Problem 8

Exercise 10.8 (Another semicircular contour)   Integrate

$$\int_{-\infty}^\infty \frac{\cos x}{x^2+a^2}dx$$ (10.40)

Proof. To evaluate this, we integrate the complexified function

$$f(z) = \frac{e^{iz}}{z^2+a^2}$$ (10.41)

along a semicircular contour which captures the pole $z=ai$. Letting $R\to\infty$ will capture the desired integral in the real part. The residue theorem implies

$$\int_{-R}^R \frac{e^{ix}}{x^2+a^2}dx + \int_{\text{arc}} f(z)dz = 2\pi i \operatorname{Res}(f,ai)$$ (10.42)

The arc integral can be shown to vanish

$$\left\vert \int_{\text{arc}} f(z)dz \right\vert \leq \int \frac{\vert e^{iz}\vert}{\vert z\vert^2-a^2}dz$$ (10.43)

When $z=a+bi$, we know $\vert e^{iz}\vert=\vert e^{i(a+bi)}\vert=e^{-b}$, so in the upper half-plane, we know $\sup \vert e^{iz}\vert=1$. Therefore,

$$\int \frac{\vert e^{iz}\vert}{\vert z\vert^2-a^2}dz \leq \int \frac{1}{R^2-a^2}dz = \frac{\pi R}{R^2-a^2} \to 0$$ (10.44)

As $R\to\infty$ we are left with the residue

$$\int_{-\infty}^\infty \frac{e^{ix}}{x^2+a^2}dx = 2\pi i \operatorname{Res}(f,ai)$$ (10.45)

$$= 2\pi i \lim_{z\to ai}\frac{(z-ai)e^{iz}}{(z-ai)(z+ai)}$$ (10.46)

$$= 2\pi i \lim_{z\to ai}\frac{e^{iz}}{z+ai}$$ (10.47)

$$= 2\pi i e^{-a}/2ai$$ (10.48)

$$= \frac{\pi e^{-a}}{a}$$ (10.49)

Taking the real part shows

$$\int_{-\infty}^\infty \frac{\cos x}{x^2+a^2}dx = \frac{\pi e^{-a}}{a}$$ (10.50)

$\qedsymbol$