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 $\dfrac{\pi}{2}$.

Answer

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
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

This leads us to

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Now the right hand side can be found easily, using integration by parts.

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

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