Problem 8

Exercise 5.8 (Sinc Integral!)   Integrate

$$\int_{-\infty}^\infty \frac{\sin x}{x}$$ (5.43)

Proof. Consider the complexified function $f(z)=e^{iz}/z$. Then

$$\int_{-\infty}^\infty \frac{\sin z}{z}= \Im \int_{-\infty}^\infty \frac{e^{iz}}{z}$$ (5.44)

By writing $e^{iz}$ as a Taylor series, we can divide to argue that the residue at zero equals one:

$$\frac{e^{iz}}{z} = \frac{1 + iz - z^2/2 + \cdots}{z}=\frac{1}{z} + i + \cdots$$ (5.45)

Integrate $f$ around a semicircular arc with radius $R$ and a dimple in the lower half-plane centered at the origin with radius $\epsilon$. Then the residue theorem says this path integral captures the pole, so that

$$\int_{\text{dimple}} + \int_\epsilon^R + \int_{\text{arc}} + \int_{-R}^{-\epsilon} = 2\pi i$$ (5.46)

The arc integral vanishes as $R\to\infty$ in the upper half-plane. Set $z=Re^{i\theta}$. Then

$$e^{iz}=e^{iRe^{i\theta}}=e^{iR(\cos\theta+i\sin\theta)} = e^{iR\cos\theta - R\sin\theta} \implies \vert e^{iz}\vert=e^{-R\sin\theta}$$ (5.47)

Simplify the following integral

$$\int_{\text{arc}} = \int_0^{\pi} \frac{e^{iRe^{i\theta}}}{Re^{i\theta}} Rie^{i\theta}d\theta = i\int_0^{\pi} e^{iRe^{i\theta}} d\theta$$ (5.48)

Then we can apply the previous estimate

$$\left\vert\int_{\text{arc}}\right\vert \leq \int_0^{\pi} e^{-R\sin\theta}d\theta$$ (5.49)

The integrand converges to 0 because $\sin\theta$ is nonnegative on this interval. Moreover, $e^{-R\sin\theta}$ is continuous and $[0,\pi]$ is compact, so we may exchange limits. Therefore, the arc integral vanishes.

The dimple integral is handled by letting $\epsilon\to 0$. Set $z=\epsilon e^{i\theta}$. We have

$$\int_{\text{dimple}}= i\int_\pi^{2\pi} e^{i\epsilon e^{i\theta}} d\theta$$ (5.50)

Again, we can apply integral interchange because of smoothness, so the dimple integral is sent to $\pi i$ as $\epsilon\to 0$.

Therefore, as $R\to\infty$ and $\epsilon\to 0$, we see

$$\int_{-\infty}^\infty \frac{e^{ix}}{x} = \pi i$$ (5.51)

Taking, the imaginary part, we see:

$$\int_{-\infty}^\infty \frac{\sin x}{x} = \pi$$ (5.52)

$\qedsymbol$