TITLE: Methods to derive a closed form for $I_n=\int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}dx$
QUESTION [8 upvotes]: I've stumbled onto this general integral that has closed form values for the $n\in \Bbb{Z^+}$ $$I_n=\int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}dx$$
Obviously $I_0=0$ but higher values of $n$ yield interesting results. The first few are:
$$I_1=\frac{\pi^2}{6}$$
$$I_2=0$$
$$I_3=\frac{7\pi^4}{60}$$
$$I_4=0$$
$$...$$
So it appears that $I_{2n}=0$ and that $I_{2n+1}=C_{2n+1}\zeta\left(2(n+1)\right)$ where $C_{n}$ is some rational number that is equal to $0$ when $n\in2\Bbb{Z^+}$. Can anyone explain why $I_{2n}=0$ and how one can derive $I_{2n+1}$?

REPLY [5 votes]: $I_n=\int_0^\infty\frac {ln^n(x+1)-ln^n(x)}{x+1}dx=\int_1^\infty\frac{ln^n(x)}{x(x+1)}dx-\int_0^1\frac{ln^n(x)}{1+x}dx$, using $x+1 \to x$ in the $ln^n(x+1)$ term.
Let $y=\frac{1}{x}$ in the first integral and it becomes $\int_0^1\frac{(-ln(y))^n}{1+y}dy$.
Net result $I_n=0$ for even $n$ and $I_n=-2\int_0^1\frac{ln^n(x)}{1+x}dx$ for odd $n$.
This last integral can be found in Gradshteyn and Ryzhik Table of Integrals...