Here we will solve the Dirichlet Integral using two different techniques. The Dirichlet Integral is defined as $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx $$
Feynman Integration
We employ Feynman’s Integration trick here by first defining the function $$ I(\alpha)=\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx $$
and notice that because \(|\frac{\sin(x)}{x}| \leq 1 \)
$$ \lim_{\alpha \rightarrow \infty} |\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx| \leq \int_{0}^{\infty} e^{-\alpha x} dx = \frac{1}{\alpha} $$
so \(I(\alpha) \) vanishes as \(\alpha \rightarrow \infty \) and thus we can continue with Fenyman’s technique: $$ \frac{\partial I(\alpha)}{\partial \alpha} = \frac{\partial}{\partial \alpha} \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx = -\int_{0}^{\infty} x \frac{\sin(x)}{x} e^{-\alpha x} dx = -\int_{0}^{\infty} \sin(x) e^{-\alpha x} dx $$ now by integration by parts, we have that $$ -\int_{0}^{\infty} \sin(x) e^{-\alpha x} dx = -1+\alpha \int_{0}^{\infty} \cos(x) e^{-\alpha x} dx $$
using integration by parts once more, we have that $$ -1+\alpha \int_{0}^{\infty} \cos(x) e^{-\alpha x} dx= -1+\alpha^2 \int_{0}^{\infty} \sin(x) e^{-\alpha x} dx = -1 - \alpha^2 I’(\alpha) $$ and thus $$ I’(\alpha)=-1 - \alpha^2 I’(\alpha) \implies I’(\alpha)(1+\alpha^2)=-1 \implies I’(\alpha)=-\frac{1}{1+\alpha^2} $$
Next, observe that
$$ \int I’(\alpha) d\alpha = \int -\frac{1}{1+\alpha^2} d\alpha = -\arctan(\alpha) + C = I(\alpha) $$
by the fundamental theorem of Calculus. In solving for C, we have
$$ I(\infty)=\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\infty x} dx = 0 = -\arctan(\infty) + C \implies C=\arctan(\infty)=\frac{\pi}{2} $$
finally, we have that
$$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx = I(0)= -\arctan(0)+\frac{\pi}{2}=\frac{\pi}{2} $$
and thus $$ \boxed{\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \frac{\pi}{2}} $$
Contour Integration
Here, we will use Cauchy’s Residue Theorem, which states that
$$ \oint_{C} f(z) dx = 2\pi i Res(f,C) $$
where C is a simple closed curve. Let us define \(f(z)=e^{iz}/z\) and note that \(f\) has a singularity at \(z=0\) . Thus, in constructing a countor for \(f\) ,
we will have a path which is a semicircle of radius R traveling counterclockwise, then on the real axis across an smaller semicircle radius \(\epsilon\)
the result of the integral along this path C is
$$ \oint_C f(z) dz = \int_{\Gamma} f(z) dx + \int_{\gamma} f(z) dx + \int_{-R}^{-\epsilon} f(z) dz + \int_{\epsilon}^{R} f(z) dz $$
We want to examine the case where \(R \rightarrow \infty\) and \(\epsilon \rightarrow 0\) since these are the bounds of our integral. Firstly, note that because \(f\) has no singularities within the contour C, we observe that $$ \oint_C f(z) dz = \oint_C \frac{e^{iz}}{z} dz = 2\pi i Res(f,C) = 2\pi i (0) = 0 $$ now solving for each integral, we have that
$$ \int_{\Gamma} f(z) dx = \int_{\Gamma} \frac{e^{iz}}{z} dz = \int_{0}^{\pi} \frac{e^{iRe^{i\theta}}}{Re^{i\theta}} iRe^{i\theta} d\theta = \int_{0}^{\pi} ie^{iRe^{i\theta}} d\theta $$
and now note that $$ |\int_{\Gamma} f(z) dx| \leq \int_{0}^{\pi} |ie^{iRe^{i\theta}}| d\theta = \int_{0}^{\pi} |ie^{iR\cos(\theta)-R\sin(\theta)}| d\theta \leq \int_{0}^{\pi} |e^{-Rsin\theta}| d\theta $$
and as \(R \rightarrow \infty , e^{-Rsin\theta} \rightarrow 0\)
so we have that $$ \int_{\Gamma} f(z) dx \leq \int |e^{-Rsin\theta}| d\theta \rightarrow_{R \rightarrow \infty} 0 \implies \int_{\Gamma} f(z) = 0 $$
Next,
$$ \int_{\gamma} f(z) dx = \int_{\pi}^{0} \frac{e^{i \epsilon e^{i\theta}}}{\epsilon e^{i\theta}} i \epsilon e^{i\theta} d\theta = -\int_{0}^{\pi} i e^{i \epsilon e^{i\theta}} d\theta $$
and note that as \(\epsilon \rightarrow 0\) , \(e^{i \epsilon e^{i\theta}} \rightarrow 1\) so
$$ \int_{\gamma} f(z) dx = -\int_{0}^{\pi} i e^{i \epsilon e^{i\theta}} d\theta = -\int_{0}^{\pi} i d\theta = -i \pi $$
so far, we have that $$ \oint_C f(z) dz = \int_{\Gamma} f(z) dx + \int_{\gamma} f(z) dx + \int_{-R}^{-\epsilon} f(z) dz + \int_{\epsilon}^{R} f(z) dz \implies 0=0-i \pi + \int_{-R}^{-\epsilon} f(z) dz + \int_{\epsilon}^{R} f(z) dz $$ now as \(R \rightarrow \infty\) and \(\epsilon \rightarrow 0\) , we have $$ i \pi = \int_{-\infty}^{0} f(z) dz + \int_{0}^{\infty} f(z) dz = \int_{-\infty}^{\infty} \frac{e^{iz}}{z} dz $$ Further, $$ \mathnormal{Im}(i \pi)= \pi =\mathnormal{Im}(\int_{-\infty}^{\infty} \frac{e^{iz}}{z} dz)=\int_{-\infty}^{\infty} \mathnormal{Im}(\frac{e^{iz}}{z}) dz= \int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx $$ so we conclude that $$ \pi= \int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx \implies \boxed{\frac{\pi}{2} = \int_{0}^{\infty} \frac{\sin(x)}{x} dx} $$ by the symmetry of $$ f(x)=\frac{\sin(x)}{x} $$