# Evaluating $\int_0^{\infty} \frac{\ln (x)}{1-x^n}\text{d}x$

This is just for recreational purposes.

I've been wondering how to evaluate the following integral:

$$\int_0^{\infty} \frac{\ln (x)}{1-x^n}\text{d}x$$

Because I noticed that certain values for $$n$$ lead to some nice rational multiples of $$\pi^2$$ when evaluated in WolframAlpha. This leads me to believe we should somehow get it into series form to turn it into something Basel-esque. However, I've been struggling with this. I managed to put into series form a related integral, to at least show some kind of context/effort, which is:

$$\int_0^{1} \frac{\ln (x)}{1-x^n}\text{d}x$$

When inputting the series expansion of $$1-x^n$$, one simply needs to evaluate an integral of the form $$x^{kn} \ln(x)$$, which is simple, and the following series emerges:

$$\int_0^{1} \frac{\ln (x)}{1-x^n}\text{d}x = \sum_{k=0}^{\infty} \frac{1}{(kn+1)^2}$$

However, this approach doesn't work with bounds from $$0$$ to $$\infty$$ for obvious reasons. Could someone give me some help?

• I believe that the infinite series evaluation of the integral should be negative. Aug 14, 2021 at 20:43
• In Mathematica: Assuming[n \[Element] Integers \[And] n > 1, Integrate[Log[x]/(1-x^n),{x,0,\[Infinity]}]]. $$- \frac{\pi^2 \csc^{-1} (\pi/n)}{n^2}$$ Aug 14, 2021 at 21:19

It turns out that $$\int_0^1\frac{\ln(x)}{1-x^a}\mathrm{d}x=-\frac{1}{a^2}\sum_{n=0}^{\infty}\frac{1}{(n+1/a)^2}$$ (I think you dropped a minus sign). Enforcing the substitution $$x\rightarrow 1/x$$ yields $$\int_1^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x=\int_0^1\frac{x^{a-2}\ln(x)}{1-x^a}\mathrm{d}x=\sum_{n=0}^{\infty}\int_0^1\ln(x)x^{an+a-2}\mathrm{d}x=-\frac{1}{a^2}\sum_{n=0}^{\infty}\frac{1}{\big(n+1-1/a\big)^2}$$ Putting both pieces together yields $$\begin{eqnarray*}\int_0^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x&=&-\frac{1}{a^2}\Bigg[\sum_{n=0}^{\infty}\frac{1}{(n+1/a)^2}+\sum_{n=0}^{\infty}\frac{1}{(n+1-1/a)^2}\Bigg] \\ &=& -\frac{1}{a^2}\Bigg[\sum_{n=0}^{\infty}\frac{1}{(n+1/a)^2}+\sum_{n=1}^{\infty}\frac{1}{(-n+1/a)^2}\Bigg] \\ &=& -\frac{1}{a^2}\sum_{n=-\infty}^{\infty}\frac{1}{(n+1/a)^2} \end{eqnarray*}$$ It is a known result in complex analysis (using residues to evaluate sums of series) that for any $$x\notin \mathbb{Z}$$ we have $$\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2}=\Big(\pi \csc(\pi x)\Big)^2$$ Taking $$x$$ to be $$1/a$$ yields $$\int_0^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x=-\frac{\pi^2}{a^2}\csc^2(\pi/a)$$

• Excellent! This was actually recently done in a complex analysis course over the Summer, so its quite nice to see that pop up. Aug 14, 2021 at 22:02

Considering the other half of the integral: $$I = \int_1^\infty {\frac{\ln(x)}{1-x^n}} \;dx$$

Let $$x = u^{-1} \implies dx = -u^{-2} \; du$$.

$$I = \int_1^0 {\frac{-\ln(u)}{1-u^{-n}}\cdot(-u^{-2})} \; du$$

$$= \int_0^1 {\frac{u^{n-2}\ln(u)}{1-u^{n}}} \; du$$

$$= \int_0^1 {\ln(u)\cdot\Bigg(u^{n-2}+u^{2n-2}+\dots\Bigg) \;du}$$

$$= -\sum_{k=1}^{\infty} \frac{1}{(kn-1)^2}$$

which is divergent for $$n=1$$. (I would be interested to know if both sums can be put together to get the closed form with $$\csc$$.)

@user429040 Already has an excellent answer. I'd like to expand a little bit on the sum formula that he has presented: $$\sum_{k\in\Bbb{Z}}\frac{1}{(k+z)^2}=\pi^2\csc^2(\pi z)$$ How do we prove this formula? Well, first we expand it using partial fractions into two different sums: $$\sum_{k\in\Bbb{Z}}\frac{1}{(k+z)^2}=\sum_{k=0}^\infty\frac{1}{(k+z)^2}+\sum_{k=0}^\infty\frac{1}{\big(k+(1-z)\big)^2}$$ Now we consider the polygamma function: $$\psi^{(n)}(z):=\mathrm D^{n+1}(\log \Gamma)(z)=(-1)^{n+1}n!\sum_{k=0}^\infty \frac{1}{(k+z)^{n+1}}$$ The sum formula is proven in this excellent video from Flammable Maths. We can write this as $$\sum_{k\in\Bbb{Z}}\frac{1}{(k+z)^2}=\psi^{(1)}(z)+\psi^{(1)}(1-z)$$ Let's let $$f_n(z)=\psi^{(n)}(z)-(-1)^n\psi^{(n)}(1-z)$$ Now let $$\mathrm I$$ be the integral operator. Ignoring constants of integration for now (it can be shown they all vanish), we can see using the derivative definition of the polygamma that $$\mathrm I^{n+1} f_n(z)=\log \Gamma(z)+\log\Gamma(1-z)$$ Using the laws of logarithms we have $$\mathrm I^{n+1}f_n(z)=\log\big(\Gamma(z)\Gamma(1-z)\big)$$ But using Euler's reflection formula which is also proven by Flammable Maths we have $$\mathrm I^{n+1}f_n(z)=\log\left(\frac{\pi}{\sin(\pi z)}\right)=\log(\pi)-\log(\sin(\pi z))$$ Taking a derivative on both sides, $$\mathrm I^n f_n(z)=-\frac{1}{\sin(\pi z)}\cdot \cos(\pi z)\cdot \pi=-\pi\cot( \pi z)$$ Hence, $$f_n(z)=\mathrm D^n(s\mapsto -\pi \cot(\pi s))(z)$$ Therefore $$f_1(z)=\mathrm D(s\mapsto -\pi\cot(\pi s))(z)=\pi^2\csc^2 (\pi z)$$