A calculus problem by Aly Ahmed

Note that $\begin{aligned} F(u) & = \; \int_0^\infty \frac{x^u}{e^x + 1}\,dx \; = \; \sum_{n=0}^\infty (-1)^n \int_0^\infty x^u e^{-(n+1)x}\,dx \; =\; \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^{u+1}} \int_0^\infty y^u e^{-y}\,dy \\ & = \; \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^{u+1}}\,\Gamma(u+1) \; = \; (1 - 2^{-u})\zeta(u+1)\Gamma(u+1) \end{aligned}$ for $u > 0$ , so that $\int_0^\infty \frac{x \ln x}{e^x + 1}\,dx \; =\; F'(1) \; = \; \tfrac12\ln2 \,\zeta(2)\Gamma(2) + \tfrac12\zeta'(2)\Gamma(2) + \tfrac12\zeta(2)\Gamma'(2) \; = \; \tfrac{1}{12}\pi^2\ln2 + \tfrac12\zeta'(2) + \tfrac{1}{12}\pi^2\Gamma'(2)$ Now standard results give $\begin{aligned} \Gamma'(2) & = \; 1 - \gamma \\ \zeta'(2) & = \; \tfrac16\pi^2(\gamma + \ln2\pi - 12\ln A) \end{aligned}$ where $\gamma$ is the Euler-Mascheroni constant and $A$ is the Glaisher-Kinkelin constant, and hence $\int_0^\infty \frac{x \ln x}{e^x + 1}\,dx \; = \; \tfrac{1}{12}\pi^2(1 + \ln4\pi - 12\ln A) \; = \; \boxed{0.449043}$