10
$\begingroup$

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?

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

$\endgroup$
  • $\begingroup$ @Supera I am finding $$I''(a) = \frac{2 x}{\left(x^2+1\right) (a-x)^2}-\frac{2 x \log (a-x)}{\left(x^2+1\right) (a-x)^2}$$ $\endgroup$ – Brad Jul 20 '14 at 22:05
  • $\begingroup$ See here and follow the links and you will find useful techniques. $\endgroup$ – Mhenni Benghorbal Jul 22 '14 at 10:34
5
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{x\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x ={35 \over 32}\,\zeta\pars{3} + {1 \over 24}\,\ln^{3}\pars{2} -{5 \over 96}\,\pi^{2}\ln\pars{2}:\ {\large ?}}$.

\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.

$\endgroup$
  • $\begingroup$ That's a cool way to go about it, Felix. $\endgroup$ – Cody Jul 27 '14 at 14:11
  • $\begingroup$ @Cody PolyLogarithms are quite useful. In that way, we don't have to "reinvent the wheel". Thanks. $\endgroup$ – Felix Marin Jul 27 '14 at 17:14
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This question addresses a series similar to $\sum_{n=1}^{\infty}(-1)^n\frac{(H_{2n})^2}{2n}$: math.stackexchange.com/questions/275643/…. $\endgroup$ – David H Jul 20 '14 at 22:36
  • $\begingroup$ 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 $$\pi csc(\pi z)(\gamma+\psi(-2z))^{2}$$ $\endgroup$ – Cody Jul 21 '14 at 21:59
  • 2
    $\begingroup$ 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. $\endgroup$ – Cody Jul 24 '14 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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