}$$

$$(A)(k+2)^3 + (k+1)(k+2)^2(B) + (k+1)(k+2)(C) + (k+1)(D) = 2H_k$$

I was thinking $k=-1$ but I am not sure if $H_{-1}$ exists.

 

[alyafey22]

We had the following,

$$\frac{2H_k}{(k+1)(k+2)^3}$$

$$\frac{A}{k+1}+\frac{B}{k+2}+\frac{C}{(k+2)^2}+\frac{D}{(k+2)^3} = \frac{2H_k}{(k+1)(k+2)^3}$$

$$(A)(k+2)^3 + (k+1)(k+2)^2(B) + (k+1)(k+2)(C) + (k+1)(D) = 2H_k$$

I was thinking $k=-1$ but I am not sure if $H_{-1}$ exists.

Ignore $H_k$ , just find the partial fraction decomposition for

$$\frac{1}{(k+1)(k+2)^3}$$

 

[Amad27]

Ignore $H_k$ , just find the partial fraction decomposition for

$$\frac{1}{(k+1)(k+2)^3}$$

That is simply, (skipping steps):

$$\frac{1}{(k+1)(k+2)^3} = -(\frac{1}{k+2} + \frac{1}{(k+2)^2} + \frac{1}{(k+2)^3}) + \frac{1}{k+1}$$

$$\frac{2H_k}{(k+1)(k+2)^3} = \frac{-2H_k}{k+2} - \frac{2H_k}{(k+2)^2} - \frac{2H_k}{(k+2)^3} + \frac{2H_k}{k+1}$$

But the theorem you stated still won't apply as the denominator is still a binomial...

 

[alyafey22]

Hint

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

You can use the same concept to find

$$\sum_{k=1}^\infty \frac{H_k}{(k+2)^q}$$

 

[Amad27]

Hint

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

You can use the same concept to find

$$\sum_{k=1}^\infty \frac{H_k}{(k+2)^q}$$

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

 

[alyafey22]

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

I cannot follow , could you rewrite that!

 

[Amad27]

I cannot follow , could you rewrite that!

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}$

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

 

[alyafey22]

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}$

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

Yes , but remember that in our problem $k$ starts from 1 not 2 , so you have to take care of that.

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

Ok , good luck with a nice proof.

 

[Amad27]

Yes , but remember that in our problem $k$ starts from 1 not 2 , so you have to take care of that.

Ok , good luck with a nice proof.

FIRST QUESTION:
Where do you find such formulas in the first place? I mean, I tried searching online but couldn't find traces of that formula, so how did you discover it and where?

NEXT:

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

We know: $H_k = H_{k-1} + \frac{1}{k}$

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

Let $n = k + 1$, the starting point is then: $k = 0$

\(\displaystyle \sum_{k=0}^\infty \frac{H_{k} + \frac{1}{(k+1)}}{(k+1)^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z)\)

\(\displaystyle \sum_{k=0}^\infty \frac{H_{k} + \frac{1}{(k+1)}}{(k+1)^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z)\)

\(\displaystyle \sum_{k=0}^\infty \frac{H_{k}}{(k+1)^q} + \frac{1}{(k+1)^{q+1}}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z)\)

\(\displaystyle \therefore \sum_{k=1}^\infty \frac{H_{k}}{(k+1)^q} = \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z) -\sum_{k=0}^\infty \frac{1}{(k+1)^{q+1}}\)

Let $q = 1$

$$= (3/2)\zeta(2) - (1/2)(\sum_{z=1}^{-1} \zeta(z+1)\zeta(0)) - \sum_{k=0}^{\infty}\frac{1}{(k+1)^{q+1}}$$

There is an issue here:

$$(\sum_{z=1}^{-1} \zeta(z+1)\zeta(0))$$ does not exist.