7
$\begingroup$

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?

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

3 Answers 3

4
$\begingroup$

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 $$\int_0^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x=-\frac{\pi^2}{a^2}\csc^2(\pi/a)$$

$\endgroup$
1
  • $\begingroup$ Excellent! This was actually recently done in a complex analysis course over the Summer, so its quite nice to see that pop up. $\endgroup$
    – Luna145
    Aug 14, 2021 at 22:02
3
$\begingroup$

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 $\csc$.)

$\endgroup$
2
$\begingroup$

@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 $$f_1(z)=\mathrm D(s\mapsto -\pi\cot(\pi s))(z)=\pi^2\csc^2 (\pi z)$$

$\endgroup$

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.