Can any person solve this integral and write the detailed answer:
$$\int_0^\infty \frac{\ln(x)}{e^x+1}~dx$$
I tried to change the variable with supposed $x=t+1$ so I have now
$$\int_{-1}^\infty \frac{\ln(t+1)}{e^{(t+1)}+1}~dt$$
Then I add a constant a before t for using Feynman technique and I have in finally
$$\int_{-1}^\infty \frac{a}{(e^{t+1}+1)(at+1)}~dt$$
But i don't know how to continue, I supposed I must use complex analysis to continue but it doesn't work, any help?
$$I=\int_0^\infty\frac{\ln x}{e^x+1}dx=\lim_{\epsilon\to 0}I_\epsilon$$ where $$I_\epsilon=\int_\epsilon^\infty\frac{\ln x}{e^x+1}dx\overset{IBP}{=}\ln x\ln(1+e^{-x})\bigg|_\epsilon^\infty-\int_\epsilon\frac{\ln(1+e^{-x})}xdx$$ $$=-\ln\epsilon\ln(1+e^{-\epsilon})-\int_\epsilon^\infty\frac{\ln(1-e^{-2x})}xdx+\int_\epsilon^\infty\frac{\ln(1-e^{-x})}xdx$$ $$=-\ln\epsilon\ln(1+e^{-\epsilon})+\int_\epsilon^{2\epsilon}\frac{\ln(1-e^{-x})}xdx$$ Here we can use the mean value theorem, or just make the substitution $x=\epsilon t$ $$I_\epsilon=-\ln\epsilon\ln(1+e^{-\epsilon})+\int_1^2\frac{\ln(1-e^{-\epsilon t})}tdt$$ $$=-\ln\epsilon\ln(1+e^{-\epsilon})+\int_1^2\frac{\ln(\epsilon t)}tdt+O(\epsilon)$$ $$=-\ln\epsilon\ln(1+e^{-\epsilon})+\ln\epsilon\ln2+\frac12\ln^22+O(\epsilon)$$ $$I=\lim_{\epsilon\to 0}I_\epsilon=\frac12\ln^22$$
$$I=\int_0^\infty\frac{e^{-x}\ln x}{1+e^{-x}}dx=\sum_{n=0}(-1)^n\int_0^\infty e^{-(n+1)x}\ln x~ dx=-\sum_{k=1}(-1)^k\int_0^\infty e^{-kx}\ln x~ dx$$
Let $t=kx$
$$\int_0^\infty e^{-kx}\ln x~ dx=\frac1k\int_0^\infty e^{-t}(\ln t-\ln k)~ dt=-\frac{\gamma+\ln k}k$$
hence $$I=\sum_{k=1}(-1)^k\cdot\frac{\gamma+\ln k}k=-\gamma\ln2+\sum_{k=1}(-1)^k\cdot\frac{\ln k}k$$
Use the result
$$\sum_{k=1}(-1)^k\cdot\frac{\ln k}k=\gamma \ln 2- \frac{1}{2}(\ln 2)^2$$
hence
$$\boxed{I=- \frac{1}{2}(\ln 2)^2}$$
We shall use the Feynman trick. Instead of the integral in question
$$I=\int_{0}^{\infty}\frac{\log(x)}{e^x+1}\;dx$$
consider the integral
$$i(s)=\int_{0}^{\infty}\frac{x^{s-1}}{e^x+1}\;dx$$
We can generate the log by taking the drivative with respect to $s$ at $s=1$, indeed
$$I = \frac{d}{ds}i(s)|_{s\to 1}$$
Writing
$$\frac{x^{s-1}}{e^x+1}=\frac{e^{-x} x^{s-1}}{1+e^{-x}}=- x^{s-1}\sum_{n=1}^ {\infty} (-1)^n e^{-n x}$$
the integral of the summand becomes
$$\int_{0}^{\infty}x^{s-1}e^{-n x}\;dx \overset{nx\to y}=\frac{1}{n^s}\int_{0}^{\infty}y^{s-1}e^{-y}\;dy=\frac{\Gamma(s)}{n^s}$$
summing over $n$ gives the integral
$$i(s) = \Gamma(s)\sum_{n=1}^ {\infty} (-1)^{n+1} \frac{1}{n^s}=\Gamma(s)\left(1-2^{1-s}\right) \zeta (s)$$
Now we calculate the derivative, close to $s=1$
$$\begin {align}&\frac{d}{ds}\left(\Gamma(s)\left(1-2^{1-s}\right) \zeta (s)\right) \\& =\left(\Gamma'(s)\left(1-2^{1-s}\right) \zeta (s)\right)+\left(\Gamma(s)\left(1-2^{1-s}\right)' \zeta (s)\right)+\left(\Gamma(s)\left(1-2^{1-s}\right) \zeta' (s)\right)\\ &\simeq \left(-\gamma \log (2)\right)+\left(\frac{\log (2)}{s-1}-\log ^2(2)\right)+\left(\frac{\log ^2(2)}{2}+\gamma \log (2)-\frac{\log (2)}{s-1}\right)+O\left((s-1)^1\right)\\ &= -\log ^2(2)+\left(\frac{\log ^2(2)}{2}\right)+O\left((s-1)^1\right)\\ &\overset{s \to 1} = -\frac{1}{2}\log ^2(2)\end{align}$$
which is the desired result.
Discussion
Consider more generally
$$I_n = \int_0^{\infty } \frac{\log ^n(x)}{e^x+1} \, dx$$
It is given by the $n$-th derivative of $i(s)$ at $s=1$.
We have for instance
$$I_2 = \frac{\log ^3(2)}{3}-2 \gamma _1 \log (2)+\gamma ^2 \log (2)+\frac{1}{6} \pi ^2 \log (2)$$
In general in $I_n$ there appears a term $\frac{\log ^{n+1}(2)}{n+1}$. Can you prove this?
