# Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?

A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:
$$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$

Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of $$\sin x$$:
$$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t – \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$
but I don’t see how $$\pi$$ comes here, since we need the answer to be equal to $$\dfrac{\pi}{2}$$.

I believe this can also be solved using double integrals.

It is possible (if I remember correctly) to justify switching the order of integration to give the equality:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Notice that
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} – y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} – y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\ &= \frac{-ye^{-xy}\sin x – e^{-xy}\cos x}{1+y^2}. \end{align*}
Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$
Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$