### A problem asked by Swati Jain

Prove that $\int_0^{\infty} \frac{\sin x}{x} \; dx = \frac{\pi}{2}$

$\int_0^{\infty} e^{-xy} \sin x \; dy$

$= \frac{-\sin x}{x} e^{-xy} \biggr. \biggr|_{y=0}^{\infty}$

$= \frac{\sin x}{x}$

So,

$\int_0^{\infty} \frac{\sin x}{x} \; dx = \int_0^{\infty} \int_0^{\infty} e^{-xy} \sin x \; dy \; dx = \int_0^{\infty} \int_0^{\infty} e^{-xy} \sin x \; dx \; dy$

Let $I = \int e^{-xy} \sin x \; dx$

$= e^{-xy} (-\cos x) - \int (-y)e^{-xy} (-\cos x) \; dx$

$= -e^{-xy} \cos x - y \int e^{-xy} \cos x \; dx$

$= -e^{-xy} \cos x - y \left( e^{-xy} \sin x - \int (-y) e^{-xy} \sin x\; dx \right)$

$= -e^{-xy} \cos x - y e^{-xy} \sin x - y^2 \int e^{-xy} \sin x \; dx$

So, $(1 + y^2)I = -e^{-xy} \cos x - y e^{-xy} \sin x$

Or, $I = \frac{-e^{-xy} \cos x - y e^{-xy} \sin x}{1+y^2}$

And $\int_0^{\infty} e^{-xy} \sin x \; dx = \frac{-e^{-xy} \cos x - y e^{-xy} \sin x}{1+y^2} \biggr. \biggr|_{x=0}^{\infty} = \frac{1}{1+y^2}$

So, $\int_0^{\infty} \frac{\sin x}{x} \; dx$

$= \int_0^{\infty} \int_0^{\infty} e^{-xy} \sin x \; dx \; dy$

$= \tan^{-1} y \biggr. \biggr|_{y=0}^{\infty}$

$= \frac{\pi}{2}$