Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can we prove

$$ \int_0^1 \frac{\ln x \cdot \ln(1+x)}{1+x}dx=-\frac{\zeta(3)}{8}? $$

This has been one of the integrals that came out of an integral from another post on here, but no solution to it.

I am not sure how to use a taylor series expansion for the $\ln(1+x)\cdot(x+1)^{-1}$ term, thus I can not simple reduce this integral to the form $$ \int_0^1 x^n \ln x dx $$ I think if I can get the integral in this form, I will be able to recover the zeta function series which is given by $$ \zeta(3)=\sum_{n=0}^\infty \frac{1}{(n+1)^3}. $$ Thanks

share|improve this question
1  
Did you try to write down that the power series for $\frac{{\rm ln}(1+x)}{1+x}$ is the Cauchy product of the power series for ${\rm ln}(1+x)$ and the power series of $\frac1{1+x}$? –  Etienne Feb 25 at 4:57
    
@Etienne No what is that? I am not sure how to perform the convolution since it is a discrete sum. For continuous sums (integrals) I don't mind inverse transforms –  Integrals Feb 25 at 5:00
    
See here: en.wikipedia.org/wiki/Cauchy_product –  Etienne Feb 25 at 5:09
1  
You're welcome! –  Etienne Feb 25 at 5:15
1  
My name is Integral. Definite Integral. :-) –  Lucian Feb 25 at 9:01
show 4 more comments

5 Answers 5

up vote 3 down vote accepted

I played around with this using parts because it looks like an integral that involves polylogs. Many of these can be done with parts or multiple use of parts.

$$\int\frac{log(x)log(1+x)}{1+x}dx$$

Let $$u=x+1$$

$$\int\frac{log(u-1)log(u)}{u}du=\int\frac{log(u)}{u}\left(log(u)+log(1-1/u)\right)du$$

$$=\frac{log^{3}(u)}{3}+\int\frac{log(u)log(1-1/u)}{u}du$$

Now, use parts on this last integral:

$u=log(u), \;\ dv=\frac{log(1-1/u)}{u}, \;\ du=\frac{1}{u}du, \;\ v=Li_{2}(1/u)$

(as a note, $\int\frac{log(1-1/u)}{u}du=Li_{2}(1/u)$ is a rather famous integral related to the dilog).

$$\int\frac{log(u)log(1-1/u)}{u}du=log(u)Li_{2}(1/u)-\int\frac{Li_{2}(1/u)}{u}du$$

Also, note this last integral is simply $$-Li_{3}(1/u)$$

Now, back sub $u=x+1$, and put it altogether using the integration limits 0 to 1.

Hence, we arrive at:

$$ \left|1/3log^{3}(x+1)+log(x+1)Li_{2}\left(\frac{1}{x+1}\right)+Li_{3}\left(\frac{1}{1+x}\right)\right|_{0}^{1}$$

$$=1/3log^{3}(2)+log(2)Li_{2}(1/2)+Li_{3}(1/2)-Li_{3}(1).........(1)$$

Note the identities:

$$Li_{2}(1/2)=\frac{\pi^{2}}{12}-1/2log^{2}(2)$$

$$Li_{3}(1/2)=7/8\zeta(3)+1/6log^{3}(2)-\frac{\pi^{2}}{12}log(2)$$

sum up (1):

$$1/3log^{3}(2)+log(2)\left(\frac{\pi^{2}}{12}-1/2log^{2}(2)\right)+\left(7/8\zeta(3)+1/6log^{3}(2)-\frac{\pi^{2}}{12}log(2)\right)-\zeta(3)$$

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

share|improve this answer
add comment

I think this ties together the aforementioned ideas quite nicely:

Step 1: Integrate by parts. Let $u=\log{x}$ and $dv=\frac{\log(1+x)}{1+x}$. We obtain $v=\frac{1}{2} [\log(1+x)]^2$. Being somewhat careful with the limits, we see that the integral itself is equal to $$ -\frac{1}{2} \int_0^1 \frac{[\log(1+x)]^2}{x}\,dx $$

Step 2: Expand $\log(1+x)$ and $\log(1+x)/x$ into their Taylor series and combine. $$ -\frac{1}{2} \int_0^1\left(\sum_{j=1}^{\infty} (-1)^{j+1} \frac{x^j}{j}\right)\left(\sum_{i=0}^\infty (-1)^i \frac{x^i}{i+1}\right)\,dx = -\frac{1}{2} \sum_{j=1}^\infty \sum_{i=0}^\infty \frac{(-1)^{i+j+1}}{j(i+1)(i+j+1)} $$

Step 3: There are a few ways to go here, but I like $k=i+j+1$ followed by a partial fraction decomposition. Then, $$ -\frac{1}{2} \sum_{k=2}^\infty \frac{(-1)^k}{k} \sum_{j=1}^{k-1} \frac{1}{j(k-j)} = -\sum_{k=2}^\infty \frac{(-1)^k}{k^2} H_{k-1} $$

Step 4: ??? It is not clear to me why this quantity is the desired one, but prior responses seem to indicate as such. Anybody else with thoughts?

[edit] I had an $H_k$ that should have been an $H_{k-1}$. Fixed now.

[edit 2] A more direct approach from the generating function (http://en.wikipedia.org/wiki/Harmonic_number#Generating_functions) of the harmonic sequence: Since $-\sum_{k=1}^\infty H_k (-x)^k = \frac{\log(1+x)}{1+x}$, we have $$ -\int_0^1 \log(x) \sum_{k=1}^\infty (-1)^k H_k x^k\,dx = \sum_{k=1}^\infty \frac{(-1)^k}{(k+1)^2} H_k $$ Definitely simpler, but requires a priori knowledge of the generating function.

share|improve this answer
    
Very nice and clear to follow. Thanks a lot @Jason –  Integrals Feb 25 at 8:46
add comment

The integral can have the form

$$ I = -\sum_{k=1}^{\infty}\frac{(-1)^k\,H_{k}}{k^2}-\frac{3}{4}\zeta(3), $$

$H_k$ are the harmonic numbers. Try to work out above sum. See a related technique.

share|improve this answer
    
from that result, how are you getting $-\zeta(3)/8$? I will try and work this sum out now...What did you do though to get the harmonic number sum? How do you manipulate the integrand to get that result. Thanks! –  Integrals Feb 25 at 6:13
1  
@Jeff: Mathematica can give you a closed form for this sum. –  Mhenni Benghorbal Feb 25 at 6:17
    
how did you find that form for the Integrand though? I see mathematica can evaluate the harmonic sum, thanks –  Integrals Feb 25 at 6:19
add comment

A handy thing to note for evaluating $$\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{2}}$$ is to use $$\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{2}}=\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-\zeta(3)$$............[1]

The first sum on the right can be shown in various ways and evaluates to $2\zeta(3)$. If you look around, I am sure it has already been done on the site.

Contours is a fun way to evaluate many Euler sums. A method published by Flajolet and Salvy in their paper "Euler sums and contour integral representations". Use the 'kernel' $\frac{1}{2}\pi\cot(\pi z)(\psi(-z))$ and note the residues for the pole at 0, the positive integers, n, and the negative integers, -n.

The pole at the negative integers is simple and the residue is

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

The residue at the positive integers is order 2 and is:

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

The residue at the pole at 0 is $$\frac{-1}{2}\zeta(3)$$

summing these and setting to 0 gives:

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

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}=0$$

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

share|improve this answer
add comment

Your idea of writing $$\frac{\log (x) \log (x+1)}{x+1}=\sum _{n=1}^{\infty } a_n x^n \log (x)$$ by a Taylor expansion looks good to me almost when you take into account that, for value of $n$ greater or equal to $0$, $$ \int_0^1 x^n \ln x dx=-\frac{1}{(n+1)^2} $$ So $$ \int_0^1 \frac{\ln x \cdot \ln(1+x)}{1+x}dx=-\sum _{n=1}^{\infty } \frac{a_n}{(n+1)^2} $$ But, at this point, I am stuck with the $a_n$ and then with the summation. I made some numerical evaluations and observed that the convergence is not very fast.

I shall wait for answers to learn more.

Thanks for the interesting problem.

share|improve this answer
    
$a_n$ is proportional to $H_n$ where $H_n$ is the harmonic number. I am still working on proving this. Not sure why, but I observe that $$ \sum_{n=1}^\infty \frac{H_n}{(n+1)^2}=\zeta(3) $$ Thanks!! let me know –  Integrals Feb 25 at 7:47
    
@Jeff. The $a_n$ are $(-1)^{n-1} H_n$ !! So, it works. –  Claude Leibovici Feb 25 at 8:01
    
yes except I am looking for a proof that doesn't rely on knowing the answer. Thanks still! –  Integrals Feb 25 at 8:03
    
@Jeff&Claude This is indeed a very interesting question. I did the same computations as Claude and got stuck at the same point. Why is it so that $\sum_1^\infty \frac{H_n}{(n+1)^2}=\zeta(3)$?? –  Etienne Feb 25 at 8:15
    
@Etienne Yes, I am not sure why...If we can prove that, we are done! –  Integrals Feb 25 at 8:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.