How can I evaluate $\displaystyle\int_{0}^{\infty}\frac{x\log(x)}{1+e^x}\,dx$? My attempt :
Evaluate a more general case,
$$F(a) = \int_{0}^{\infty}\frac{x\log x}{1+e^x}\cdot a^{1+e^x} \,dx$$
$$F'(a) = \int_{0}^{\infty}x\cdot \log(x)\cdot a^{e^x} \,dx$$
Is there any way to take it from here without using by parts multiple times ? Will Feynman's trick be of any help here ? Please don't give away the entire solution.
Thanks :)
 A: Hint: Start by the following 
$$F(s)=\int^ \infty_0 \frac{x^{s-1}}{e^x+1}\,dx = \Gamma(s) (1-2^{1-s})\zeta(s)$$
By differentiation and $s=2$
$$F'(2) = \frac{1}{12} (6 ζ'(2) + π^2 - \gamma π^2 + π^2 \log(2))
$$
Which reduces to 
$$\int^ \infty_0 \frac{x\log(x)}{e^{x}+1}\,dx = \frac{1}{12} π^2 (1 + \log(4) - 12 \log(A) + \log(π))$$
where $A$ is Glaisher–Kinkelin constant.

Note by the functional equation 
$$\zeta(s)=2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)$$
By differentiation 
$$2\pi^2\zeta'(-1)=-\log(2\pi)\zeta(2)+\psi(2)\zeta(2)+ \zeta'(2)$$
And since 
$$\zeta'(-1) = \frac{1}{12}-\log A$$
The result follows. 
Reference 
How can we prove that $2e^2\int_{0}^{\infty} {x\ln x\over 1-e^{2e\pi{x}}}\mathrm dx=\ln A?$
A: I just typed this question on Google and got a hit on Quora.com , so i'm just pasting that guys answer here , if it's not allowed i'll delete it 
$$I(s)=\displaystyle\int_{0}^{\infty}\dfrac{x^{s-1}}{e^x+1}\,dx=\eta(s)\Gamma(s) \tag*{} $$
Then required integral is nothing but $I'(s)$ at $s=2$
$$\begin{align}I'(s)\big{|}_{s=2}=I'(2)&=\displaystyle\int_{0}^{\infty}\dfrac{x^{s-1}\cdot \ln x}{1+e^x}\,dx \\&=\displaystyle\int_{0}^{\infty}\dfrac{x \ln x}{e^x+1}\,dx\\&=\{\eta'(2)\Gamma (2)+\Gamma'(2)\eta(2)\}\bigg{|}_{s=2}\tag{1}\\&=\eta'(2)+\dfrac{\pi^2}{12}\Gamma'(2) \tag{2}\\&=\eta'(2)+\dfrac{\pi^2}{12}(1-\gamma) \tag{3}\\&=\dfrac{\pi^2}{12}\ln(\pi)+\dfrac{\pi^2}{6}\ln(2)+\dfrac{\pi^2}{12}\,\gamma-\pi^2\ln(\mathcal{A})+\dfrac{\pi^2}{12}(1-\gamma)\tag{4}\\&=\dfrac{\pi^2}{12}\ln(\pi)+\dfrac{\pi^2}{6}\ln(2)+\dfrac{\pi^2}{12}-\pi^2\ln(\mathcal{A})\end{align}$$
Explanation:
(1) $I(s)=\displaystyle\int_{0}^{\infty}\dfrac{x^{s-1}}{e^x+1}\,dx=\eta(s)\Gamma(s) \tag*{} $
$I'(s)=\eta'(s)\Gamma(s)+\eta(s)\Gamma'(s) \tag*{} $
(2) Dirichlet eta function  is defined as $\implies \eta(s)=\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n^s} $
And it's relation with zeta function is : $\implies \eta(s)=\left(1-2^{1-s}\right)\zeta(s)$
Put $s=2$ , we'll have $\eta(2)=\dfrac{1}{2}\zeta(2)=\dfrac{\pi^2}{12}$
(3) The Digamma function  can be written as
$\begin{align}\Gamma'(s)&=\Gamma(s) \,\psi_{0}(s)\\&=\Gamma(s)\left\{\displaystyle\sum_{k=1}^{s-1}\dfrac{1}{k}-\gamma\right\}\end{align} \tag*{} $
$ $$\Gamma'(2)=\Gamma(2)\{\mathcal{H}_{1}-\gamma\}=1-\gamma \tag*{} $
Here $H_{n} \implies$Harmonic number  , $\gamma \implies$ Euler–Mascheroni constant 
(4) We have $\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$ Differentiating  it at s=2
$\begin{align}\eta'(2)&=\left(1-2^{1-s}\right)\zeta'(s)+2^{1-s}\ln(2)\zeta(s)\bigg{|}_{s=2}\\&=\dfrac{1}{2}\zeta'(2)+\dfrac{\pi^2}{12}\ln(2)\\&=\dfrac{\pi^2}{12}\left\{\gamma-12\ln(\mathcal{A})+\ln(2\pi)\right\}+\dfrac{\pi^2}{12}\ln 2\\&=\dfrac{\pi^2}{12}\ln \pi+\dfrac{\pi^2}{12}\gamma+\dfrac{\pi^2}{6}\ln 2-\pi^2 \ln(\mathcal{A}) \end{align}  \tag*{} $
So our integral is finally
$$J=\dfrac{\pi^2}{12}\ln(\pi)+\dfrac{\pi^2}{6}\ln(2)+\dfrac{\pi^2}{12}-\pi^2\ln(\mathcal{A}) \tag*{} $$
Where $\mathcal{A} \implies $ Glaisher–Kinkelin constant 
