Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ 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 $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$
 A: Writing
$$\frac 1{x^3-1}=\frac{1}{(a-1) (a-b) (x-a)}+\frac{1}{(b-1) (b-a) (x-b)}+\frac{1}{(a-1) (b-1) (x-1)}$$ where
$$a=-\frac{1}{2}-\frac{i \sqrt{3}}{2} \qquad \text{and} \qquad b=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$$ we face three integrals
$$I(c)=\int \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{x}{c}\right)+\log (x) \log \left(1-\frac{x}{c}\right)$$
$$J(c)=\int_0^t \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{t}{c}\right)+\log (t) \log \left(1-\frac{t}{c}\right)$$
Recombining the three terms, computing at the bounds and using the values of the polylogarithms leads to
$$\int_0^\infty\frac{\ln(x)}{x^3-1}dx=\frac{5 \pi ^2}{54}-\left(-\frac{\pi ^2}{18} \right)=\frac{4 \pi ^2}{27}$$
A: Carrying on from where the OP left off, we have
$$\begin{align}
\int_{-\infty}^\infty{ue^u\over e^{3u}-1}du
&=\int_0^\infty{ue^u\over e^{3u}-1}du+\int_{-\infty}^0{ue^u\over e^{3u}-1}du\\
&=\int_0^\infty{ue^{-2u}\over1-e^{-3u}}du+\int_\infty^0{ue^{-u}\over e^{-3u}-1}du\\
&=\int_0^\infty{ue^{-2u}\over1-e^{-3u}}du+\int_0^\infty{ue^{-u}\over1-e^{-3u}}du\\
&=\int_0^\infty u(e^{-2u}+e^{-5u}+e^{-8u}+\cdots+e^{-u}+e^{-4u}+e^{-7u}+\cdots)du\\
&={1\over2^2}+{1\over5^2}+{1\over8^2}+\cdots+1+{1\over4^2}+{1\over7^2}+\cdots\\
&=\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots \right)-\left({1\over3^2}+{1\over6^2}+{1\over9^2}+\cdots \right)\\
&=\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots \right)-{1\over9}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots \right)\\
&={8\over9}\zeta(2)\\
&={8\over9}{\pi^2\over6}\\
&={4\pi^2\over27}
\end{align}$$
Remark: The final step(s) require knowing that $\zeta(2)=\pi^2/6$, which may not be in the OP's toolbox. All other steps are standard manipulations of (improper) integrals and infinite series, with the geometric series rearing its beautiful head in the middle.
A: I thought it might be instructive to present an approach that relies on straightforward contour integration.  To that end, we now proceed.
Let $I$ be the integral given by
$$I=\int_0^\infty \frac{\log(x)}{x^3-1}\,dx$$
Now, moving to the complex plane, we analyze the contour integral $J$ given by
$$J=\oint_C \frac{\log^2(z)}{z^3-1}\,dz$$
where $C$ is the classical keyhole contour with a semi-circular deformation at $z=1+i0^-$.  That is, $J$ is given by
$$\begin{align}
J&=\int_0^\infty \frac{\log^2(x)}{x^3-1}\,dx-\text{PV}\int_0^\infty\frac{(\log(x)+i2\pi)^2}{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I+4\pi^2\text{PV}\int_0^\infty \frac1{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I-\frac{4\pi^3}{3\sqrt3}+i\frac{4\pi^3}3\tag1
\end{align}$$
Rearranging $(1)$ reveals that the integral of interest can by written in terms of $J$ as
$$I=i\frac1{4\pi}J+i \frac{\pi^2}{3\sqrt 3}+\frac{\pi^2}{3}\tag2$$
Now, applying the residue theorem we see that $\frac{iJ}{4\pi}$ is given by
$$\begin{align}
\frac{iJ}{4\pi}&=-\frac12 \left(\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i2\pi/3}\right)+\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i4\pi/3}\right)\right)\\\\
&=-\frac{4\pi^2}{27}\left(\frac54+i\frac{3\sqrt{3}}{4}\right)\tag3
\end{align}$$
Using $(3)$ in $(2)$ yields the coveted result
$$I=\frac{4\pi^2}{27}$$
as expected!
A: 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 $J_{2,3}=\frac{4\pi^2}{27}$
A: Split the integral to simplify as follows
\begin{align}\int_0^\infty\frac{\ln x}{x^3-1}{d}x
= &\int_0^1\frac{\ln x}{x^3-1}{d}x + \int_1^\infty\overset{x\to \frac1x}{\frac{\ln x}{x^3-1}}{d}x
=\int_0^1\frac{(1+x)\ln x}{x^3-1}{d}x \\
= &\int_0^1\frac{(x^2+x+1)\ln x}{x^3-1}{d}x
 - \int_0^1\overset{x^3\to x}{\frac{x^2\ln x}{x^3-1}}{d}x\\
=&\int_0^1\frac{\ln x}{x-1}{d}x-\frac19 \int_0^1\frac{\ln x}{x-1}{d}x
=\frac89 \int_0^1\frac{\ln x}{x-1}{d}x\\
=&\frac89\cdot \frac{\pi^2}6
=\frac{4\pi^2}{27}
\end{align}
Integrate $\int_0^1\frac{\ln x}{x-1}{d}x=\frac{\pi^2}6$
A: Using Psi-Gamma function:
$$I=\int_{0}^{\infty} \frac{\log x}{x^3-1}dx$$
Let $x=e^t$, then
$$I=\int_{-\infty}^{\infty} \frac{t e^t}{e^{3t}-1} dt= \int_{-\infty}^{0}  \frac{t e^t}{e^{3t}-1} dt + \int_{0}^{\infty} \frac{t e^t}{e^{3t}-1} dt$$
In the first one take $t=-u$, then
$$I=\int_{0}^{\infty} \frac{ue^{-u}}{1-e^{-3u}} du+\int_{0}^{\infty} \frac{t e^{-2t}}{1-e^{-3t}}dt.$$
Using IGP, we get
$$I=\int_{0}^{\infty} \sum_{k=0}^{\infty} u e^{-(3k+1)u}+\sum_{0}^{\infty} t e^{-(3k+2)t} dt=\sum_{k=0}^{\infty}  \frac{1}{(3k+1)^2}+\sum_{k=0}^{\infty}\frac{1}{(3k+1)^2}.$$
In terms of Psi-Gamma (Poly Gamma) functions:https://en.wikipedia.org/wiki/Polygamma_function,
we can write $$I=\frac{1}{9}[\psi^1(1/3)+\psi^1(2/3)]$$
Using $$(-1)^m \psi^m(1-z)-\psi^m(z)=\pi^2 \frac{d^m}{dz^m} \cot (\pi z)$$
we get $$I=\frac{1}{9} \pi^2 \csc^2(\pi/3)= \frac{4 \pi^2}{27}.$$
A: By my post, the general integral is
$$\int_{0}^{\infty} \frac{\ln x}{x^{n}-1} d x =\left(\frac{\pi}{n}\right)^{2} \csc ^{2}\left(\frac{\pi}{n}\right)$$
Putting $n=3$, we have
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
