TITLE: Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$
QUESTION [12 upvotes]: Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

REPLY [2 votes]: I will present two methods to evaluate this integral.
Method 1:-
Consider the integral $$I(m,n)=\int_{0}^{\infty}\frac{x^{m-1}}{x^{n}-1}dx=\frac{-\pi}{n}\cot\frac{m\pi}{n}$$  where $m<n$
Differentiate both sides w.r.t  $m$
$$I^{'}(m,n)=\int_{0}^{\infty}\frac{x^{m-1}\ln(x)}{x^{n}-1}dx= \frac{\pi^2}{n^2}\csc^{2}\frac{m\pi}{n}$$
Let $m=1$ and $n=3$ , we obtain our required integral as $\frac{4\pi^2}{27}$
Method 2:-
Let $$J_{k,a}=\int_{0}^{\infty}\frac{\ln^{k-1}(z)}{z^{a}-(-1)^k}dz$$  where $k\in N$ and
$a>1$
$$J_{k,a}=\int_{0}^{1}\frac{\ln^{k-1}(z)}{z^{a}-(-1)^k}dz +\int_{1}^{\infty}\frac{\ln^{k-1}(z)}{z^{a}-(-1)^k}dz$$
Let $k=\frac{1}{u}$ in second integral
$$J_{k,a}=\int_{0}^{1}\frac{\ln^{k-1}(z)}{z^{a}-(-1)^k}dz +\int_{0}^{1}\frac{\ln^{k-1}(u)}{u^{-a}-(-1)^k}\frac{1}{u^2}du$$
Using Geometric series and Changing order of Integration and Summation (by Monotone convergence theorem) and after applying repeated integration by parts we get,
$$J_{k,a}=(k-1)!\sum_{n=0}^{\infty}\frac{(-1)^{nk}}{(an+1)^k} + (k-1)!\sum_{n=1}^{\infty}\frac{(-1)^{nk}}{(1-an)^k}$$
$$J_{k,a}=(k-1)!\sum_{n=0}^{\infty}\frac{(-1)^{nk}}{(an+1)^k} + (k-1)!\sum_{n= -\infty}^{-1}\frac{(-1)^{nk}}{(an+1)^k}$$
$$J_{k,a}=(k-1)!\sum_{n= -\infty}^{\infty}\frac{(-1)^{nk}}{(an+1)^k}$$
Required integral is $J_{2,3}$
$$J_{2,3}=\sum_{n= -\infty}^{\infty} \frac{1}{(3n+1)^2}$$
Using $$\sum_{n= -\infty}^{\infty} \frac{1}{(an+1)^2}=\frac{\pi^2}{a^2}\csc^{2}\frac{\pi}{a}$$
Therefore $J_{2,3}=\frac{4\pi^2}{27}$