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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.

Contour

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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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

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

Now,

$$\int_{\gamma_3}f(z)\,dz=-\int_{-R}^R\frac{e^{4/3(t+i\pi)}}{1+e^{2(t+i\pi)}}\,dt=-e^{4/3i\pi}\int_{\gamma_1}f(z)\,dz$$

so finally

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

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  • $\begingroup$ How have you chosen the height of the contour rectangle? $\endgroup$ – Jsevillamol Jan 16 '17 at 17:56
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    $\begingroup$ 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
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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*}

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  • $\begingroup$ This one is great, but I was hoping for a solution based on contour techniques. $\endgroup$ – Jsevillamol Jan 16 '17 at 12:39
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    $\begingroup$ 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
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    $\begingroup$ 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

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