I want to show that :
$$\int_{0}^{\infty}\ln^2\Big(e^{-x}+1\Big)dx=\frac{\zeta(3)}{4}$$
I have tried substitution like $y=e^{-x}$ it gives :
$$\int_{0}^{1}\frac{\ln^2(y+1)}{y}dy=\frac{\zeta(3)}{4}$$
Next I was working on the power series :
$$\ln^2(1 + x) = \Big(\sum_{k=1}^{\infty}\frac{(-1)^kx^k}{k}\Big)^2 ,\quad|x|<1$$
And now I'm stuck because I don't know how to make zeta appear.
Thanks in advance .
This is shown by Rutledge and Douglass [Am Math Monthly 45 (1938) p 525] in Table of definite integrals,
$$f(y)=\int\frac{\log^2(y+1)}{y}\,dy$$ Make one integration by parts first $$f(y)=\log (y) \log ^2(y+1)-2\int \frac{ \log (y) \log (y+1)}{y+1}\,dy$$ $$\int \frac{ \log (y) \log (y+1)}{y+1}\,dy=2 \text{Li}_3(y+1)-2 \text{Li}_2(y+1) \log (y+1)-i \pi \log ^2(y+1)$$ $$f(1)-f(0)=-2(\frac 78 \zeta(3)-\zeta(3))=\frac {\zeta(3)}4$$
