Few minutes ago I was playing with Wolfram Alpha about integrals like this $$-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx.$$

Previous integral is convergent, and Wolfram Alpha online calculator provide me a closed-form for the indefinite integral

int -log(1-cosh(x))x^2e^(-x)dx

Notice that my idea was to choose the integration limits from $0$ to $\infty$ with the purpose to get $\zeta(3)$ as a summand in the output. But I don't know how to evaluate all terms, specially those involving the polylogarithm (I know that the other terms are tedious, and one could to calculate those with a CAS).

Question. Is it possible to get a closed-form of $$-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx.$$ in terms of particular values of of well-known special functions? Many thanks.

If you prefer different calculations of previous explanation (get a closed-form and evaluate the integration limits as did Wolfram Alpha), feel free to tell us your approach.

  • $\begingroup$ WA can find the anti-derivative, so all we have to do now is take the limits. $\endgroup$ – Simply Beautiful Art Jul 10 '17 at 23:30
  • $\begingroup$ @SimplyBeautifulArt You are right. I know that there are some terms that are tedious, and I should do this task myself, but now I don't know how to evaluate the terms involving $L_2(y)$. Additionally maybe some user know a different approach. Any case I should have calculated the easy terms. $\endgroup$ – user243301 Jul 10 '17 at 23:33
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    $\begingroup$ Well, we certainly have known things, such as $\operatorname{Li}_s(1)=\zeta(s),\operatorname{Li}_s(-1)=\eta(s)=(1-2^{1-s})\zeta(s),\operatorname{Li}_s(0)=0$. $\endgroup$ – Simply Beautiful Art Jul 10 '17 at 23:36
  • $\begingroup$ (Honestly too many terms for me to be taking the limit of atm, simply don't got the time for that stuff $\ddot\frown$) $\endgroup$ – Simply Beautiful Art Jul 10 '17 at 23:37
  • $\begingroup$ Many thanks for your contribution @SimplyBeautifulArt $\endgroup$ – user243301 Jul 10 '17 at 23:39

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &-\int_{0}^{\infty}\ln\pars{1 - \cosh\pars{x}}\,{x^{2} \over \expo{x}}\,\dd x = -\int_{0}^{\infty}\bracks{\ln\pars{\cosh\pars{x} - 1} + \ic\pi}\, x^{2}\expo{-x}\,\dd x \\[5mm] = &\ -2\pi\ic - \int_{0}^{\infty}\ln\pars{{\expo{x} + \expo{-x} \over 2} - 1} \,x^{2}\expo{-x}\,\dd x \\[5mm] \stackrel{\substack{x\ =\ -\ln\pars{t}\\[0.25mm] t\ =\ \expo{-x}}\\ }{=}\,\,\,& -2\pi\ic - \int_{1}^{0}\ln\pars{{1/t + t \over 2} - 1}\ln^{2}\pars{t}\, t\,\,{\phantom{-}\dd t \over -t} \\[5mm] = &\ -2\pi\ic - \int_{0}^{1}\ln\pars{\bracks{1 - t}^{2} \over 2t}\ln^{2}\pars{t}\, \dd t \\[5mm] = &\ -2\pi\ic - 2\ \underbrace{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t} _{\ds{-6 + {\pi^{2} \over 3} + 2\,\zeta\pars{3}}}\ +\ \ln\pars{2}\ \underbrace{\int_{0}^{1}\ln^{2}\pars{t}\,\dd t}_{\ds{2}}\ +\ \underbrace{\int_{0}^{1}\ln^{3}\pars{t}\,\dd t}_{\ds{-6}}\label{1}\tag{1} \\[5mm] = &\ \bbx{6 + 2\ln\pars{2} - {2\pi^{2} \over 3} - 4\zeta\pars{3} - 2\pi\ic} \approx -4.0015 + 6.2832\,\ic \end{align}

Note that the integrals in \eqref{1} are evaluated as follows:

$$ \left\{\begin{array}{rcl} \ds{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t} & \ds{=} & \ds{\left.\partiald[2]{}{\mu}\partiald{}{\nu} {\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 2}}\right\vert_{\ \mu =\ \nu\ =\ 0}} \\[5mm] \ds{\int_{0}^{1}\ln^{k}\pars{t}\,\dd t} & \ds{=} & \ds{\left.\partiald[k]{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t\, \right\vert_{\ \mu\ =\ 0} = \pars{-1}^{k}\, k!} \end{array}\right. $$

  • $\begingroup$ Many thanks, I am going to study your great answer. $\endgroup$ – user243301 Jul 11 '17 at 7:26
  • $\begingroup$ @user243301 Thanks. $\left]\bullet\quad\bullet\atop\smile\right\}$. $\endgroup$ – Felix Marin Jul 11 '17 at 23:38

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