$$\int_0^\infty \frac{\ln(1+u)}{u+u^\phi}\mathrm du$$
I saw this problem posted before, however it was deleted.
I forgot the solution to this integral but I don't know how to begin. Maybe I should try expanding $\ln(1+u)$ but I would be forced to split the integral from 0 to 1 and 1 to $\infty$ with $u \to 1/u$.
It would be easier if the $u$'s on the bottom were in brackets, cuz then I could use the beta function. I tried Feynman's trick with ln(1+au) but that didn't help.
Best Answer
\begin{eqnarray*} \int_0^\infty \frac{\ln(1+u)}{u+u^\phi}du &=& \int_0^1 \frac{\ln(1+u)}{u(1+u^{\phi-1})}du + \int_1^\infty \frac{\ln(1+u)}{u(1+u^{\phi-1})}du \\ &=& \int_0^1 \frac{\ln(1+u)}{u(1+u^{\phi-1})}du + \underbrace{\int_0^1 \frac{\ln\left(\frac{1+u}{u}\right)u^{\phi-1}}{u(1+u^{\phi-1})}du}_{u\mapsto \frac{1}{u}}\\ &=& \int_0^1 \frac{\ln(1+u)}{u(1+u^{\phi-1})}du+ \int_0^1 \frac{\ln(1+u)u^{\phi-1}}{u(1+u^{\phi-1})}du - \int_0^1 \frac{\ln(u)u^{\phi-1}}{u(1+u^{\phi-1})}du\\ &=& \int_0^1 \left[\frac{\ln(1+u)}{u(1+u^{\phi-1})}+ \frac{\ln(1+u)u^{\phi-1}}{u(1+u^{\phi-1})} \right]du - \int_0^1 \frac{\ln(u)u^{\phi-1}}{u(1+u^{\phi-1})}du\\ &=& \int_0^1 \frac{\ln(1+u)}{u}du - \int_0^1 \frac{\ln(u)u^{\phi-1}}{u(1+u^{\phi-1})}du\\ &=& \frac{\pi^2}{12}- \sum_{j=0}^{\infty}(-1)^j\int_0^1 \ln(u)u^{\left[j(\phi-1)+\phi-2\right]}\\ &=& \frac{\pi^2}{12}- \lim_{w \to 0+} \frac{d}{dw}\sum_{j=0}^{\infty}(-1)^j\int_0^1 u^wu^{\left[j(\phi-1)+\phi-2\right]}\\ &=& \frac{\pi^2}{12}- \lim_{w \to 0+} \frac{d}{dw}\sum_{j=0}^{\infty}(-1)^j\int_0^1 u^{\left[j(\phi-1)+\phi-2+w\right]}\\ &=& \frac{\pi^2}{12}- \lim_{w \to 0+} \frac{d}{dw}\sum_{j=0}^{\infty}(-1)^j\frac{1}{(j(\phi-1)+\phi-1+w)}\\ &=& \frac{\pi^2}{12}+ \sum_{j=0}^{\infty}(-1)^j\frac{1}{(j(\phi-1)+\phi-1)^2}\\ &=& \frac{\pi^2}{12}+ \frac{1}{(\phi-1)^2}\sum_{j=0}^{\infty}\frac{(-1)^j}{(j+1)^2} \quad \phi\neq 1\\ &=& \frac{\pi^2}{12}\left[1+\frac{1}{(\phi-1)^2}\right] \end{eqnarray*}
$$\boxed{\int_0^\infty \frac{\ln(1+u)}{u+u^\phi}du = \frac{\pi^2}{12}\left[1+\frac{1}{(\phi-1)^2}\right], \phi > 1 } $$