# Proving that $\int_0^1\frac{x \log^2(1-x)}{1+x^2} \ dx = \frac{35}{32}\zeta(3)+\frac{1}{24}\log^3(2) -\frac{5}{96} \pi^2 \log(2)$

Could we possibly prove this result without using the polylogarithm? I know how to do it by polylogarithm means, but I want a different way. Is that possible?

$$\int_0^1\frac{x \log^2(1-x)}{1+x^2} \ dx = \frac{35}{32}\zeta(3)+\frac{1}{24}\log^3(2) -\frac{5}{96} \pi^2 \log(2)$$

• @Supera I am finding $$I''(a) = \frac{2 x}{\left(x^2+1\right) (a-x)^2}-\frac{2 x \log (a-x)}{\left(x^2+1\right) (a-x)^2}$$ – Brad Jul 20 '14 at 22:05
• See here and follow the links and you will find useful techniques. – Mhenni Benghorbal Jul 22 '14 at 10:34


\begin{align}&\color{#c00000}{% \int_{0}^{1}{x\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x} =\Re\int_{0}^{1}{\ln^{2}\pars{1 - x} \over \ic + x}\,\dd x =\Re\int_{0}^{1}{\ln^{2}\pars{x} \over \ic + 1 - x}\,\dd x \\[3mm]&=\Re\int_{0}^{1/\pars{1 + \ic}} {\ln^{2}\pars{\bracks{1 + \ic}x} \over 1 - x}\,\dd x =\Re\int_{0}^{\pars{1 - \ic}/2}\ln\pars{1 - x} \bracks{2\ln\pars{\bracks{1 + \ic}x}\,{1 \over x}}\,\dd x \\[3mm]&=-2\Re\int_{0}^{\pars{1 - \ic}/2}{{\rm Li}_{1}\pars{x} \over x}\, \ln\pars{\bracks{1 + \ic}x}\,\dd x =-2\Re\int_{0}^{\pars{1 - \ic}/2}{\rm Li}_{2}'\pars{x} \ln\pars{\bracks{1 + \ic}x}\,\dd x \\[3mm]&=2\Re\int_{0}^{\pars{1 - \ic}/2}{{\rm Li}_{2}\pars{x} \over x}\,\dd x =2\Re\int_{0}^{\pars{1 - \ic}/2}{\rm Li}_{3}'\pars{x}\,\dd x \end{align}

$$\color{#c00000}{% \int_{0}^{1}{x\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x} =2\,\Re{\rm Li}_{3}\pars{1 - \ic \over 2}$$

With one of the MW formulas in group $\pars{1}$: \begin{align} &\overbrace{{\rm Li}_{3}\pars{\half - {\ic \over 2}} +{\rm Li}_{3}\pars{\half + {\ic \over 2}}} ^{\ds{2\,\Re{\rm Li}_{3}\pars{1 - \ic \over 2}}} +{\rm Li}_{3}\pars{1 - {2 \over 1 - \ic}} \\[3mm]&=\zeta\pars{3} + {1 \over 6}\,\ln^{3}\pars{1 - \ic \over 2} + {1 \over 6}\,\pi^{2}\ln\pars{1 - \ic \over 2} -\half\,\ln^{2}\pars{1 - \ic \over 2}\ln\pars{1 + \ic \over 2} \end{align}

I trust you can take from here.

• That's a cool way to go about it, Felix. – Cody Jul 27 '14 at 14:11
• @Cody PolyLogarithms are quite useful. In that way, we don't have to "reinvent the wheel". Thanks. – Felix Marin Jul 27 '14 at 17:14

My thoughts on the problem (too long for comment):

Prove that the definite integral $I$ has the stated closed form value: $$I:=\int_{0}^{1}\frac{x\log^2{(1-x)}}{1+x^2}\mathrm{d}x;\\ I=\frac{35}{32}\zeta{(3)}-\frac{5}{16}\zeta{(2)}\log{(2)}+\frac{1}{24}\log^3{(2)}.$$

Define the auxiliary function $f(\mu)$ for all $\mu\in\mathbb{R}^+$ by the definite integral,

$$f(\mu):=\int_{0}^{1}\frac{x(1-x)^{\mu-1}}{1+x^2}\mathrm{d}x.$$

Differentiating with respect to $\mu$,

\begin{align} \frac{d^2}{d\mu^2}f(\mu) &=\int_{0}^{1}\frac{\partial^2}{\partial\mu^2}\frac{x(1-x)^{\mu-1}}{1+x^2}\mathrm{d}x\\ &=\int_{0}^{1}\frac{x(1-x)^{\mu-1}\log^2{(1-x)}}{1+x^2}\mathrm{d}x. \end{align}

Evaluating the second derivative of $f(\mu)$ at $\mu=1$ then yields the integral $I$:

$$\frac{d^2}{d\mu^2}f(\mu)\bigg{|}_{\mu=1}=\int_{0}^{1}\frac{x(1-x)^{0}\log^2{(1-x)}}{1+x^2}\mathrm{d}x=\int_{0}^{1}\frac{x\log^2{(1-x)}}{1+x^2}\mathrm{d}x=:I.$$

The function $f(\mu)$ itself can be represented as an alternating series of beta functions:

\begin{align} f(\mu)&=\int_{0}^{1}\frac{x(1-x)^{\mu-1}}{1+x^2}\mathrm{d}x\\ &=\int_{0}^{1}x(1-x)^{\mu-1}\sum_{n=0}^{\infty}(-1)^nx^{2n}\,\mathrm{d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^{2n+1}(1-x)^{\mu-1}\,\mathrm{d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\operatorname{B}{(2n+2,\mu)}\\ &=-\sum_{n=1}^{\infty}(-1)^n\operatorname{B}{(2n,\mu)}. \end{align}

The second derivative of $\operatorname{B}{(2n,\mu)}$ with respect to $\mu$ at $\mu=1$ is (courtesy of WolframAlpha),

$$\frac{\partial^2}{\partial\mu^2}\operatorname{B}{(2n,\mu)}\bigg{|}_{\mu=1}=\frac{(H_{2n})^2}{2n}+\frac{\zeta{(2)}}{2n}-\frac{\psi_1{(2n+1)}}{2n}.$$

Hence,

\begin{align} I &=\frac{d^2}{d\mu^2}f(\mu)\bigg{|}_{\mu=1}\\ &=-\sum_{n=1}^{\infty}(-1)^n\frac{\partial^2}{\partial\mu^2}\operatorname{B}{(2n,\mu)}\bigg{|}_{\mu=1}\\ &=-\sum_{n=1}^{\infty}(-1)^n\left[\frac{(H_{2n})^2}{2n}+\frac{\zeta{(2)}}{2n}-\frac{\psi_1{(2n+1)}}{2n}\right]\\ &=\sum_{n=1}^{\infty}(-1)^n\left[\frac{\psi_1{(2n+1)}}{2n}-\frac{(H_{2n})^2}{2n}-\frac{\zeta{(2)}}{2n}\right]\\ &=\sum_{n=1}^{\infty}(-1)^n\left[\frac{\psi_1{(2n+1)}}{2n}-\frac{(H_{2n})^2}{2n}\right]-\frac{1}{2}\zeta{(2)}\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\\ &=\sum_{n=1}^{\infty}(-1)^n\frac{\psi_1{(2n+1)}}{2n}-\sum_{n=1}^{\infty}(-1)^n\frac{(H_{2n})^2}{2n}+\frac{1}{2}\zeta{(2)}\log{(2)}. \end{align}

...Aaand that's as far as I've gotten.

• This question addresses a series similar to $\sum_{n=1}^{\infty}(-1)^n\frac{(H_{2n})^2}{2n}$: math.stackexchange.com/questions/275643/…. – David H Jul 20 '14 at 22:36
• residues would be a good way to tackle those two remaining sums. The left one involving trigamma can probably be done using the kernel $$\pi \csc(\pi z)\psi'(-2z)$$. There should be residues at the positive integers, the positive half-integers,0, and a simple pole at the negative integers. I am just taking a stab here, but the Euler sum in the middle can probably be done using $$\pi csc(\pi z)(\gamma+\psi(-2z))^{2}$$ – Cody Jul 21 '14 at 21:59
• Just adding to my previous comment. To add to David's fine solution, one can use the methods I mentioned above to find that $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi'(2n+1)}{n}=\frac{-\pi}{2}G-\frac{3\pi^{2}}{16}\log(2)+2\zeta(3)$$ and $$\sum_{n=1}^{\infty}\frac{(-1)^{n}(H_{2n})^{2}}{n}=\frac{-3}{16}\zeta(3)-\frac{\pi}{2}G-\frac{1}{12}\log^{3}(2)+\frac{\pi^{2}}{12}$$. Combining these with the other portion of his solution gives the required result. – Cody Jul 24 '14 at 14:21