Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$

Question

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 $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

Answer 1

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 $\int_0^1\frac{\ln x}{x-1}{d}x=\frac{\pi^2}6$

Answer 2

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 $$\int_0^\infty\frac{\ln(x)}{x^3-1}dx=\frac{5 \pi ^2}{54}-\left(-\frac{\pi ^2}{18} \right)=\frac{4 \pi ^2}{27}$$

Answer 3

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.

Answer 4

$ \newcommand{\Res}[1]{\underset{#1}{\operatorname{Res}}} $For completeness, I feel that I should post a complex solution. The exponential substitution in the question makes this problem more amenable to contour integration, because it means that we don't have to work around branch cuts.

Let $f(z)=\dfrac{z^2e^z}{e^{3z}-1}$ (yes, $z^2$ and not $z$). We'll integrate $f(z)$ around the indented rectangular contour shown below:

We'll take the limit as the left and right sides $L$ and $R$ of the rectangle move out to infinity and the indentations $I_1$ and $I_2$ shrink to zero radius. The top and bottom sides ($T_+,T_-,B_+,B_-$) will always have imaginary parts $\pm 2\pi$.

It's not too hard to see that $\int_L f(z) \, dz$ and $\int_R f(z) \, dz$ vanish in the limit. By a standard indented contour argument, we have $\int_{I_1} f(z) \, dz \to -\pi i \Res{z=2\pi i} f(z)$ and $\int_{I_2} f(z) \, dz \to -\pi i \Res{z=-2\pi i}f(z)$ (because we're going clockwise around both poles). Finally, in the limit we have

$$ \int_{T_-+T_+} f(z) \, dz \to -\textrm{p.v.}\int_{-\infty}^\infty \frac{(u+2\pi i)^2e^u}{e^{3u}-1} \, du $$

and

$$ \int_{B_-+B_+} f(z) \, dz \to \textrm{p.v.}\int_{-\infty}^\infty \frac{(u-2\pi i)^2e^u}{e^{3u}-1} \, du $$

By the residue theorem we have therefore have:

$$ \textrm{p.v.}\int_{-\infty}^\infty \left[\frac{(u-2\pi i)^2e^u}{e^{3u}-1} -\frac{(u+2\pi i)^2e^u}{e^{3u}-1}\right] \, du=\pi i \left[\Res{z=2\pi i} f(z)+\Res{z=-2\pi i} f(z)\right]+2\pi i \left[\Res{z=2\pi i/3} f(z)+\Res{z=-2\pi i/3} f(z)+\Res{z=4\pi i/3}f(z)+\Res{z=-4\pi i/3}f(z)\right] \, . $$

Canceling terms on the left-hand side gives

$$ -8 \pi i \int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1} \, du=\pi i \left[\Res{z=2\pi i} f(z)+\Res{z=-2\pi i} f(z)\right]+2\pi i \left[\Res{z=2\pi i/3} f(z)+\Res{z=-2\pi i/3} f(z)+\Res{z=4\pi i/3}f(z)+\Res{z=-4\pi i/3}f(z)\right] $$

Note that this integral is legitimately convergent so we no longer need to be considering principal values. It follows that

$$ \int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\, du=-\frac{1}{8}\left[\Res{z=2\pi i} f(z)+\Res{z=-2\pi i} f(z)\right]-\frac{1}{4} \left[\Res{z=2\pi i/3} f(z)+\Res{z=-2\pi i/3} f(z)+\Res{z=4\pi i/3}f(z)+\Res{z=-4\pi i/3}f(z)\right] \, . $$

So we just need to compute these residues! Letting $\omega=e^{2\pi i/3}$ be a primitive cube root of unity and letting $k$ be a nonzero integer (so that $f$ has a simple pole at $\dfrac{2\pi i k}{3}$), we have:

$$ \begin{align*} \Res{z=2\pi i k / 3} f(z)&=\lim_{z \to 2\pi i k / 3} \left(z-\frac{2\pi i k}{3}\right)\frac{z^2e^z}{e^{3z}-1}\\ &=\left(\frac{2\pi i k}{3}\right)^2\omega^k\lim_{z \to 2\pi i k/3} \frac{z-2 \pi i k/3}{e^{3z}-1}&&\\ &=\left(\frac{2\pi i k}{3}\right)^2\omega^k\left(\frac{1}{3}\right)&&\text{(by L'Hôpital)}\\ &=-\frac{4}{27}\pi^2 k^2 \omega^k \, . \end{align*} $$

Now, the residues in the expression for the integral can be combined in pairs symmetric about $z=0$:

\begin{align*} \Res{z=2\pi i} f(z)+\Res{z=-2\pi i}f(z)&=-\frac{4}{27}\pi^2 \cdot 9\omega^3-\frac{4}{27}\pi^2 \cdot 9\omega^3\\ &=-\frac{8}{3}\pi^2&&\text{(}\omega\text{ is a cube root of unity)}\\ \Res{z=2\pi i/3} f(z)+\Res{z=-2\pi i/3} f(z)&=-\frac{4}{27}\pi^2\omega-\frac{4}{27}\pi^2\omega^{-1}\\ &=-\frac{4}{27}\pi^2(\omega+\omega^{-1})\\ &=\frac{4}{27}\pi^2&&\text{(}\omega+\omega^{-1}=-1\text{)}\\ \Res{z=4\pi i/3} f(z)+\Res{z=-4\pi i/3} f(z)&=-\frac{4}{27}\pi^2\cdot 4\omega^2-\frac{4}{27}\pi^2\cdot 4\omega^{-2}\\ &=-\frac{16}{27}\pi^2(\omega^{-1}+\omega)&&\text{(}\omega^2=\omega^{-1}\text{)}\\ &=\frac{16}{27}\pi^2 \, . \end{align*}

Putting all this together, we have

\begin{align*} \int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1} \, du &= -\frac{1}{8}\left(-\frac{8}{3}\pi^2\right)-\frac{1}{4}\left(\frac{4}{27}\pi^2\right)-\frac{1}{4}\left(\frac{16}{27}\pi^2\right)\\ &=\frac{1}{3}\pi^2-\frac{1}{27}\pi^2-\frac{4}{27}\pi^2\\ &=\frac{4}{27}\pi^2 \end{align*}

as desired.

Answer 5

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!

Answer 6

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 $J_{2,3}=\frac{4\pi^2}{27}$

Answer 7

Let $\displaystyle I = \int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x= \int_0^1\frac{\ln x}{x^3-1}\mathrm{d}x + \int_1^\infty\frac{\ln x}{x^3-1}\mathrm{d}x.$

Let $\displaystyle x \mapsto \frac{1}{x}$ in the last integral then:

$\displaystyle I = \int_0^1\frac{\ln x}{x^3-1}\mathrm{d}x - \int_0^1\frac{\ln \frac{1}{x}}{1/x^3-1} \cdot \frac{1}{x^2}\mathrm{d}x =\int_0^1\frac{\ln x}{x^3-1}\mathrm{d}x - \int_0^1\frac{\ln x}{\frac{1}{x}-x^2} \mathrm{d}x$

$\displaystyle I =\int_0^1\frac{\ln x}{x^3-1}\mathrm{d}x - \int_0^1\frac{\ln x}{\frac{1}{x}-x^2} \,\mathrm{d}x = \int_0^1\frac{(x+1)\ln x}{x^3-1}\,\mathrm{d}x$.

Consider the geometric series expansion:

$\displaystyle \frac{(x+1)\ln{x}}{x^3-1} = (x+1) \ln{x} \sum_{k \ge 0} x^{3k} = \sum_{k \ge 0} x^{3k+1}\ln{x}+\sum_{k \ge 0} x^{3k}\ln{x}$

Since $x \in (0, 1)$ we can write $I = \displaystyle \sum_{k \ge 0} \int_0^1 x^{3k+1}\ln{x}\,\mathrm{d}x+\sum_{k \ge 0} \int_0^1 x^{3k}\ln{x}\,\mathrm{d}x$

Consider $\displaystyle f(m) = \int_0^1 x^m \,{dx} = \frac{1}{1+m}.$ Then $\displaystyle f^{(n)}(m) = \int_0^1 x^m \ln^{n}{x} \,{dx} = \frac{(-1)^n n! }{(1+m)^{n+1}}.$

Applying this result with $n=1, ~m=3k, ~m=3k+1$ we have

$\displaystyle I = \sum_{k \ge 0} \frac{1}{(2 + 3 k)^2}+\sum_{k \ge 0} \frac{1}{(1 + 3 k)^2} := S_1 + S_2 $; separating odd/even terms:

$$\displaystyle S_1 = \sum_{k \ge 0} \frac{1}{(2 + 3 k)^2}= \sum_{k \ge 0} \frac{1}{(2 + 6 k)^2}+\sum_{k \ge 0} \frac{1}{(5 + 6 k)^2} \\ $$

$$\displaystyle S_2 = \sum_{k \ge 0} \frac{1}{(1 + 3 k)^2}= \sum_{k \ge 0} \frac{1}{(1+6k)^2}+\sum_{k \ge 0} \frac{1}{(4 + 6 k)^2} \\ $$

Therefore $\displaystyle I =\sum_{k \ge 0} \frac{1}{(2+ 6 k)^2}+\sum_{k \ge 0} \frac{1}{(5 + 6 k)^2}+\sum_{k \ge 0} \frac{1}{(1 + 6 k)^2}+\sum_{k \ge 0} \frac{1}{(4 + 6 k)^2}$

Let $\displaystyle \mu(m) = \sum_{k \ge 0} \frac{1}{(m+6k)^2}.$ Then $\displaystyle \zeta(2) = \sum_{k \ge 0} \frac{1}{(1+k)^2} $ can be rewritten as $\displaystyle \sum_{1 \le m\le 6}\mu(m)$.

But we seek this sum, with terms for $m=3,6$ removed, which tells us that

$$\begin{aligned} \displaystyle I & = \sum_{k \ge 0}\frac{1}{(k+1)^2}-\sum_{k \ge 0}\frac{1}{(6k+3)^2} -\sum_{k \ge 0}\frac{1}{(6k+6)^2} \\& = \left(1-\frac{1}{36}\right)\sum_{k \ge 0}\frac{1}{(k+1)^2}-\frac{1}{9}\sum_{k \ge 0}\frac{1}{(2k+1)^2} \end{aligned}$$

But $\displaystyle \sum_{k \ge 0} \frac{1}{(2k+1)^2} = \sum_{k \ge 0} \frac{1}{(k+1)^2}- \sum_{k \ge 0} \frac{1}{(2k+2)} = \sum_{k \ge 0} \frac{1}{(k+1)^2}- \frac{1}{4}\sum_{k \ge 0} \frac{1}{(k+1)^2}$

Putting this altogether $\displaystyle I = \left(1-\frac{1}{36}\right)\zeta(2)-\frac{1}{9}\cdot \left(\zeta(2)-\frac{1}{4} \zeta(2)\right) = \frac{8}{9} \zeta(2).$

Therefore $$I = \frac{4\pi^2}{27}.$$

Answer 8

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 $$I=\frac{1}{9} \pi^2 \csc^2(\pi/3)= \frac{4 \pi^2}{27}.$$

Answer 9

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 $$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$