# 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. Dec 30, 2017 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 $s=1$. Dec 30, 2017 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 Dec 30, 2017 at 22:21
• @SangchulLee by expanding the exponential in geometric series? The denominator huh Dec 30, 2017 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, 2017 at 22:58

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 $$\int^\infty_0\frac{\log x}{1+e^x}\,dx=-\frac12\log^22.$$

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. Dec 30, 2017 at 23:12
• @Shashi: typo fixed, and the mentioned integral representation is $$\gamma=\int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx.$$ Dec 30, 2017 at 23:18
• many thanks!! I'll remember this one! Dec 30, 2017 at 23:21
• @JackD'Aurizio thanks for this beautiful answer! I have learned some new tricks :) Dec 31, 2017 at 9:06
• @JackD'Aurizio How did you get $$\int_{0}^{+\infty}\frac{\log(1+e^{-x})-e^{-x}\log 2}{x}\,dx=\gamma\log(2)-\frac{1}{2}\log^2(2)$$. I integrated by parts this integral and got $$\int_{0}^{+\infty}\frac{\log(1+e^{-x})-e^{-x}\log 2}{x}\,dx=\gamma \log(2)+\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx$$ which does´t add any information. Thank you very much! Oct 15, 2021 at 19:40

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.

We have the lemma： If f(x) is differentiable on intervals $$(0,+\infty)$$,and $$f'(x)$$ is monotonic function ,$$\lim\limits_{x\to\infty}f'(x)=0$$,then exist the limit： $$\lim\limits_{n\to\infty}\Big[\frac{1}{2}f(1)+f(2)+f(3)+\cdots+f(n-1)+\frac{1}{2}f(n)-\int_1^nf(x)dx\Big]=l$$ Pro:Let the $$F(x)=\int_1^xf(t)dt$$Use the Taylor formula ,then exist $$\xi_k,\eta_k:k<\xi_k,meet $$F(k+\frac{1}{2})-F(k)=\frac{1}{2}F'(k)+\frac{1}{8}F''(\xi_k)=\frac{1}{2}f(k)+\frac{1}{8}f'(\xi_k)-F(k+\frac{1}{2})+F(k+1)$$ $$=\frac{1}{2}f(k+1)-\frac{1}{8}f'(\eta_k)$$ Sum Up the above formula from $$k=1$$ to $$n-1$$,we have $$\frac{1}{2}f(1)+f(2)+\cdots+f(n-1)+\frac{1}{2}f(n)-F(n)$$ $$=\frac{1}{8}\Big[f'(\eta_1)-f'(\xi_1)+f'(\eta_2)-f'(\xi_2)+\cdots+f'(\eta_{n-1})-f'(\xi_{n-1})\Big]$$ Use the Leibniz test,On the right side of the above formula is exist. For $$f(x)=\frac{\ln x}{x}$$,we have $$\sum_{n=1}^\infty(-1)^{n}\frac{\ln n}{n}=l$$ let $$S_n=\sum_{k=1}^n(-1)^n\frac{\ln k}{k}$$ so \begin{align} \lim\limits_{n\to\infty}S_{2n}=&\lim_{n\to\infty}\Big(-\frac{\ln1}{1}+\frac{\ln 2}{2}+\cdots+\frac{\ln 2n}{2n}\Big)\\ =&\lim_{n\to\infty}\Big[-(\frac{\ln1}{1}+\frac{\ln2}{2}+\cdots+\frac{\ln2n}{2n}-\frac{(\ln2n)^2}{2})\\ &+{}2(\frac{\ln2}{2}+\frac{\ln4}{4}+\cdots+\frac{\ln2n}{2n})-\frac{(\ln2n)^2}{2}\Big]\\ =&-l+\lim_{n\to\infty}\Big(\frac{1}{1}+\frac{\ln2}{2}+\cdots+\frac{\ln n}{n}-\frac{(\ln n)^2}{2}\Big)\\ &-\lim_{n\to\infty}\Big(\frac{\ln1}{1}+\frac{\ln2}{2}+\cdots+\frac{\ln2}{n}-\ln2\ln n\Big)-\frac{(\ln2)^2}{n}\\ =&\ln2\Big(\gamma-\frac{\ln2}{2}\Big) \end{align} where $$\gamma$$ is Euler-Mascheroni constant:$$\gamma=\lim\limits_{n\to\infty}\Big(\sum\limits_{k=1}^n\frac{1}{k}-\ln n\Big)$$

so by Jack D'Aurizio's work: $$\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx = \sum_{n\geq 1}\frac{(-1)^n\log n}{n}+\underbrace{\Gamma'(1)}_{-\gamma}\underbrace{\int_{0}^{+\infty}\frac{dx}{e^x+1}}_{\log 2}$$ we have $$\int_0^\infty\frac{\ln x}{e^x+1}=\gamma\ln2-\frac{1}{2}\ln^22-\gamma\ln2=-\frac{1}{2}\ln^22$$ Alos,when $$\Re(v)>0$$,we have $$\int_0^\infty\frac{x^{v-1}\ln x}{e^x+1}dx=\Gamma(v)\Big(\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^v}[\psi(v)-\ln k]\Big)$$

• Haha,sorry，I have some typos ,but not affect understanding. Jan 7 at 13:25