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

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

Anyone an idea on how to prove this?

• Can you tell us what you've tried? Jan 19, 2012 at 2:44
• To me it looks like a "typical" complex plane integral. Work out where the poles are, choose the right contour integral, show parts of the contour integral vanish, count the residues at the surrounded poles... and you've got your answer :) Fun part is picking the right contour. Jan 19, 2012 at 2:46
• The tricky part is finding the right integrand. I've been playing with this for an hour or more and can't make it quite work. I've been trying $$\frac{(\ln z)^2}{e^z + 1}$$ and making a Riemann cut on the positive real axis. This looked promising. But the trouble with this approach is the sum of the infinite number of residues up and down the imaginary axis doesn't converge. Jan 19, 2012 at 17:59
• I considered $$\frac{\log (z)}{e^z+1}$$ Poles are located at $$z_{n} = j(2n+1)\pi$$ residues are equal to $$-\log ((2n+1) i \pi )$$
– wnvl
Jan 19, 2012 at 18:07
• The other approach I've been playing around with is trying to tie this result back to an integral like $$\int_0^\infty \frac{x^s}{e^x -1} dx$$ which is standard. The integral in question is equal to $dJ/ds(0)$ where $$J(s) = \int_0^\infty \frac{x^s}{e^x+1} dx$$ as $$\frac{d \ }{ds} \int_0^\infty \frac{x^s}{e^x+1} dx = \int_0^\infty \frac{\partial \ }{\partial s} \frac{x^s}{e^x+1} dx = \int_0^\infty \frac{ (\ln x)x^s}{e^x+1} dx$$ It wasn't too hard to write down an expression for J(s) but upon differentiating it and evaluating it at $s = 0$, I ran into trouble again with convergence. Jan 19, 2012 at 18:22

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}

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 $$\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)\tag{9}$$

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.

• $$\frac{1}{1+e^x}=\frac{e^{-x}}{1+e^{-x}}=e^{-x}-e^{-2x}+e^{-3x}-e^{-4x}+\dots$$ which converges for $x\in(0,\infty)$.
– robjohn
Jan 21, 2012 at 9:42
• @Peter: It's not your fault - until you have 50 reputation points, you can only comment on your own questions and answers. Jan 23, 2012 at 1:08
• @Zev: I think that Peter had intended his comment to be for Kirill's answer. However, it is interesting to have a comment to my answer posted before I answered :-)
– robjohn
Jan 25, 2012 at 0:01
• Hi robjohn, I know it's an old post but looking at most of your answers I'm really astonished by your huge range of techniques and knowledge! Could you please give (a little) guide (for me as an undergraduate student) as well. I'm also kind of into integration, series, etc exactly the topics that most of your answers are written brilliantly. Jul 4, 2021 at 7:20

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.

• Very nice. I didn't know how to handle $$\sum \frac{(-1)^n}{n^s}$$ Could you expand a little more on the derivation of the expansion of $$(1-2^{-s})\zeta(1+s)$$ Jan 19, 2012 at 22:07
• Thanks. I had to look up the $\eta$, Gamma and Riemann zeta functions and their expansions on MathWorld. The series expansion $$\zeta(1+s)=1/s+\gamma+O(s)$$ is standard, and $$1-2^{-s} = (\log 2)s - \frac12(\log2)^2s^2 + O(s^3)$$ is just Taylor series. Jan 19, 2012 at 22:28