Problem 8

Exercise 9.8 (Sector-based contour integral)   Evaluate

$$\int_0^\infty \frac{x^{1/3}}{1+x^2}$$ (9.47)

Proof. Substitute $z=x^{1/3}$ to transform the integral into

$$3\int_0^R \frac{z^3}{1+z^6}$$ (9.48)

Find the pole at $e^{i\pi/6}$ and integrate around the sector of radius $R$ and $0\leq \theta \leq \pi/3$. The integral can be written as

$$\int_0^R f(z) + \int_{\text{arc}} f(z)+\int_R^0 f(re^{i\pi/3})e^{i\pi/3} = 2\pi i\operatorname{Res}(f,e^{i\pi/6})$$ (9.49)

This equation is analyzed in three stages: find the residue, express the backwards integral in terms of the forwards integral, and show the middle integral vanishes as $R\to\infty$.

For the residue:

$$\operatorname{Res}(f,e^{i\pi/6}) = \lim_{z\to e^{i\pi/6}} \frac{(... ...})z^3}{1+z^6} = \lim_{z\to e^{i\pi/6}} \frac{z^3}{6z^5}=\frac{1}{6}e^{-\pi i/3}$$ (9.50)

For the backwards integral

$$e^{i\pi/3}\int_R^0 f(re^{i\pi/3}) =e^{i\pi/3}\int_R^0\frac{r^3e^{i\pi}}{1+r^6} = e^{i\pi/3}\int_0^R\frac{r^3}{1+r^6}=e^{i\pi/3}\int_0^R f(z)$$ (9.51)

For the middle integral

$$\int_{\text{arc}}f(z) \leq \int_{\vert z\vert=R}f(z) \leq \int \frac{\vert z\vert^3}{\vert z\vert^6-1} = 2\pi R\frac{R^3}{R^6-1} \to 0$$ (9.52)

Letting $R\to\infty$ shows that

$$(1+e^{i\pi/3})\int_0^\infty \frac{z^3}{1+z^6} = 2\pi i \frac{1}{6}e^{-\pi i/3}$$ (9.53)

Therefore,

$$\int_0^\infty \frac{z^3}{1+z^6} = \frac{2\pi i}{6}\frac{e^{-\pi i/3}}{1+e^{i\pi/3}}= \frac{2\pi}{... ...{2\pi}{6}\frac{1}{e^{-i\pi/6}+e^{i\pi/6}}= \frac{2\pi}{6}\frac{1}{2\cos(\pi/6)}$$ (9.54)

$$= \frac{\pi}{3\sqrt{3}}$$ (9.55)

This means the original integral equals

$$\int_0^\infty \frac{x^{1/3}}{1+x^2} = \frac{\pi}{\sqrt{3}}$$ (9.56)

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