Calculating $\int_0^{\infty} \frac{\log^2(1 - e^{-x})\:x^5}{e^x - 1} \: dx $

Question

I am having trouble calculating the following improper integral: $$\displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx $$

Can someone give me a way that I can calculate this?

Answer 1

Given $$I = \displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx $$ then let $t = e^{-x}$ to obtain \begin{align} I &= - \int_{1}^{0} ( - \ln t)^{5} \, \ln^{2}(1-t) \, \frac{dt}{1-t} \\ &= - \int_{0}^{1} \frac{\ln^{5}t \, \ln^{2}(1-t)}{1-t} \, dt \\ &= \int_{0}^{1} \ln^{5}t \, \left( \frac{1}{3} \, \frac{d}{dt} \ln^{3}(1-t)\right) \, dt \\ &= \left[ \frac{1}{3} \ln^{5}t \, \ln^{3}(1-t) \right]_{0}^{1} - \frac{5}{3} \, \int_{0}^{1} \frac{\ln^{3}(1-t) \, \ln^{4}t}{t} \, dt \\ &= - \frac{5}{3} \, \int_{0}^{1} \frac{\ln^{3}(1-t) \, \ln^{4}t}{t} \, dt \end{align}

From here consider the Beta function in the form \begin{align} B(x,y) = \int_{0}^{1} t^{x-1} \, (1-t)^{y-1} \, dt \end{align} for which \begin{align} I &= - \frac{5}{3} \, \partial_{x}^{4} \partial_{y}^{3} \, \left. B(x,y) \right|_{x=0, y=1} \\ &= 5! \, (\zeta(4) \, \zeta(3) + \zeta(2) \, \zeta(5) - 3 \, \zeta(7)) \end{align}

Answer 2

Hint. Here is an approach using the Euler beta function.

By the change of variable, $u=e^{-x}$, you get $$ \begin{align} \int_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx&=-\int_0^1\frac{\log^2(1-u)\log^5u}{1-u}\:du\\\\ &=-\left.\partial_b^2\partial_a^5\left(\int_{0}^{1} {u}^{a-1}(1-u)^{b-1}du\right)\right|_{a=1,b=0}\\\\ &=-\left.\partial_b^2\partial_a^5\left(\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\right)\right|_{a=1,b=0}\\\\ &=\frac{4}{3} \pi ^4 \zeta(3)+20 \pi ^2 \zeta(5)-360 \zeta (7). \end{align} $$