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I need to use residues to evaluate $\int_0^\infty \frac{x^{1/3}}{1+x^2} \, dx$. So first,

$$\int_0^\infty \frac{x^{1/3}}{1+x^2} \, dx = \int_{C_R} \frac{z^{1/3}}{1+z^2} + \int_{-R}^R \frac{x^{1/3}}{1+x^2} \, dx$$

We first observe that $f(z)$ has two simple poles at $z= \pm i$. Since only $z=i$ will be interior to $C_R$, then we assert that

$$\int_0^\infty \frac{x^{1/3}}{1+x^2} \, dx = \int_{C_R} \frac{z^{1/3}}{1+z^2} + \int_{-R}^R \frac{x^{1/3}}{1+x^2} \, dx = 2\pi i \cdot \mathrm{Res}_{z=i} f(z)$$

Now, let $p(z) = z^{1/3}$ and $q(z) = 1+z^2$. Then

$$\mathrm{Res}_{z=i} f(z) = \frac{p(i)}{q'(i)} = \frac{i^{1/3}}{2i} = \frac{1}{2}i^{-2/3}$$

So we have in fact that

$$\int_0^\infty \frac{x^{1/3}}{1+x^2} \, dx = 2 \pi i \cdot \frac{1}{2}i^{-2/3} = \pi i^{1/3}$$

Note: it is not hard to show that $\int_{-R}^R f(x) \, dx \to 0$ as $R \to \infty$.

Does this seem correct? Did I make any mistakes anywhere?

Grigor Hakobyan
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    Which value does $i^{1/3}$ take? Note that the original integral is clearly a real number. – Gary Apr 29 '22 at 22:18
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    What is $C_R$? You should try a keyhole contour. – J.G. Apr 29 '22 at 22:20
  • @Gary Well, since $i = \cos (\pi/2) + i \sin(\pi/2)$, would we say $i^{1/3} = \cos(\frac{\pi}{6} + i \sin \frac{\pi}{6})$? – Grigor Hakobyan Apr 29 '22 at 22:20
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    In general, it's often best to remove the multivalued aspect first before using complex methods (unless you explicitly desire it for some application). In this case, your integral is (up to a scalar) given by $\int_0^\infty \frac{x^3}{1+x^6}$, which is solvable via a wedge contour. – Brevan Ellefsen Apr 30 '22 at 00:51
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    There's a fun trick here worth mentioning here, namely that under the map $x \mapsto x^{-1}$ we have $\int_0^\infty \frac{x^3}{x^6+1} dx = \int_0^\infty \frac{x}{x^6+1} dx$. Averaging these yields $\frac{1}{2}\int_0^\infty \frac{x + x^3}{x^6+1} dx$. Letting $x \mapsto x^{1/2}$ turns this into $\frac{3}{4}\int_0^\infty \frac{x + 1}{x^3+1} dx$, and simplifying the integrand yields $\frac{3}{4}\int_0^\infty \frac{1}{x^2-x+1} dx$ which can be solved using elementary techniques. – Brevan Ellefsen Apr 30 '22 at 02:26
  • See THIS ANSWER and THIS ONE for evaluating a more general class of integrals. – Mark Viola Oct 12 '22 at 18:23

1 Answers1

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Doing this without massaging the contour first is not advised; see my comments above. Nevertheless, we can fill in the gaps in your attempted proof.

Your integrand isn't actually holomorphic at the origin so we are forced to use a keyhole contour, where we travel forward along $\gamma_1$ and backward along $\gamma_2$ (paths from $\epsilon$ to $\rho$ and from $\rho$ to $\epsilon$ respectively, where $\epsilon \to 0$ and $\rho \to \infty$). Using the branch cut with $0 < \theta < 2\pi$ we can define the integrand $f(z)$ by $\exp(\tfrac{1}{3}(\operatorname{Log} r + i \theta))$, where $z = r \exp(i \theta)$. By continuity, this means that along $\gamma_2$ we must have $f(z) = \exp(2\pi i /3)\frac{z^{1/3}}{z^2+1}$ (where $z^{1/3}$ is the value defined along $\gamma_1$) so that $\int_{\gamma_1} f(z) dz + \int_{\gamma_2} f(z) dz = \int_\epsilon^\rho \frac{z^{1/3}}{z^2+1} dz - \int_\epsilon^\rho \exp(2\pi i /3)\frac{z^{1/3}}{z^2+1} dz = (1-\exp(2\pi i /3))\int_{\gamma_1} \frac{z^{1/3}}{z^2+1} dz$

We now calculate residues within the contour, being careful when the argument passes the branch cut: $$\text{Res}(i) = \lim_{z \to i}(z-i)f(z) = \lim_{\;r \to 1\\\theta \to \color{red}{\pi/2}}\frac{\exp(\tfrac{1}{3}(\text{Log}(r) + i \theta)}{r\exp(i \theta)+i} = \frac{1}{2i} \exp(i \pi /6)$$ $$\text{Res}(-i) = \lim_{z \to -i}(z+i)f(z) = \lim_{\;r \to 1\\\theta \to \color{red}{3\pi/2}}\frac{\exp(\tfrac{1}{3}(\text{Log}(r) + i \theta)}{r\exp(i \theta)+i} = \frac{1}{-2i} \exp(i \pi /2)$$

Putting this together with the Residue Theorem (along with the observation that the integral vanishes around the larger and small circles of the contour in the limits because of the form of the integrand) we have: $$(1-\exp(2\pi i/3))\int_0^\infty \frac{z^{1/3}}{z^2+1} dz = 2\pi i\sum \text{Res}(\pm i) $$ $$= \pi(\exp(i \pi /6) - \exp(i \pi /2)) = \pi \exp(-i \pi /6)$$

Where this last equality follows from $\exp(i \pi /6) - \exp(-i \pi /6) = 2i \sin(\pi/6) = i = \exp(i \pi /2)$. We finally conclude:

$$\int_0^\infty \frac{z^{1/3}}{z^2+1} dz = \pi\frac{\exp(-i \pi /6)}{1-\exp(2\pi i/3)} = \pi\frac{\exp(3i \pi /6)}{\exp(4i \pi /6)-\exp(8\pi i/6)} $$ $$= \frac{\pi}{2}\frac{2i}{\exp(4i \pi /6)-\exp(-4\pi i/6)} = \frac{\pi}{2} \frac{1}{\sin(2\pi/3)} = \frac{\pi}{2} \frac{2}{\sqrt 3} = \frac{\pi}{\sqrt 3}$$

For alternative proofs, see my above comments and the same calculation but using a wedge contour and a number of proofs of the general case.

Brevan Ellefsen
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