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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 $I_{2n+1}$?

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$I_n=\int_0^\infty\frac {ln^n(x+1)-ln^n(x)}{x+1}dx=\int_1^\infty\frac{ln^n(x)}{x(x+1)}dx-\int_0^1\frac{ln^n(x)}{1+x}dx$, 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...

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  • $\begingroup$ Can you list what the value for the integral is, because I cannot find it in any tables. Or can you link to where the value is stated? $\endgroup$ – aleden May 2 at 1:15
  • $\begingroup$ Nevermind, Wolfram can evaluate it and the values match up. Thank you for your answer! $\endgroup$ – aleden May 2 at 3:59
  • $\begingroup$ From the Table I referenced: $\int_0^1\frac{(lnx)^{2n-1}}{1+x}dx=\frac{1-2^{2n-1}}{2n}\pi^{2n}|B_{2n}|$ where $B_{2n}$ are Bernoulli numbers. If you can access that table, look for section 4.271. $ $\endgroup$ – herb steinberg May 2 at 19:38
  • $\begingroup$ I see, I also realized that the integral you had $I_n$ reduced to can be evaluated by a simple substitution to yield the answer in terms of the Gamma and Dirichlet Eta Function. $\endgroup$ – aleden May 2 at 19:40

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