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 $I_{2n+1}$?


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.