# Evaluate $\int_0^{\infty}\frac{\log x}{1+e^x}\,dx$

Evaluate $$\int_0^{+\infty}\frac{\log x}{1+e^x}\,dx.$$

I have tried using Feynman's Trick (in several ways, but for example by introducing a variable $a$ such that $I(a)=\int_0^{+\infty}\frac{\log ax}{1+e^x}\,dx$), but that doesn't seem to work. Also integration by parts and all kinds of substitutions make things worse (I have no idea how to substitute such that $\log$ and $\exp$ both become simpler.

(Source: Dutch Integration Championship 2013 - Level 5/5)

• $$I(a) = \int_0^{+\infty} \frac{x^a}{1+e^x}\,dx$$ would be the candidate for differentiation under the integral. – Daniel Fischer Dec 30 '17 at 22:15
• Another hint: $$\int_{0}^{\infty}\frac{x^{s-1}}{1+e^x}\,dx=\Gamma(s)(1-2^{1-s})\zeta(s).$$ Now you may compute the derivative w.r.t. $s$ at $s=1$. – Sangchul Lee Dec 30 '17 at 22:18
• @SangchulLee may I ask how one gets that? Do you have a reference for it? It is beyond me and I would like to learn how that gets done. It is okay if you don't have a reference in hand range – Shashi Dec 30 '17 at 22:21
• @SangchulLee by expanding the exponential in geometric series? The denominator huh – Shashi Dec 30 '17 at 22:22
• @rae306 For Daniel Fischer's approach, you would differentiate with respect to $a$, obtaining a factor of $\log(x)$, and then set $a=0$. – Ian Dec 30 '17 at 22:58

By the inverse Laplace transform $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx$$ and by differentiating both sides with respect to $s$ $$\sum_{n\geq 1}\frac{(-1)^n \log n}{n^s} = -\frac{\Gamma'(s)}{\Gamma(s)^2}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx + \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}\log(x)}{e^x+1}\,dx$$ so by evaluating at $s=1$ $$\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx = \sum_{n\geq 1}\frac{(-1)^n\log n}{n}+\underbrace{\Gamma'(1)}_{-\gamma}\underbrace{\int_{0}^{+\infty}\frac{dx}{e^x+1}}_{\log 2}$$ and it just remains to crack the mysterious series $\sum_{n\geq 1}\frac{(-1)^n\log n}{n}$. On the other hand by Frullani's integral, the inverse Laplace transform or Feynman's trick we have $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx$, so $$\sum_{n\geq 1}\frac{(-1)^n\log n}{n}=\int_{0}^{+\infty}\frac{\log(1+e^{-x})-e^{-x}\log 2}{x}\,dx=\gamma\log(2)-\frac{1}{2}\log^2(2)\tag{J}$$ where the last identity follows from the integral representation for the Euler-Mascheroni constant, got by applying the inverse Laplace transform to the series definition $\gamma=\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$. Summarizing, we simply have $$\int_{0}^{+\infty}\frac{\log(x)}{e^x+1}\,dx = \color{red}{-\frac{1}{2}\log^2(2)}.$$ It is possible to prove the equality between the LHS and the RHS of $(J)$ by summation by parts and Euler sums, too.

• This answer add new things to my math knowledge (+1). However due to being new to Euler Mascheroni I do not understand how you got (J), can you please tell which integral representation? I looked up on the Internet and I could not find which one you meant. Thanks in advance. Moreover I think in the second line the second Gamma in the denominator should not be squared although that won't change anything in the rest of the elaboration. – Shashi Dec 30 '17 at 23:12
• @Shashi: typo fixed, and the mentioned integral representation is $$\gamma=\int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx.$$ – Jack D'Aurizio Dec 30 '17 at 23:18
• many thanks!! I'll remember this one! – Shashi Dec 30 '17 at 23:21
• @JackD'Aurizio thanks for this beautiful answer! I have learned some new tricks :) – rae306 Dec 31 '17 at 9:06

There's a way avoiding special functions and transforms, similar to the device for calculating Frullani integrals: Let's calculate $\displaystyle\int^\infty_\epsilon\frac{\log x}{1+e^x}\,dx$ up to terms $o(1)$ for $\epsilon\to0+$. With the identity $$\frac1{e^x+1}=\frac1{e^x-1}-\frac2{e^{2x}-1},$$ we have \begin{align}\int^\infty_\epsilon\frac{\log x}{1+e^x}\,dx&=\int^\infty_\epsilon\frac{\log x}{e^x-1}\,dx-\int^\infty_\epsilon\frac{2\log x}{e^{2x}-1}\,dx\\&=\int^\infty_\epsilon\frac{\log x}{e^x-1}\,dx-\int^\infty_{2\epsilon}\frac{\log x-\log2}{e^x-1}\,dx \\&=\int^{2\epsilon}_\epsilon\frac{\log x}{e^x-1}\,dx+\int^\infty_{2\epsilon}\frac{\log2}{e^x-1}\,dx \end{align} Using $\displaystyle\frac1{e^x-1}=\frac1x+O(1)$ and $\displaystyle\int^{2\epsilon}_\epsilon|\log x|\,dx=o(1)$, we see $$\int^{2\epsilon}_\epsilon\frac{\log x}{e^x-1}\,dx=\int^{2\epsilon}_\epsilon\frac{\log x}{x}\,dx+o(1)=\frac12\log^22+\log2\log\epsilon+o(1)$$ and $$\int^\infty_{2\epsilon}\frac{\log2}{e^x-1}\,dx=\log2\log\frac1{1-e^{-2\epsilon}}=-\log^22-\log2\log\epsilon+o(1),$$ so $$\int^\infty_\epsilon\frac{\log x}{1+e^x}\,dx=-\frac12\log^22+o(1),$$ and our integral is $$\int^\infty_0\frac{\log x}{1+e^x}\,dx=-\frac12\log^22.$$

As pointed out in the comments, let $I(a)$ be the following integral: $$I(a)=\int\limits_0^{\infty} \frac{x^a}{1+e^x}\,dx$$ We are then looking for $I'(0)$.

\begin{align} I(a)&=\int\limits_0^{\infty}e^{-x}x^a\frac{1}{1+e^{-x}}\,dx\\ I(a)&= \int\limits_0^{\infty}e^{-x}x^a \sum_{n=0}^{+\infty}(-1)^n e^{-nx}\,dx\\ I(a)&=\sum_{n=0}^{+\infty}(-1)^n\int_\limits0^{\infty}e^{-x(n+1)}x^a\,dx\\ I(a)&=\sum_{n=1}^{+\infty}(-1)^{n-1}\int\limits_0^{\infty}e^{-nx}x^a\,dx &u=nx\\ I(a)&=\sum_{n=1}^{+\infty}(-1)^{n-1}\int\limits_0^{\infty} e^{-u}\frac{u^a}{n^a}\frac{du}{n}\\ I(a)&=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n^{a+1}} \int\limits_0^{\infty}e^{-u}u^{a}\,du\\ I(a)&=\eta(a+1)\Gamma(a+1) \end{align} Where $\eta(s)$ is the Dirichlet eta function and $\Gamma(s)$ is the Gamma function. Taking the derivative of both sides, \begin{align} I'(a)&=\eta'(a+1)\Gamma(a+1)+\eta(a+1)\psi(a+1)\Gamma(a+1)\\ I'(0)&=\eta'(1)\Gamma(1)+\eta(1)\psi(1)\Gamma(1)\\ I'(0)&=\log(2)\gamma-\frac{1}{2}\log^2(2)-\log(2)\gamma\\ I'(0)&=-\frac{1}{2}\log^2(2) \end{align} Thus,

$$\int\limits_0^{\infty} \frac{\log(x)}{1+e^x}\,dx=-\log^2\left(\sqrt{2}^{\sqrt{2}}\right)$$

$\log(x)/(1+e^x)=e^{-x} \log(x)/(1+e^{-x})$ then expand $1/(1+e^{-x})$ using the geometric series. Up to changes of variables you are then left to integrate $\log(x) e^{-x}$ using any method, and then computing a certain infinite series.