Problem 7

Exercise 8.7 (Funky sine integral)   Evaluate

$$\int_0^\pi \frac{d\theta}{2+\sin(2\theta)}$$ (8.30)

Proof. Substitute $z=e^{i\theta}$ into the integral and find $d\theta=dz/2iz$, so that the integral may be interpreted as a contour integral of a rational function about the unit circle

$$\int_0^\pi \frac{d\theta}{2+\sin(2\theta)} = \int_0^\pi \frac{d\t... ... - 1/z} = \int_\gamma \frac{dz}{z(4i+z-1/z)} = \int_\gamma \frac{dz}{4iz+z^2-1}$$ (8.31)

The poles are found by the quadratic formula

$$4iz + z^2 - 1 = 0 \implies z = -2i \pm \sqrt{3}i$$ (8.32)

But only one of them lies within the unit disk, namely, $z_0=-2i+\sqrt{3}i$. To compute the residue, we evaluate a limit with l'Hôpital's rule:

$$\operatorname{Res}(f,z_0) = \lim_{z\to z_0} \frac{z-z_0}{4iz + z^2 - 1} = \lim_{z\to z_0} \frac{1}{4i + 2z} = \frac{1}{2\sqrt{3}i}$$ (8.33)

Then we can find by the Residue Theorem that

$$\int_0^\pi \frac{d\theta}{2+\sin(2\theta)} = \int_\gamma \frac{dz... ...i\operatorname{Res}(f,z_0) = 2\pi i \frac{1}{2\sqrt{3}i} = \frac{\pi}{\sqrt{3}}$$ (8.34)

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