7
$\begingroup$

Prove using contour integration that $\displaystyle \int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$

I am at a loss at how to start this problem and which contour to pick. I have been trying to get the sector with angle $2\pi/3$ to work with a bump around the pole at $e^{i2\pi/3}$ and the origin, but I am getting 5 or 6 different integrals and it is not really getting me anywhere.

$\endgroup$
  • $\begingroup$ I don't have much to contribute, but this is more or less the way I would approach it as well. Logarithms are pretty fugly in contours and you often end up having to do several pieces. Your professor is pretty tough to ask a question like this! Also don't forget the pole at $x=1$! $\endgroup$ – Cameron Williams Oct 23 '14 at 4:18
  • $\begingroup$ Maybe, do not bump at the pole and use the residue theorem? You'll get less integrals. $\endgroup$ – Aram Oct 23 '14 at 4:35
  • $\begingroup$ The pole at 1 is removable atleast. So that one we dont need to worry about. $\endgroup$ – anonymous Oct 23 '14 at 4:49
  • 1
    $\begingroup$ @Aram, anonymous: ignoring the pole at $z=1$ using a contour integration approach in this case may lead to an incorrect result. There is in fact a nonzero contribution from an infinitesimal bump under the positive axis in the neighborhood of $z=1$. $\endgroup$ – Ron Gordon Oct 24 '14 at 1:15
7
$\begingroup$

Actually, we do need to worry about the pole at $x=1$ if we intend to use contour integration, for reasons that are a bit subtle. I will demonstrate below.

The standard way to treat integrals of rational functions times logs over $[0,\infty)$ in complex analysis is to consider a keyhole contour, and an integral over that contour of the next higher power of log. In this case, the integral is

$$\oint_C dz \frac{\log^2{z}}{z^3-1}$$

$C$, however, is a modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. The modification lies on small semicircular bumps above and below $z=1$ of radius $\epsilon$, and we will consider the limits as $\epsilon \to 0$ and $R\to\infty$.

Let's evaluate this integral over the contours. There are $8$ pieces to evaluate, as follows:

$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log^2{x}}{x^3-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1+\epsilon}^R dx \frac{\log^2{x}}{x^3-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^2{\left (R e^{i \theta}\right )}}{R^3 e^{i 3 \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^3-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(\log{\left (1+\epsilon e^{i \phi}\right )}+i 2 \pi)^2}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^3-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (\epsilon e^{i \phi}\right )}}{\epsilon^3 e^{i 3 \phi}-1} $$

(To see this, draw the contour out, including the bumps about $z=1$.)

As $R \to \infty$, the fourth integral vanishes as $\log^2{R}/R^2$. As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^3)$, while the eighth integral vanishes as $\epsilon \log^2{\epsilon}$. This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become

$$PV \int_0^{\infty} dx \frac{\log^2{x} - (\log{x}+i 2 \pi)^2}{x^3-1} + i \frac{4 \pi^3}{3}$$

EDIT

It should be appreciated that, in the fifth, sixth, and seventh integrals, the $i 2 \pi $ factor appears because, on the lower branch of the real axis, we write $z=x \, e^{i 2 \pi}$. In the sixth integral, in fact, $z = e^{i 2 \pi} + \epsilon \, e^{i \phi + 2 \pi}$.

END EDIT

The $PV$ denotes the Cauchy principal value of the integral. As it stands, the integral does not actually converge. Nevertheless, we are not actually considering the integral straight through the pole at $z=1$, but a very small detour around the pole. Thus, in the limit, we get the Cauchy PV. A little rearranging cancels the $\log^2$ term, and we now have two integrals to evaluate:

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^3-1} + 4 \pi^2 PV \int_0^{\infty} \frac{dx}{x^3-1} + i \frac{4 \pi^3}{3}$$

Note we could remove the $PV$ on the first integral because the pole is a removable singularity.

The contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles. The poles here are at $z=e^{i 2 \pi/3}$ and $z=e^{i 4 \pi/3}$. Note that the pole at $z=1$ is not inside the contour $C$ because of the detour around that "pole". It should be appreciated that the poles must have their arguments between $[0,2 \pi]$ because of the way we defined $C$.

In any case, we now have that the above 1D integrals over the positive real line are equal to

$$i 2 \pi \left [\frac{-4 \pi^2/9}{3 e^{i 4 \pi/3}} + \frac{-16 \pi^2/9}{3 e^{i 8 \pi/3}} \right ] = -\frac{4 \pi ^3}{3 \sqrt{3}}+i \frac{20 \pi ^3}{27} $$

Equating real and imaginary parts, we find that

$$ \int_0^{\infty} dx \frac{\log{x}}{x^3-1} = \frac{4 \pi^2}{27} $$

$$ PV \int_0^{\infty} \frac{dx}{x^3-1} = -\frac{\pi}{3 \sqrt{3}} $$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The easiest and certainly most general way is to compute $$PV\int_0^{\infty} dx \frac{x^a}{1-x^b},$$ and then take the derivative w.r.t. $a$ of the result. To do the first integral, use a circular sector ('pizza slice contour').

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This isn't quite complex analytic, but first denote your as $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$

Consider the double integral:

$$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x}{(x^2+y^3)(1+x^2)}dydx.$$

We intend to evaluate $J$ and relate $J$ to $I.$
To evaluate, $J$ we integrate with respect to $y.$ You can proceed in two ways. The long way is to integrate by partial fractions, but the short way is to let $y=ux^\frac{2}{3}$ so that $dy=x^\frac{2}{3} du.$ Then we get: $$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x^\frac{5}{3}}{(x^2+x^2u^3)(1+x^2)}dudx=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x^\frac{-1}{3}}{(1+u^3)(1+x^2)}dudx.$$ Using the nice formula $$\int_{0}^{\infty} \frac{t^m}{1+t^n} dt=\frac{\pi}{n}\csc\left(\frac{\pi(m+1)}{n}\right),$$ we get $$J=\frac{2\pi^2}{9}.$$

Now use Fubini's Theorem on $J$ as such: $$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x}{(x^2+y^3)(1+x^2)}dxdy.$$

We will need partial fractions to integrate with respect to $x.$ Omitting the details of this computation, $$\frac{x}{(x^2+y^3)(1+x^2)}=\frac{1}{y^3-1} \left(\frac{x}{x^2+1}-\frac{x}{x^2+y^3}\right).$$ Now integrating with respect to $x$, and plugging in the endpoints, we get

$$J=\int_{0}^{\infty} \frac{1}{y^3-1}\lim_{x\rightarrow \infty} \left(\frac{\ln(x^2+1)}{2}-\frac{\ln(x^2+y^3)}{2}\right)-\frac{1}{y^3-1}\lim_{x\rightarrow 0} \left(\frac{\ln(x^2+1)}{2}-\frac{\ln(x^2+y^3)}{2}\right)dy.$$ We get $$J=\int_{0}^{\infty} \frac{\ln(y^3)}{2(y^3-1)}dy=\int_{0}^{\infty} \frac{3\ln(y)}{2(y^3-1)}dy=\frac{3}{2}I.$$ Thus we have $$I=\frac{2}{3}J=\frac{2}{3}\left(\frac{2\pi^2}{9}\right)=\frac{4\pi^2}{27}.$$

Addendum Consider $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$ Split the integral into two parts: $$\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx = \int_{0}^{1} \frac{\ln(x)}{x^3-1} dx+ \int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$ We can convert $\frac{\ln(x)}{x^3-1}$ into a geometric series as such (assuming $0<x<1.$) $$ \frac{\ln(x)}{x^3-1}=-\sum_{n=0}^{\infty}\ln(x)x^{3n}.$$ If we do integration by parts, $$\int_{0}^{1} \frac{\ln(x)}{x^3-1} dx= \int_{0}^{1} -\sum_{n=0}^{\infty}\ln(x)x^{3n}dx=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^2}.$$ For,$$\int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx,$$ perform a $u$ substitution $x=\frac{1}{u}, dx=\frac{-1}{u^2} du,$ and convert transformed integrand into a geometric series. You will see: $$\int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^2}=\sum_{n=-\infty}^{-1} \frac{1}{(3n+1)^2},$$ and the latter equality can be seen by simply writing out the terms of each summation. Now, add the two computed series together and see:

$$\sum_{n=-\infty}^{\infty} \frac{1}{(3n+1)^2}= \frac{4 \pi^2}{27.}$$

| cite | improve this answer | |
$\endgroup$

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.