Find the value of: $ J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(x)\,dx $

Question

I'm trying to find to value of: $$ J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(x)\,dx $$

Here's what I've done:

$$ \ln(x)=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}(x-1)^n=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sum_{m=0}^{\infty}{n\choose m} (-1)^{n-m}x^m $$ Therefore: $$ J=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sum_{m=0}^{\infty}{n\choose m} (-1)^{n-m}\int_{0}^{\infty}\frac{x^{m+3}}{e^x-1}\,dx \\= -\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sum_{m=0}^{\infty}{n\choose m} (-1)^{n-m} \Gamma(m+4) \zeta(m+4)\\ = -\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=0}^{\infty}{n\choose m} (-1)^{m}(m+3)! \zeta(m+4)\\ =-\sum_{n=1}^{\infty}(n-1)!\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(n-m)!}(m+3)(m+2)(m+1) \zeta(m+4)\\ $$ And I'm stuck. If someone could point me in the right direction, that would be great. Mathematica gives the answer as $\frac{\pi^4}{90}(11-6\gamma)+6\zeta'(4)$.

Answer 1

A related problem. Here is a solution based on the well known result

$$ \zeta(s)\Gamma(s)= \int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}\,dx \implies \frac{d}{ds}(\zeta(s)\Gamma(s))= \int_{0}^{\infty}\frac{x^{s-1} \ln(x)}{e^x-1}\,dx $$

So, your integral can be evaluated by taking $s=4$ after performing the derivative with respect to $s$, that is

$$ \frac{d}{ds}(\zeta(s)\Gamma(s))|_{s=4}= 6\,\zeta'\left( 4 \right) +\frac{{\pi }^{4}}{15}\, \left( {\frac {11}{6}}- \gamma \right).$$

Added: To evaluate $\Gamma'(x)$, recall the fact

$$ \psi(x)=\frac{\Gamma'(x)}{\Gamma(x)},$$

which implies

$$\Gamma'(x)=\Gamma(x)\psi(x). $$