Integration of $\int_{0}^{\infty} \frac{\ln(x)}{1+e^{x}} dx$

Question

Integration of $$\int_{0}^{\infty} \frac{\ln(x)}{1+e^{x}} dx$$

I'd tried using integration by parts but both the LHS and RHS cancels that term of integration itself. So what should be the approach ?

Answer 1

Let $$I=\int_0^{\infty}\frac{\ln(x)}{1+e^x}\text{d}x$$

$$=\int_0^{\infty}\frac{e^{-x}\cdot\ln(x)}{1+e^{-x}}\text{d}x$$

$$=\int_0^{\infty}e^{-x}\cdot\ln(x)\sum_{n=0}^{\infty}e^{-nx}\text{d}x$$

$$=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}e^{-(n+1)x}\ln(x)\text{d}x.$$

Let $u=(n+1)x$, then we have

$$I=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^{\infty}e^{-u}\ln\left(\frac{u}{n+1}\right)\text{d}u$$

$$=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^{\infty}\left[e^{-u}\ln(u)-e^{-u}\ln(n+1)\right]\text{d}u$$

$$=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\left[\int_0^{\infty}e^{-u}\ln(u)\text{d}u-\int_0^{\infty}e^{-u}\ln(n+1)\text{d}u\right]$$

$$=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\left(-\gamma-\ln(n+1)\right)$$

$$=-\gamma\ln(2)-\sum_{n=0}^{\infty}\frac{(-1)^n\ln(n+1)}{n+1},$$

where $\gamma$ is the Euler-Mascheroni constant. Now,

$$I=-\gamma\ln(2)+\sum_{n=1}^{\infty}\frac{(-1)^n\ln(n)}{n}$$

$$=-\gamma\ln(2)+\eta'(1),$$

where $\eta(s)$ is the Dirichlet eta function. Standard techniques allow us to calculate $\eta'(1)=\gamma\ln(2)-\frac{1}{2}\ln^2(2)$, and so we have

$$I=-\gamma\ln(2)+\gamma\ln(2)-\frac{1}{2}\ln^2(2)=-\frac{1}{2}\ln^2(2).$$