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invertible if and only if bijective

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.

- Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$
- Compute $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$
- Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$.
- Contour integration of $\int \frac{dx} {(1+x^2)^{n+1}}$
- Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
- Complex analysis: contour integration

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

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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):

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

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

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.

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.

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:

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

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

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

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.

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