# Is it possible to evaluate $-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx$?

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.

• WA can find the anti-derivative, so all we have to do now is take the limits. – Simply Beautiful Art Jul 10 '17 at 23:30
• @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. – user243301 Jul 10 '17 at 23:33
• Well, we certainly have known things, such as $\operatorname{Li}_s(1)=\zeta(s),\operatorname{Li}_s(-1)=\eta(s)=(1-2^{1-s})\zeta(s),\operatorname{Li}_s(0)=0$. – Simply Beautiful Art Jul 10 '17 at 23:36
• (Honestly too many terms for me to be taking the limit of atm, simply don't got the time for that stuff $\ddot\frown$) – Simply Beautiful Art Jul 10 '17 at 23:37
• Many thanks for your contribution @SimplyBeautifulArt – user243301 Jul 10 '17 at 23:39

$$\left\{\begin{array}{rcl} \ds{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t} & \ds{=} & \ds{\left.\partiald[2]{}{\mu}\partiald{}{\nu} {\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 2}}\right\vert_{\ \mu =\ \nu\ =\ 0}} \\[5mm] \ds{\int_{0}^{1}\ln^{k}\pars{t}\,\dd t} & \ds{=} & \ds{\left.\partiald[k]{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t\, \right\vert_{\ \mu\ =\ 0} = \pars{-1}^{k}\, k!} \end{array}\right.$$
• @user243301 Thanks. $\left]\bullet\quad\bullet\atop\smile\right\}$. – Felix Marin Jul 11 '17 at 23:38