A calculus problem by Aly Ahmed

Consider the integral:

$F(s) = \int_{0}^{\infty} \mathrm{e}^{-sx}\sin{x} \ dx$

This is a standard integral with a result of:

$\int_{0}^{\infty} \mathrm{e}^{-sx}\sin{x} \ dx = \frac{1}{1+s^2}$

Now, integrating both sides from some constant $p$ to $\infty$ with respect to $s$ gives:

$\int_{p}^{\infty} \int_{0}^{\infty} \mathrm{e}^{-sx}\sin{x} \ dx \ ds = \int_{p}^{\infty}\frac{ds}{1+s^2}$

Order of integration can be changed (in LHS) as limits of integration are constant:

$\int_{0}^{\infty} \int_{p}^{\infty} \mathrm{e}^{-sx}\sin{x} \ ds \ dx = \frac{\pi}{2}-\arctan{p}$ $\int_{0}^{\infty} \frac{\mathrm{e}^{-px}\sin{x}}{x} \ dx = \frac{\pi}{2}-\arctan{p}$

Substituting $p=0$ gives:

$\boxed{\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}}$

The integral can be solved using Feynman's integration trick differentiation under the integral sign as below:

$\begin{aligned} I(a) & = \int_0^\infty \frac {e^{-ax}\sin x}x dx \\ \frac {\partial I(a)}{\partial a} & = \int^0_\infty e^{-ax}\sin x\ dx & \small \blue{\text{By integration by parts}} \\ & = -e^{-ax}\cos x - \int a e^{ax} \cos x \ dx \bigg|^0_\infty \\ & = -e^{-ax}\cos x - a e^{-ax} \sin x - a^2 \int e^{-ax} \sin x \ dx \bigg|^0_\infty \\ & = - \frac {e^{-ax}(a\sin x +\cos x)}{1+a^2} \ dx \bigg|^0_\infty \\ & = - \frac 1{1+a^2} \\ \implies I(a) & = - \tan^{-1} a + \blue C & \small \blue{\text{where }C \text{ is the constant of integration.}} \\ I (\infty) & = - \frac \pi 2 + C = 0 & \small \blue{\implies C = \frac \pi 2} \\ \implies I(a) & = \frac \pi 2 - \tan^{-1} a \\ \implies I(0) & = \int_0^\infty \frac {\sin x}x dx = \frac \pi 2 \approx \boxed{1.57} \end{aligned}$