$\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)$ How to show?

Question

$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$

Anyone an idea on how to prove this?

Answer 1

By the recursive relation $\Gamma(x+1)=x\Gamma(x)$, we get $$ \small{\log(\Gamma(x))=\log(\Gamma(n+x))-\log(x)-\log(x+1)-\log(x+2)-\dots-\log(x+n-1)}\tag{1} $$ Differentiating $(1)$ with respect to $x$, evaluating at $x=1$, and letting $n\to\infty$ yields $$ \begin{align} \frac{\Gamma'(1)}{\Gamma(1)}&=\log(n)+O\left(\frac1n\right)-\frac11-\frac12-\frac13-\dots-\frac1n\\ &\to-\gamma\tag{2} \end{align} $$ Next, apply $(2)$ to the following: $$ \begin{align} \int_0^\infty\log(t)\;e^{-t}\;\mathrm{d}t &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\int_0^\infty t^x\;e^{-t}\;\mathrm{d}t\right]_{x=0}\\ &=\Gamma'(1)\\ &=-\gamma\tag{3} \end{align} $$ Then, a simple change of variables yields $$ \int_0^\infty\log(t)\;e^{-nt}\;\mathrm{d}t=-\frac{\gamma+\log(n)}{n}\tag{4} $$ Since $\dfrac{1}{e^t+1}=e^{-t}-e^{-2t}+e^{-3t}-e^{-4t}+\dots$, by applying $(4)$ to this result, we have that $$ \begin{align} \int_0^\infty\frac{\log(t)}{e^t+1}\mathrm{d}t &=\int_0^\infty\sum_{n=1}^\infty(-1)^{n-1}\log(t)\;e^{-nt}\;\mathrm{d}t\\ &=\sum_{n=1}^\infty(-1)^n\frac{\gamma+\log(n)}{n}\\ &=-\frac12\log(2)^2\tag{5} \end{align} $$

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More about $\mathbf{(2)}$:

The fact that $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)$ relies on the log-convexity of $\Gamma(x)$; that is, $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))$ is monotonically increasing. By the recursive relation for $\Gamma(x)$, we have that $$ \log(\Gamma(x))-\log(\Gamma(x-1))=\log(x-1)\tag{6} $$ and that $$ \log(\Gamma(x+1))-\log(\Gamma(x))=\log(x)\tag{7} $$ The Mean Value Theorem and $(6)$ imply that $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(\xi_1))=\log(x{-}1)$ for some $\xi_1{\in}(x{-}1,x)$.

The Mean Value Theorem and $(7)$ imply that $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(\xi_2))=\log(x)$ for some $\xi_2{\in}(x,x{+}1)$.

By the monotonicity of $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))$, we get that $$ \log(x-1)\le\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))\le\log(x)\tag{8} $$ Since $\log(x)-\log(x-1)=O\left(\frac1x\right)$, $(8)$ implies that $$ \frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)\tag{9} $$

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Log-Convexity of $\mathbf{\Gamma(x)}$:

If $\frac{\mathrm{d}^2}{\mathrm{d}x^2}f(x)\ge0$, then $f$ is convex at $x$. Thus, if $\dfrac{f(x)f''(x)-f'(x)^2}{f(x)^2}=\frac{\mathrm{d}^2}{\mathrm{d}x^2}\log(f(x))\ge0$, then $f$ is log-convex. So we need to show that $\Gamma(x)\Gamma''(x)\ge\Gamma'(x)^2$. That is, $$ \int_0^\infty t^{x-1}\;e^{-t}\;\mathrm{d}t \int_0^\infty\log(t)^2\;t^{x-1}\;e^{-t}\;\mathrm{d}t \ge \left(\int_0^\infty\log(t)\;t^{x-1}\;e^{-t}\;\mathrm{d}t\right)^2\tag{10} $$ Dividing both sides of $(10)$ by $\int_0^\infty t^{x-1}\;e^{-t}\;\mathrm{d}t$, $(10)$ becomes $$ \int\log(t)^2\;\mathrm{d}\mu \ge \left(\int\log(t)\;\mathrm{d}\mu\right)^2\tag{11} $$ where $\mathrm{d}\mu=\dfrac{t^{x-1}\;e^{-t}\;\mathrm{d}t}{\int_0^\infty t^{x-1}\;e^{-t}\;\mathrm{d}t}$ is a unit measure on $[0,\infty)$. Thus, $(11)$ is simply Jensen's inequality.

Strictly speaking:

Note that $$ \log(t)^2 + a^2 \ge 2a\log(t)\tag{12} $$ with equality if and only if $\log(t)=a$. Integrating $(12)$ w.r.t. the unit measure $\mathrm{d}\mu$, yields $$ \int\log(t)^2\;\mathrm{d}\mu + a^2 \ge 2a\int\log(t)\;\mathrm{d}\mu\tag{13} $$ with equality in $(13)$ if and only if $\log(t)=a$ a.e. $\mathrm{d}\mu$. Let $a=\int\log(t)\;\mathrm{d}\mu$, then $(13)$ becomes $$ \int\log(t)^2\;\mathrm{d}\mu \ge \left(\int\log(t)\;\mathrm{d}\mu\right)^2\tag{14} $$ with equality if and only if $\log(t)$ is constant a.e. $\mathrm{d}\mu$. Since the $\mathrm{d}\mu$ in $(11)$ is absolutely continuous and $\log(t)$ is strictly increasing on $(0,\infty)$, the inequality in $(11)$ is strict. Therefore, $\Gamma$ is strictly log-convex.

Answer 2

Start with $J(s)$ given by $$ J(s) = \int_0^\infty \frac{x^s}{1+e^x}dx. $$ Expand the denominator using geometric series, like so: $$ J(s) = \sum_{k\geq0}\int_0^\infty (-1)^k x^s e^{-(1+k)x}dx$$ $$ = \sum_{k\geq1} \frac{(-1)^{k+1}}{k^{s+1}} \int_0^\infty x^s e^{-x}dx$$ Now, the sum is the Dirichlet eta function, related to the Riemann zeta function like so, $$ \sum_{k\geq1}\frac{(-1)^{k+1}}{k^{s+1}} = (1-2^{-s})\zeta(s+1), $$ and the integral is $\Gamma(1+s)$. Thus $$ J(s) = (1-2^{-s})\zeta(1+s) \Gamma(1+s). $$

To find the derivative at $s=0$ we need the Laurent series for each of these functions at $s=0$, ($\zeta(1+s)$ is singular at $s=0$, but $1-2^{-s}$ has a zero there, so $J$ is regular), they are $$ (1-2^{-s})\zeta(1+s) = \log2 + (\gamma \log 2 - \frac{(\log 2)^2}{2})s + O(s^2), $$ $$ \Gamma(1+s) = 1 - \gamma s + O(s^2), $$ where $\gamma$ is Euler's constant. Multiplying the two series and taking the coefficient of $s$, we get $$ \frac{d J}{ds}(0) = -\frac12 (\log 2)^2, $$ which is the integral you were looking for.