Find the value of $\displaystyle \int^{\infty}_{0}\frac{x\ln x}{1+e^x}dx$
What I tried. $$I = \int^{\infty}_{0}\frac{x\ln x}{1+e^x}dx = \int^{\infty}_{0}x\ln x\frac{e^{-x}}{1+e^{-x}}dx$$ By parts $$I =-x\ln x\ln |1+e^{-x}|\bigg|^{\infty}_{0}+\int^{\infty}_{0}(1+\ln x)\ln|1+e^{-x}|dx$$ How do i solve it?
On
\begin{align*} I&=\left.\frac{\partial}{\partial s}\right|_{s=2}\int_0^\infty\frac{x^{s-1}}{e^x+1}\,\mathrm dx\\ &=\left.\frac{\partial}{\partial s}\right|_{s=2}\big((1-2^{1-s})\zeta(s)\Gamma(s)\big) \end{align*} Then use the result $$\zeta'(2)=\frac{\pi^2}6(-12\ln A+\gamma+\ln(2\pi)),\quad\Gamma'(2)=1-\gamma$$
We have that $$\begin{align} I&:=\int^{\infty}_{0}\frac{x\ln(x)}{1+e^x}dx=\int^{\infty}_{0}\frac{e^{-x}x\ln(x)}{1+e^{-x}}dx\\&=\int^{\infty}_{0}x\ln (x)\sum_{k=1}^{\infty}(-1)^{k-1}e^{-kx}\,dx\\&=\sum_{k=1}^{\infty}(-1)^{k-1}\int^{\infty}_{0}x\ln(x)e^{-kx}\,dx. \end{align}$$ Moreover, after integrating by parts, $$\begin{align}\int^{\infty}_{0}x\ln(x)e^{-kx}\,dx&=\frac{1}{k}\int^{\infty}_{0}(1+\ln(x))e^{-kx}\,dx\\&=\frac{1-\ln(k)}{k^2}+\frac{1}{k^2}\int^{\infty}_{0}\ln(t)e^{-t}\,dt\\&=\frac{1-\ln(k)-\gamma}{k^2}.\end{align}$$ where $\gamma$ is the Euler–Mascheroni constant. Hence, $$\begin{align} I&=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{1-\ln(k)-\gamma}{k^2}\\ &=( 1-\gamma)\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}+\frac{\ln(2)}{2}\sum_{k=2}^{\infty}\frac{1}{k^2}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}\\ &=\frac{\pi^2}{12}\left(1-\gamma+\ln(2)\right)-\frac{\pi^2}{12}\left(12\ln(A)-\gamma-\ln(2\pi)\right)\\ &=\frac{\pi^2}{12}\left(1+\ln(4\pi)-12 \log (A)\right) \end{align}.$$ where $A$ is the Glaisher–Kinkelin constant and we used Evaluating a sum $-\zeta'(2)$ .