Question

Integrate:
\[I_n=\int_1^\infty \frac{\ln(x+1)}{x^n}\,dx\]
for \(n\ge2\). Mathematica is giving some interesting patterns, so a closed form would be interesting to see.

Answer 1

I've realized soon enough that IBP is the way to go...

Answer 2

IBP?

Answer 3

Integrating By Parts. Incredibly useful, yet it always seems to be the last approach I have in mind.

Answer 4

Setting the following
\[\begin{matrix}u=\ln(x+1)&&&dv=\frac{dx}{x^n}\\
du=\frac{dx}{x+1}&&&v=-\frac{1}{(n+1)x^{n+1}}\end{matrix}\]
so
\[\begin{align*}I_n&=-\frac{1}{n+1}\left[\frac{\ln(x+1)}{x^{n+1}}\right]_1^\infty+\frac{1}{n+1}\int_1^\infty\frac{dx}{(x+1)x^{n+1}}\\\\
&=\frac{\ln2}{n+1}+\frac{1}{n+1}\int_1^\infty\frac{dx}{(x+1)x^{n+1}}
\end{align*}\]

Answer 5

Cool! I'm not sure about this but I would like to see anybody solving this up. I will have an eye over this question now. :) Nice question, btw! :)

Answer 6

\[\begin{align}
\int_1^\infty \frac{\ln(x+1)}{x^n}\,dx &\stackrel{IBP}{=} \frac{\ln 2}{n-1} + \frac{1}{n-1}\int \limits_1^{\infty} \dfrac{dx}{x^{n-1}(1+x)}\\~\\
&= \frac{\ln 2}{n-1} + \frac{1}{n-1}\int \limits_0^{1} \dfrac{u^{n-2}}{u+1}du\\~\\
&= \frac{\ln 2}{n-1} + \frac{1}{n-1}\int \limits_0^{1} u^{n-2} \sum\limits_{k=0}^{\infty} (-1)^ku^kdu\\~\\
&= \frac{\ln 2}{n-1} + \frac{1}{n-1}\sum\limits_{k=0}^{\infty} (-1)^k\int \limits_0^{1} u^{k+n-2}du\\~\\
&= \frac{\ln 2}{n-1} + \frac{1}{n-1}\sum\limits_{k=0}^{\infty} \frac{(-1)^k}{ k+n-1}\\~\\

\end{align}\]

Answer 7

looks it is difficult to get into a more nicer closed form than that
http://mathworld.wolfram.com/LerchTranscendent.html

Answer 8

Ah alright. I'm curious to see if there's a generalization to be made if we introduce a parameter, like
\[I_{m,n}=\int_1^\infty \frac{(\ln(x+1))^m}{x^n}\,dx\]
I'll have to work on that...

Answer 9

for m>1 i think we can use beta function

Answer 10

I ran some computations in Mathematica, using \(1\le m\le5\) and \(2\le n\le6\). The integral seems to converge for any finite values of \(m,n\) so long as \(n\ge2\), which goes against my tuition.

Here's my impression:
\[\ln(x+1)=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}=x+O(x^2)\]
so it must be that
\[\frac{(\ln(x+1))^m}{x^n}=\frac{x^m+O(x^{m+1})}{x^n}\stackrel{\large\color{red}?}{=}\frac{O(1)}{x^{n-m-1}}\]
(I think; I may be abusing the big O notation) This leads me to think that the integral should converge only for \(n-m-1\ge2\).

Computational results attached.