TITLE: The source of the Integral
QUESTION [7 upvotes]: Wolfram alpha calculates the integral
$$\int\limits_0^\infty \frac{x^2\ln{x}}{e^x-1}dx=2\zeta^\prime(3)+3\zeta(3)-2\gamma\zeta(3).$$
However, I need to cite the source of this identity (the table of integrals, or the article where this integral was calculated). Could you indicate any?

REPLY [18 votes]: One way to get the claimed value of the given integral, $J$ in notation below, is by starting from the standard relation
$$
\begin{aligned}
\zeta(s)  
&= \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx\ ,\qquad
\text{ so }\\
\zeta'(s)
&= 
\frac\partial{\partial s}
\left(\ 
\frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx
\ \right)
\\
&= 
\frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}\; \ln x}{e^x-1}\; dx
-
\underbrace{\frac{\Gamma'(s)}{\Gamma(s)}}_{=\psi(s)}\zeta(s)\ ,\qquad\text{leading to }
\\
\zeta'(3)&=\frac 1{\Gamma(3)}\underbrace{\int_0^\infty \frac {x^2\; \ln x}{e^x-1}\; dx}_{\text{our integral }J}
-
\psi(3)\zeta(3)\ .
\\[2mm]
&\qquad\text{Extracting $J$ from above,}\\[2mm]
J &=\Gamma(3)\; \zeta'(3) \ +\ \Gamma(3)\; \psi(3)\; \zeta(3)
\\
&=2\zeta'(3)+(3-2\gamma)\zeta(3)\ ,
\end{aligned}
$$
where we have used the relations $\displaystyle\psi(x+1)=\frac 1x+\psi(x)$ and $\psi(1)=-\gamma$, e.g. from here. Explicitly, this gives the value for $\displaystyle\psi(3)=\frac 12+\psi(2)=\frac 12+1+\psi(1)=\frac 12(3-2\gamma)$.
$\square$

The above shows how to get similar relations when there is an other power of $x$ instead of $x^2$ in the numerator of then integrand, and gives an interpretation of the involved coefficients.