So I want to compute $ I = \int _0^{\infty }\frac{\sqrt[3]{x}}{x^2+1\:}\:dx$.

First thing I thought of is that integrals from $0$ to $\infty$ usually take a nicer form when we apply the change $x = e^t$, so I did that: $$ I = \int _{-\infty }^{\infty }\frac{e^{\frac{4}{3}t}}{e^{2t}+1\:}\:dt $$

Now, I've seen some other integrals with a form like this one. For the other ones, I computed the contour integral of a rectangle of vertex $[-R, R, R+xi, -R+xi]$, choosing $x$ such that the integrand takes opposite values in $[-R, R]$ and $[-R+xi, R+xi]$.

If I could choose such an $x$, then I would have succeeded, since I can prove that the lateral integrals go to $0$ as $R\to \infty$, and thus the desired integrals would be equal to half the value of the residues enclosed by the rectangle.

But that requires solving the following system: $$ \begin{cases} e^{\frac{4}{3}t} = -e^{\frac{4}{3}(t+xi)} \\ e^{2t} = e^{2(t+xi)} \end{cases} \implies \begin{cases} \frac{4}{3}xi+i\pi = 2ki\pi \\ 2xi = 2ki\pi \end{cases} \implies \begin{cases} x = \frac{3}{4}(2k+1)\pi \\ x = k\pi \end{cases} $$ For $k\in \mathbb{Z}$.

But this system does not have a solution!

The first equation forces $x$ to be $q\pi$ for some $q\not\in\mathbb{Z}$, contradicting the second equation.

From here I am stuck. Can I get a hint?

EDIT: I followed @DanielFischer suggestion to use a keyhole contour.

Let $C$ be the keyhole contour formed by a big circle $\Gamma_R$ of radii $R$, a small circle $\gamma_\epsilon$ of radii $\epsilon$ and two segments connecting the two circles surrounding the positive axis, separated by a $\delta$ margin.


Then we have thanks to the estimation lemma that: $$ |\int_{\Gamma_R}|\le {\sup}_{z\in{\Gamma_R}}{\frac{\sqrt[3]{z}}{z^2+1\:}}\cdot long(\Gamma_R)\sim \frac{R^{1/3}}{R^2}\cdot 2\pi R \to 0 $$

On the other hand, when $\delta \to 0$: $$ \int_R^\epsilon\frac{\sqrt[3]{z}}{z^2+1\:} = \int_R^\epsilon\frac{e^{\log{z}/3}}{z^2+1\:} = \int_R^\epsilon\frac{e^{\frac{\log{|z| + i\arg{z}}}{3}}}{z^2+1\:} = -e^{-2\pi i/3}\int_\epsilon^R\frac{e^{\frac{\log{|z| + i\arg{z}+2\pi i}}{3}}}{z^2+1\:} = -e^{-2\pi i/3}\int_\epsilon^R $$

Thus in the limit: $$ \int_C = \int_\Gamma + \int_\gamma + \int_\epsilon^R + \int_R^\epsilon = \int_0^\infty -e^{-2\pi i/3}\int_0^\infty =\\ =(1 -e^{-2\pi i/3})\int_0^\infty = 2\pi i (Res(i)+Res(-i)) $$

The residues can be easily calculated as: $$ Res(i) = \frac{\sqrt[3]{i}}{2i\:}\\ Res(-i) = \frac{\sqrt[3]{-i}}{-2i\:} $$

Thus $\int_0^\infty = \frac{2\pi i (\frac{\sqrt[3]{i}}{2i\:}+\frac{\sqrt[3]{-i}}{-2i\:})}{(1-e^{-2\pi i/3})} = \frac{π}{2 \sqrt{3}} + \frac{i π}{2}$... which is not real as it should be.

Along the way I've also assumed that $\int_{\gamma_\epsilon}\to 0$, which seems like the case but I cannot prove it.

Therefore, I ask:

  1. How do I prove that $\int_{\gamma_\epsilon}\to 0$?
  2. Why is my result wrong?

Turns out the minus sign on the exponent was wrong: $$I = \frac{i \left(e^{\frac{i \pi }{3}}-e^{-\frac{2 i \pi }{3}} \right) \pi }{1-e^{\frac{2 i \pi }{3}}} = -2\pi / \sqrt{3}$$

This result is still wrong according to the almighty Wolfram Alpha, but at least it has a similar form! I'll keep debugging.

  • $\begingroup$ The standard way to evaluate the first integral - when using the residue theorem - is to use a keyhole contour, since $\sqrt[3]{z}$ has a branch point at $0$. $\endgroup$ – Daniel Fischer Jan 16 '17 at 10:41
  • $\begingroup$ the beta function comes in here very handy+ $\endgroup$ – tired Jan 16 '17 at 10:42
  • $\begingroup$ @DanielFischer What is a keyhole contour? What would be such a contour in this case? $\endgroup$ – Jsevillamol Jan 16 '17 at 10:49
  • 1
    $\begingroup$ @Jsevillamol en.wikipedia.org/wiki/… $\endgroup$ – b00n heT Jan 16 '17 at 10:51
  • $\begingroup$ if you substituted x = tan u, the integral becomes cube root of tan, I'm not saying to do that, but I recall this method for the cube root of tan, that you could easily adapt for your integral answers.yahoo.com/question/index?qid=20090504221503AAUg701 $\endgroup$ – Cato Jan 16 '17 at 10:52

So, after the change of variable $x=e^t$, you have to compute the integral

$$I=\int_{-\infty}^\infty\frac{e^{4/3 t}}{e^{2t}+1}\,dt.$$

Consider the contour $$\Gamma:=\underbrace{[-R,R]}_{\gamma_1}\cup\underbrace{[R,R+i\pi]}_{\gamma_2} \cup\underbrace{[R+i\pi,-R+i\pi]}_{\gamma_3}\cup \underbrace{[-R+i\pi,-R]}_{\gamma_4},$$ and note that $f(z):=\frac{e^{4/3 z}}{e^{2z}+1}$ has only a simple pole at $z=i\pi/2$ inside $\Gamma$.

It is easy to see that the integral of $f$ along $\gamma_2$ and $\gamma_4$ goes to zero, so taking the limit $R\to\infty$ we have

$$I+\lim_{R\to\infty}\int_{\gamma_3}f(z)\,dz=2\pi i\;\text{res}(f,z=i\pi)=-i\pi e^{2i\pi/3}.$$



so finally

$$I-e^{4/3i\pi}I=-i\pi e^{2i\pi/3}\implies I=\frac12\frac{\pi}{\sin(2\pi/3)}=\frac{\pi}{\sqrt{3}}.$$

  • $\begingroup$ How have you chosen the height of the contour rectangle? $\endgroup$ – Jsevillamol Jan 16 '17 at 17:56
  • 1
    $\begingroup$ I wanted the denominator to remain the same, so $i\pi$, $2i\pi$, etc. are possible choices. But while the corresponding contour for $i\pi$ contains only one pole, the others contain more than one, so if you want to avoid unnecessary calculations you should choose $i\pi$. $\endgroup$ – iqcd Jan 16 '17 at 18:21

Let $\sqrt[3]{x}=\sqrt{u}$, then $x^2=u^{3} \implies 2x\, dx=3u^2\, du \implies dx=\dfrac{3}{2} \sqrt{u}\, du$

\begin{align*} \int_{0}^{\infty} \frac{\sqrt[3]{x} \, dx}{x^2+1} &= \int_{0}^{\infty} \frac{3u\, du}{2(u^3+1)} \\ &= \int_{0}^{\infty} \left[ \frac{u+1}{2(u^2-u+1)}-\frac{1}{2(u+1)} \right] \, du \\ &= \int_{0}^{\infty} \left[ \frac{(2u-1)+3}{4(u^2-u+1)}-\frac{1}{2(u+1)} \right] \, du \\ &= \int_{0}^{\infty} \left[ \frac{2u-1}{4(u^2-u+1)}+ \frac{3}{(2u-1)^2+3}-\frac{1}{2(u+1)} \right] \, du \\ &= \left[ \frac{1}{4} \ln (u^2-u+1)+ \frac{\sqrt{3}}{2} \tan^{-1} \frac{2u-1}{\sqrt{3}}- \frac{1}{2} \ln (u+1) \right]_{0}^{\infty} \\ &= \left[ \frac{1}{4} \ln \frac{u^2-u+1}{(u+1)^2}+ \frac{\sqrt{3}}{2} \tan^{-1} \frac{2u-1}{\sqrt{3}} \right]_{0}^{\infty} \\ &= \frac{\pi}{\sqrt{3}} \end{align*}

By contour integral \begin{align*} \oint_{C} \frac{\sqrt[3]{x} \, dx}{x^2+1} &= \int_{\epsilon}^{R} \frac{\sqrt[3]{x} \, dx}{x^2+1}+ \int_{0}^{2\pi} \frac{\sqrt[3]{R}e^{i\theta/3} Rie^{\theta}\, d\theta} {R^2e^{2i\theta}+1} \\ &\quad -\int_{R}^{\epsilon} \frac{\sqrt[3]{x}e^{2i\pi/3} \, dx}{x^2+1}- \int_{0}^{2\pi} \frac{\sqrt[3]{\epsilon}e^{i\theta/3} \epsilon ie^{\theta}\, d\theta} {\epsilon^2e^{2i\theta}+1} \\ &= \left( 1-e^{2i\pi/3} \right) \int_{\epsilon}^{R} \frac{\sqrt[3]{x} \, dx}{x^2+1}+ \frac{i}{\sqrt[3]{R^2}} \int_{0}^{2\pi} \frac{e^{-2i\theta/3} \, d\theta} {1+\dfrac{e^{-2i\theta}}{R^2}}- i\sqrt[3]{\epsilon^{4}} \int_{0}^{2\pi} \frac{e^{4i\theta/3}\, d\theta} {1+\epsilon^2e^{2i\theta}} \\ &= \left( 1-e^{2i\pi/3} \right) \int_{0}^{\infty} \frac{\sqrt[3]{x} \, dx}{x^2+1} \\ \int_{0}^{\infty} \frac{\sqrt[3]{x} \, dx}{x^2+1} &= \frac{2\pi i}{1-e^{2i\pi/3}} \left[ \operatorname{Res} \left( \frac{\sqrt[3]{x}}{x^2+1}, i \right)+ \operatorname{Res} \left( \frac{\sqrt[3]{x}}{x^2+1}, -i \right) \right] \\ &= \frac{2\pi i}{1-e^{2i\pi/3}} \left( \frac{\sqrt[3]{i}}{i+i}+\frac{\sqrt[3]{-i}}{-i-i} \right) \\ &= \frac{\pi(e^{i\pi/6}-e^{i\pi/2})}{1-e^{2i\pi/3}} \\ &= \frac{\pi(e^{-i\pi/6}-e^{i\pi/6})}{e^{-i\pi/3}-e^{i\pi/3}} \\ &= \frac{\pi \sin \dfrac{\pi}{6}}{\sin \dfrac{\pi}{3}} \\ &= \frac{\pi}{\sqrt{3}} \end{align*}

  • $\begingroup$ This one is great, but I was hoping for a solution based on contour techniques. $\endgroup$ – Jsevillamol Jan 16 '17 at 12:39
  • 1
    $\begingroup$ Answer updated! $\endgroup$ – Ng Chung Tak Jan 16 '17 at 18:06
  • $\begingroup$ Thank you! Can you also sketch to me how to bound the inner integral to show that it tends to zero? $\endgroup$ – Jsevillamol Jan 16 '17 at 18:14
  • 1
    $\begingroup$ Take $\epsilon \to 0$, $\dfrac{e^{4i\theta/3}}{1+\epsilon^{2} e^{2i\theta}} \to e^{4i\theta/3}$ and $\sqrt[3]{\epsilon^4} \to 0$, hence the inner contour integral tends to zero. $\endgroup$ – Ng Chung Tak Jan 16 '17 at 18:17

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.