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

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 $$n\in \mathbb{Z}^+$$.

• 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. Commented Dec 13, 2018 at 2:02

\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.

• Hi, can you tell me which one of the integral in the third line have you integrated by parts? Commented Dec 13, 2018 at 0:40
• @Zacky The third term in the second line. 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 $$\boxed{I=\left(1+\frac{1}{n^2}\right) \frac{\pi^2}{12}}$$