Issue when solving $\int_{0}^{\infty}\frac{\ln\left(x\right)x}{\left(x^{3}-1\right)}dx$ via keyhole contour.

Question

I wanted to prove that $$I=\int_{0}^{\infty}\frac{\log\left(x\right)x}{\left(x^{3}-1\right)}dx=\frac{4 \pi^{2}}{27},$$ using a contour integral of $$f(z)=\int_{0}^{\infty}\frac{\log\left(z\right)^{2}z}{\left(z^{3}-1\right)}dz,$$ with log branch cut along the positive, real axis. let C be a positively oriented contour made up of $C_r,C_R,C_0,C_1$ After showing $\oint_{C_r}$ and $\oint_{C_R}$ approach $0$ when their radius approaches $0$ and $\infty$, we have$$\oint_C f(z) dz = 2 \pi i \sum_{n=0}^1\operatorname{Res}( f(z),a_n) = \oint_{C_0}f(z)dz+\oint_{C_1}f(z)dz,$$ by the residual theorem, where $a_0$ and $a_1$ are the non-real cube roots of $1$. Since $C_0$ and $C_1$ become paths of opposite direction infinity close to the real axis, their difference becomes the branch cut of $2\pi i$.$$=\int_{0}^{\infty}\frac{\log\left(z\right)^{2}z}{\left(z^{3}-1\right)}dz+\int_{\infty}^{0}\frac{(\log\left(z\right)+2 \pi i)^{2}z}{\left(z^{3}-1\right)}dz.$$ multiplying out the squared term and changing bounds give $$2 \pi i\sum_{n=0}^1\operatorname{Res}( f(z),a_n) =-4\pi i \int_{0}^{\infty}\frac{\log\left(z\right)z}{\left(z^{3}-1\right)}dz+4 \pi ^2\int_{0}^{\infty}\frac{z}{\left(z^{3}-1\right)}dz.$$ The residuals and rational integral come out to be $$ \frac{2\pi^2}{27} + i\frac{2\pi^2}{9 \sqrt{3}}, \frac{8\pi^2}{27} - i\frac{8\pi^2}{9 \sqrt{3}},$$ $$2 \pi i\sum_{n=0}^1\operatorname{Res}( f(z),a_n) =\frac{4 \pi^3}{3\sqrt{3}}+i \frac{20\pi^3}{27},$$$$P.V.\int_{0}^{\infty}\frac{z}{\left(z^{3}-1\right)}dz = \frac{\pi}{3 \sqrt{3}}.$$

However, using this information to solve for the original integral, we get that $$ \frac{4 \pi^3}{3\sqrt{3}}+i \frac{20\pi^3}{27}= -4 \pi i I +\frac{4\pi^3}{3 \sqrt{3}} $$$$I=-\frac{5 \pi^{2}}{27},$$ which is clearly incorrect. My question is, where did I make a mistake and what are some other methods to solve this integral?

Answer 1

During the last week-end i tried to see where the OP argument went wrong, but now i could find the time to draw and type. This is a detailed answer, but the reader in hurry may take a look to the inserted picture, then skip all computations (that double OP computations) and to straightforward to [The difference] section.

---

---

To have a clear situation, consider the following closed contour $C= C(0) \sqcup C(1) \sqcup C(2) \sqcup C(3) \sqcup C(4) \sqcup C(3') \sqcup C(2') \sqcup C(1') \sqcup $, it has many parts, and a prime suggest a reflection w.r.t. $Ox$-axis. Consider the following variables.

• $\varepsilon>0$ ("tiny" i.e. later we take first the limit $\varepsilon\searrow 0$),
• $r>\varepsilon>0$ ("small" i.e. $r<1/2$, later we take the limit $r\searrow 0$),
• $R>0$ ("big", i.e. $R>2$, and we will go with $A\nearrow\infty$).

In the picture then the segments parallel to the $Ox$-axis have distance $\varepsilon$ to it. The paths used are denoted by $C(j)$, where $j$ is a label among $0,1,2,3,4,3',2',1'$. The arcs of circles centered in $0,1$ correspond to circles with radius $r$ for indices $j\in\{0,2,2'\}$, and $R$ for $j=4$.

Let $\ln$ be the usual branch of the logarithm function which is real on the positive real ray $\Bbb R_{>0}$, and it is defined on $\Bbb C$ with a cut on the closed ray $\Bbb R_{\le 0}$.

Let $1,a,a^2$ be the roots of $Z^3-1$, $a=(-1\pm i\sqrt 3)/2$.

Let our function of interest be $f(z) = z\; \log^2 z\; /\; (z^3-1)$.

Here, $\log$ is the branch of the logarithm function which is defined inside the drawn closed contour $C$, and coincides with $\ln$ in their common part of the first quadrant, where they are both defined. Then on $C(j)$, $j=1,2,3$ we have $\log=\ln$, and on the reflected parts $C(j')$ we have $\log = \ln + 2\pi i$.

The integral of $f$ on $C(j)$ is denoted by $J(j)$.

The Cauchy PV-integral of $(z\ln z)/(z^3-1)$ on $(0,\infty)$ is denoted by $J$. This is the integral of interest.

We can already pass to the limit $\varepsilon\searrow 0$, also in the drawing, but take care which is $\log$ (either $\ln$, or $\ln + 2\pi i$) when integrating. Note that we can focus on the real part of the integrals.

Then we have the relations: $$ \begin{aligned} 2\pi i\sum_{b=a,\,a^2 }\underset{z=b}{\operatorname{Res}} \frac {z\;\log^2 z}{z^3-1} & = \sum_j \int_{C(j)}\frac {z\;\log^2 z}{z^3-1}\; dz \\ &= J(0) + (J(1)+J(3)) + (J(1')+J(3')) + (J(2)+J(2')) + J(4) \ . \\[4mm] &\qquad\text{Consider now each piece in part:} \\[4mm] \underset{z=b}{\operatorname{Res}} \frac {z\;\log^2 z}{z^3-1} &= \lim_{z\to b}(z-b)\cdot \frac {z\;\log^2 z}{(z-1)(z-b)(z-b^2)} =\frac {b\;\log^2 b}{(b-1)(b-b^2)} \ , \\ \sum_{b=a,\,a^2 }\underset{z=b}{\operatorname{Res}} f(z) &= \frac {a\;\log^2 a}{(a-1)(a-a^2)} + \frac {a^2\;\log^2 a^2}{(a^2-1)(a^2-a)} \\ &= \left(\frac{2\pi i}3\right)^2\cdot \frac 1{(a-1)(1-a)} + \left(\frac{4\pi i}3\right)^2\cdot \frac 1{(a^2-1)(1-a^2)} \\ &= \left(\frac{2\pi}3\right)^2\cdot \frac 1{(1-a)^2} \left(1 + \frac 4{(a+1)^2}\right) = \left(\frac{2\pi}3\right)^2\cdot \left(\frac 13-a\right) \ . \\[2mm] 2\pi i\sum_{b=a,\,a^2 }\underset{z=b}{\operatorname{Res}} f(z) &= 2\pi i \left(\frac{2\pi}3\right)^2\cdot \left(\frac 13-a\right) =2\pi \left(\frac{2\pi}3\right)^2\cdot \left(\frac 56i+\frac {\sqrt3}2 \right) \\ &=\frac {4\sqrt 3}{9}\pi^3 + i\frac {20}{27}\pi^3\ , \\ &\qquad\text{ in agreement with OP residue computations.} \\[4mm] \left|\int_{C(0)}f(z)\; dz\right| &\le \int_{C(0)}\frac {|z|\;|\log z|^2}{|1-z^3|}\; dz \le \operatorname{length}(C(0))\frac {r\;(\ln r + 2\pi)^2}{1-\frac 18} \\ &\le 2\pi r\cdot r\;(\ln r + 2\pi)^2 \cdot \frac 87\to 0\ , \\[2mm] \left|\int_{C(4)}f(z)\; dz\right| &\le \int_{C(4)}\frac {|z|\;|\log z|^2}{|z^3-1|}\; dz \le \operatorname{length}(C(4))\frac {R\;(\ln R + 2\pi)^2}{R^3-1} \\ &\le 2\pi R\cdot \frac R{R^3-1}\;(\ln R + 2\pi)^2 \to 0\ , \\[2mm] J(1)+J(3)+J(1')+J(3') &\to \left(\int_r^{1-r}+\int_{1+r}^R\right) \frac {z\;\ln^2 z}{z^3-1}\; dz - \left(\int_r^{1-r}+\int_{1+r}^R\right) \frac {z\;(\ln z + 2\pi i)^2}{z^3-1}\; dz \\ &= \left(\int_r^{1-r}+\int_{1+r}^R\right) \frac {z\;(2\pi)^2}{z^3-1}\; dz - \left(\int_r^{1-r}+\int_{1+r}^R\right) \frac {z\cdot 2\;\ln z(2\pi i)}{z^3-1}\; dz \\ &\to (2\pi)^2\cdot\operatorname{PV}\int_0^\infty \frac z{z^3-1}\; dz - 4\pi i \underbrace{\operatorname{PV}\int_0^\infty \frac {z\;\ln z}{z^3-1}\; dz}_{=J} \\ &= (2\pi)^2\cdot\lim_{r\nearrow 1} \left( \int_0^r\frac z{z^3-1}\; dz + \int_{1/r}^\infty\frac z{z^3-1}\; dz\right) -4\pi i\; J \\ &= (2\pi)^2 \int_0^1\frac {z-1}{z^3-1}\; dz -4\pi i\; J = (2\pi)^2\cdot \frac {\pi\sqrt 3}9 -4\pi i\; J\ , \\ &\qquad\text{ in agreement with OP P.V. computations.} \end{aligned} $$

---

---

The difference: So far everything went on the lines of the question. Where is the error in the computation?

The problem is with the integral parts on the $C(2)$ and $C(2')$, the limiting process must be taken in detail, so that a Cauchy-P.V. part and the "remainder" are explicitly exposed. Let $C(1, r)$ be the circle contour around one, radius $r$, it is (in the limit) $C(2)\sqcup C(2')$, but with opposite orientation. Then we have: $$ \begin{aligned} J(2) &= \int_{C(2)}\frac {z\;\log^2z}{z^3-1} = -\int_{\displaystyle\substack{z = 1+re^{it}\\t\in[0,\pi]}} \frac {z\;\ln^2z}{z^3-1}\; dz \\ J(2') &= \int_{C(2')}\frac {z\;\log^2z}{z^3-1} = -\int_{\displaystyle\substack{z = 1+re^{it}\\t\in[\pi,\,2\pi]}} \frac {z\;(\ln z+2\pi i)^2}{z^3-1}\; dz \\[2mm] J(2)+J(2') &= -\int_{C(1,r)} \underbrace{\frac {z\;\ln^2z}{z^3-1}}_{\text{zero residue}}\; dz -\int_\pi^{2\pi} \frac {(1+re^{it})\;(2\pi i)^2}{(1+re^{it})^3-1}\; ir\; e^{it}\; dt \\ &= (2\pi)^2\int_\pi^{2\pi} \frac {1+re^{it}}{r^2e^{2it} + 3 re^{it} + 3}\; i\; dt \\ &\to (2\pi)^2\int_\pi^{2\pi} \frac {1+0}{0 + 0 + 3}\; i\; dt \qquad\text{ for $r\to 0$} \\ &=i\frac 43\pi^3\ . \end{aligned} $$

---

We can now put all together. In the limit we obtain from $$ \begin{aligned} 2\pi i\sum_{b=a,\,a^2 }\underset{z=b}{\operatorname{Res}} \frac {z\;\log^2 z}{z^3-1} & = \sum_j \int_{C(j)}\frac {z\;\log^2 z}{z^3-1}\; dz \\ &= J(0) + \Big((J(1)+J(3)) + (J(1')+J(3'))\Big) + (J(2)+J(2')) + J(4) \\[4mm] &\text{ the following relation:} \\[4mm] \frac {4\sqrt 3}{9}\pi^3 + i\frac {20}{27}\pi^3 &= 0 + \left(\frac {4\sqrt 3}9\pi^3 -4\pi i\; J\right) + i\frac 43\pi^3 + 0\ . \end{aligned} $$ We have a perfect match of the real part. And for the imaginary part we get: $$ 4\pi i\;\frac 5{27}\pi^2 =-4\pi i\; J +4\pi i\;\frac 13\pi^2\ . $$ This gives: $$ J = \left(\frac 13 - \frac 5{27}\right)\pi^2 = \frac {9-5}{27}\pi^2 = \frac 4{27}\pi^2 \ , $$ as expected and wanted.

---

Note: The answer could have been written in a more spartan manner, but in this case all giving details may be a good benefit for the community.

Answer 2

Alternatively, we can use the positively oriented boundary of the sector $0\leq r\leq R$, $0\leq\theta\leq\frac{2\pi}3$ with arc jumps at $z=0,1, e^{\frac{2\pi i}3}$ and $f(z)=\frac{z\ln z}{z^3-1}$.

Let $I=\int_0^\infty f(x)dx.$ In the limit, the integral of $f(z)$ along the arcs at $z=0,1$ and the big arc tend to zero. So, by Residue theorem, we get $$(1-e^{\frac{4\pi i}3})I-\frac{2\pi i}3e^{\frac{4\pi i}3}\int_0^\infty\frac x{x^3-1}dx\\ =\pi i\operatorname{Res}_{z=e^{\frac{2\pi i}3}}f(z)=\pi i\frac{2\pi i}{9e^{\frac{2\pi i}3}}$$ and hence $I=\frac{4\pi^2}{27}$.

Answer 3

Use the substitution $u = 1/x$, which implies $x = 1/u$ and $dx = -1/u^2 du$. Change the limits of integration:

• As $x \to 0^+$, $u \to \infty$
• As $x \to \infty$, $u \to 0$

Substituting these into the integral, we get: $$I = \int_{\infty}^{0} \frac{\frac{1}{u} \ln\left(\frac{1}{u}\right)}{\left(\frac{1}{u}\right)^3 - 1} \left(-\frac{1}{u^2}\right) du$$ Simplify the expression inside the integral. $$I = \int_{0}^{\infty} \frac{\frac{1}{u} (-\ln u)}{\frac{1-u^3}{u^3}} \left(\frac{1}{u^2}\right) du$$

$$I = \int_{0}^{\infty} \frac{-\ln u}{u} \cdot \frac{u^3}{1-u^3} \cdot \frac{1}{u^2} du$$ $$I = \int_{0}^{\infty} \frac{-\ln u}{1-u^3} du = \int_{0}^{\infty} \frac{\ln u}{u^3-1} du$$ By renaming the variable of integration from $u$ back to $x$, we have: $$I = \int_{0}^{\infty}\frac{\ln\left(x\right)}{\left(x^{3}-1\right)}dx$$

This integral can be easily evaluated, yet it is still posted here: Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$.

Answer 4

An alternative approach is

$$ \int_{0}^{1}\frac{x\log(x)}{x^3+1}\,dx +\int_{1}^{+\infty}\frac{x\log(x)}{x^3+1}\,dx = \int_{0}^{1}\frac{x\log(x)}{x^3+1}\,dx -\int_{0}^{1}\frac{\log(x)}{x^3+1}\,dx = \int_{0}^{1}\frac{x-1}{x^3+1}\log(x)\,dx $$

leading to

$$ I = \int_{0}^{1}\frac{-1+x+x^3-x^4}{1-x^6}\log(x)\,dx = \sum_{k\geq 0}\int_{0}^{1}(x^{6k}-x^{6k+1}-x^{6k+3}+x^{6k+4})(-\log x)\,dx$$

then to

$$ I = \sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}-\frac{1}{(6k+2)^2}-\frac{1}{(6k+4)^2}+\frac{1}{(6k+5)^2}\right)=\sum_{n\geq 1}\frac{\chi_6(n)}{n^2}-\frac{1}{4}\sum_{n\geq 1}\frac{\chi_3(n)}{n^2}$$

where $\chi_6$ is the principal character $\!\!\pmod{6}$ and $\chi_3$ is the principal character $\!\!\pmod{3}$.

By the reflection formula for the trigamma function, or just by the Euler product for by $L(\chi_6,s)$ and $L(\chi_3,s)$, we have $L(\chi_6,2)=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\zeta(2)=\frac{\pi^2}{9}$ and $L(\chi_3,2)=\left(1-\frac{1}{9}\right)\zeta(2)=\frac{4\pi^2}{27}$, immediately leading to the wanted outcome.