I'm trying to evaluate the integral $\displaystyle\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$.

My book explains that to evaluate integrals of the form $\displaystyle\int_0^\infty x^\alpha R(x)dx$, with real $\alpha\in(0,1)$ and $R(x)$ a rational function, one first starts with a substitution $x=t^2$, to transform the integral to $$ 2\int_0^\infty t^{2\alpha+1} R(t^2) dt. $$ It then observes that $$ \int_{-\infty}^\infty z^{2\alpha+1}R(z^2)dz=\int_0^\infty(z^{2\alpha+1}+(-z)^{2\alpha+1})R(z^2)dz. $$ Since $(-z)^{2\alpha}=e^{2\pi i\alpha}z^{2\alpha}$, the integral then equals $$ (1-e^{2\pi i\alpha})\int_0^\infty z^{2\alpha+1}R(z^2)dz. $$ How are you suppose to apply the residue theorem to this integral if the integrand is not a rational function?

In my case, I have $$ \int_0^\infty x^{1/3}R(x)dx=2\int_0^\infty t^{5/3}R(t^2)dt. $$ Also $$ \int_0^\infty z^{5/3} R(z^2)dz=\frac{1}{1-e^{(2\pi i)/3}}\int_{-\infty}^{\infty}\frac{z^{5/3}}{1+z^4}dz. $$

I don't know what to do after that, since this last integrand still has a fractional power in the numerator. How does this work? Thanks.

(This is part (g) of #3 on page 161 of Ahlfors' Complex Analysis, part of some self-study.)


You have found, correctly, that $$\begin{eqnarray*} I &\equiv& \int_0^\infty\frac{x^{1/3}dx}{1+x^2} \\ &=& \frac{2}{1-e^{2\pi i/3}} \int_{-\infty}^\infty d z\, \frac{z^{5/3}}{1+z^4}. \end{eqnarray*}$$ The second integral has a branch cut at $z=0$ and singularities at the roots of $1+z^4$. Close the contour in the upper half-plane. Let $\gamma$ denote the contour. We will pick up residues at $e^{i\pi/4}$ and $e^{i 3\pi/4}$.

We deform the contour slightly near $z=0$ to avoid the cut. Letting $z=\epsilon e^{i\theta}$ we see that the contribution to the integral here goes like $\epsilon^{1+5/3} = \epsilon^{8/3}$ and so vanishes in the limit.

Letting $z = R e^{i\theta}$, we find the integral over the semicircle at infinity goes like $R^{1+5/3-4} = 1/R^{4/3}$. Thus, the contribution from this part of the contour is also zero. Therefore, $\int_{-\infty}^\infty d z\, z^{5/3}/(1+z^4) = \int_\gamma d z\, z^{5/3}/(1+z^4)$.

We find $$\begin{eqnarray*} I &=& \frac{2}{1-e^{2\pi i/3}} \int_\gamma d z\, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{2}{1-e^{2\pi i/3}} 2\pi i \sum\mathrm{Res} \, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{\pi}{\sqrt{3}} \end{eqnarray*}$$ where the residues are to be taken from the upper half-plane.

The residue of $f(z)$ at $z=z_0$ is just the coefficient of $(z-z_0)^{-1}$ in the Laurent series and, of course, $z^{5/3}/(1+z^4)$ has a Laurent series about the zeros of $1+z^4$. (It does not, however, have a Laurent series about $z=0$.)

A simpler solution

Recognize that the integral $I$ is already in the form $$\int_0^\infty d t\, t^\beta f(t^2).$$ But $$\begin{eqnarray*} \int_0^\infty d t\, t^\beta f(t^2) &=& \frac{1}{1+e^{\beta \pi i}} \int_{-\infty}^\infty d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} \int_\gamma d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} 2\pi i \sum \mathrm{Res} \, t^\beta f(t^2) \end{eqnarray*}$$ assuming the contributions from the contour around the branch cut at $z=0$ and the contour at infinity vanish. (Again, we have closed the contour in the upper half-plane and pick up only the residues residing there.) It is also important that the singularities of $f$ do not lie on the real line.

The contribution near $z=0$ goes like $\epsilon^{1+1/3} = \epsilon^{4/3}$. On the semicircle at infinity the integral goes like $R^{1+1/3-2} = R^{-2/3}$. Thus, $\int_{-\infty}^\infty d t\, t^{1/3}/(1+t^2) = \int_\gamma d t\, t^{1/3}/(1+t^2)$.

There is one residue, at $t=i=e^{i\pi/2}$, so we have halved our work. We find $$\begin{eqnarray*} I &=& \frac{1}{1+e^{\pi i/3}} 2\pi i \frac{e^{i\pi/6}}{2i} \\ &=& \frac{\pi}{2\cos \pi/6} \\ &=& \frac{\pi}{\sqrt{3}}. \end{eqnarray*}$$

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    $\begingroup$ Much thanks oenamen. I always appreciate the answer, even a month later! :) $\endgroup$ – Dedede Apr 20 '12 at 3:24
  • $\begingroup$ @Dedede: Sure thing, Dedede! Let me know if anything is not clear. $\endgroup$ – user26872 Apr 20 '12 at 3:42

Continue reading the next few paragraphs where it is explained how to go on. In fact, that substitution isn't really necessary if you integrate over a "keyhole contour", which is also explained a bit further down the page.

  • $\begingroup$ So do I just add the residues at the zeros of $1+z^4$ which have positive imaginary part as usual? Also, how can the "keyhole contour" be set up? Ahlfors draws a picture, but doesn't show how to actually do it. $\endgroup$ – Dedede Mar 19 '12 at 6:57
  • $\begingroup$ I did a writeup of a similar integral a few weeks ago. Have a look at: math.stackexchange.com/questions/114884/… $\endgroup$ – mrf Mar 19 '12 at 6:59
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    $\begingroup$ By the way, I'm not sure I'd recommend Ahlfors for self-study. There is nothing wrong with the text, but it contains relatively few worked examples to illustrate the ideas. $\endgroup$ – mrf Mar 19 '12 at 7:01
  • $\begingroup$ Thanks for the link. And for the first question in my first comment, is $\int_{-\infty}^{\infty}\frac{z^{5/3}}{1+z^4}dz=2\pi i(\text{Res}(\sqrt[4]{-1})+\text{Res}((-1)^{3/4}))$? where $R(z)=z^{5/3}/(1+z^4)$ by the residue theorem as usual? I find $\sqrt[4]{-1}$ and $(-1)^{3/4}$ to be the poles in the upper half plane. $\endgroup$ – Dedede Mar 19 '12 at 7:09
  • $\begingroup$ Sorry, I should have given you a more thorough answer. You need to show that the integral over the semi-circle of radius $R$ tends to $0$ as $R \to \infty$. Once you know that, the integral is given by the sum of the residues in the upper half-plane. $\endgroup$ – mrf Mar 19 '12 at 22:00

Here is the simplest I could get without introducing other variables or making strange substitutions. Everything is in complex on the main branch ($0 \le arg(z) < 2\pi$).

First we calculate a bound change : $$\begin{eqnarray*} \int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz &=& \int_{-\infty}^0 \frac{z^{1/3}}{1+z^2} dz + \int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz\\ &=& \int_{0}^\infty \left( \frac{z^{1/3}}{1+z^2} + \frac{(-z)^{1/3}}{1+(-z)^2}\right) dz \\ &=& \int_{0}^\infty \frac{z^{1/3} + (e^{i\pi}z)^{1/3}}{1+z^2} dz \\ &=& (1+e^{i\frac{\pi}{3}})\int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz \\ \end{eqnarray*}$$

So we have (1) $$\int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz = \frac{1}{1+e^{i\frac{\pi}{3}}} \int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz $$

Now we only have to find $\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz$. Because $1+z^2 = (z-i)(z+i)$, the only singularity in the upper half-plane is $i$ and by the Residue Theorem we immediatly get $$ \int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz = 2 \pi i \text{Res}_i \text{ with } \text{Res}_i = \lim_{z \to i} (z-i) \frac{e^{log(z)/3}}{(z-i)(z+i)} = \frac{e^{i\frac{\pi}{6}}}{2i} $$

So we have (2) $$ \int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz = \pi e^{i\frac{\pi}{6}} $$

Finally, by (1) and (2) we get $$ \int_{0}^\infty\frac{z^{1/3}}{1+z^2} dz = \frac{\pi e^{i\frac{\pi}{6}}}{1+e^{i\frac{\pi}{3}}} = \frac{\pi}{\sqrt{3}} $$

  • $\begingroup$ By a bound change do you mean a change in the interval of your integral? $\endgroup$ – Kieran Cooney Oct 26 '13 at 2:02
  • $\begingroup$ What is your contour? $\endgroup$ – The Substitute Jun 3 '15 at 23:08

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