Does $$I_n=\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$$ have a general form?

I tried to evaluate some small $n$s.
For $n=1$, $I_1$ is obviously $\frac16\pi^2$.
For $n=2$, see here. $I_2=\frac5{48}\pi^2$.
For $n=3$, I put it in Mathematica and get $$\small{\frac{1}{108} \left(9 \left(4 \left(\text{Li}_2\left(\frac{\sqrt[6]{-1}}{\sqrt{3}}\right)+\text{Li}_2\left(-\frac{(-1)^{5/6}}{\sqrt{3}}\right)\right)+\log ^2(3)\right)+5 \pi ^2\right)}$$ Use the result $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4,$$ I'm able to show $I_3=\frac5{54}\pi^2$.
For $n=4$, I numerically found $I_4=\frac{17}{192}\pi^2$.
I'm not able to find the general form with $n\in \mathbb{Z}^+$.

  • 1
    $\begingroup$ Sorry for not perceiving the answer in the linked question can be applied in this question. Feel free to close this question as a duplicate of that one. $\endgroup$ Commented Dec 13, 2018 at 2:02

2 Answers 2


$$\begin{aligned} I_n &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_1^\infty \frac{\ln(1+x)}{x(1+x^n)}dx \\ &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_0^1 \frac{x^n \ln(1+x)}{x(1+x^n)}dx - \int_0^1 \frac{x^{n-1} \ln x}{1+x^n}dx \\ &= \int_0^1 \frac{\ln(1+x)}{x}dx + \frac{1}{n}\int_0^1 \frac{\ln(1+x^n)}{x}dx \\ &= \int_0^1 \frac{\ln(1+x)}{x}dx+\frac{1}{n^2}\int_0^1 \frac{\ln(1+u)}{u}du = \color{blue}{(1+\frac{1}{n^2})\frac{\pi^2}{12}} \end{aligned}$$

  • Second line: $x\mapsto 1/x$
  • Third line: Integration by parts
  • Fourth line: $x=u^{1/n}$

Note that $n$ need not be an integer.

  • $\begingroup$ Hi, can you tell me which one of the integral in the third line have you integrated by parts? $\endgroup$
    – Zacky
    Commented Dec 13, 2018 at 0:40
  • 1
    $\begingroup$ @Zacky The third term in the second line. $\endgroup$
    – pisco
    Commented Dec 13, 2018 at 4:05

We first split the integral into two. $$ I=\int_0^{\infty} \frac{\ln (1+x)}{x\left(1+x^n\right)} d x = \int_0^1 \frac{\ln (1+x)}{x\left(1+x^n\right)} d x+\int_1^{\infty} \frac{\ln (1+x)}{x\left(1+x^n\right)} d x $$ For the second integral, let $x\mapsto \dfrac{1}{x} $, then $$ \int_1^{\infty} \frac{\ln (1+x)}{x\left(1+x^n\right)} d x =\int_0^1 \frac{x^{n-1}[\ln (1+x)-\ln x]}{x^n+1}dx $$ Plugging back yields

$$ \begin{aligned} I & =\int_0^1\left[\frac{\ln (1+x)}{x\left(1+x^n\right)}+\frac{x^{n-1} \ln (1+x)}{x^n+1}\right] d x-\int_0^1 \frac{x^{n-1} \ln x}{x^n+1} d x \\ & =\int_0^1 \frac{\ln (1+x)}{x} d x-\frac{1}{n} \int_0^1 \ln x d \ln \left(x^n+1\right) \\ & =\int_0^1 \frac{\ln (1+x)}{x} d x+\frac{1}{n} \int_0^1 \frac{\ln \left(x^n+1\right)}{x} d x \end{aligned} $$ For any natural number $n$, $$ \begin{aligned} \int_0^1 \frac{\ln \left(x^n+1\right)}{x} d x = & \int_0^1 \frac{1}{x} \sum_{k=0}^{\infty}(-1)^k x^{n k} \\ = & \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \int_0^1 x^{n k-1} d x \\ = & \frac{1}{n} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2} \\ = & \frac{\pi^2}{12 n} \end{aligned} $$ Now we can conclude that $$ \boxed{I=\left(1+\frac{1}{n^2}\right) \frac{\pi^2}{12}} $$


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