A note on the Riemann integral of sin(x)/x

In this post, we will prove the famous result that $$\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}.$$ We first fix some $0<a<\infty$ and consider $e^{-xy}\sin x$. We shall show that $e^{-xy}\sin x$ is integrable on $0<x< a$ and $y>0$. As $\left|\sin x\right|\le1$, one may instead try to prove the integrability of $e^{-xy}$ for this purpose. Unfortunately, the attempt would fail since $$\int_{0}^{a}\int_{0}^{\infty}e^{-xy}\,dy\,dx=\int_{0}^{a}\frac{1}{x}dx=\infty$$ diverges. Bounding $e^{-xy}\sin x$ with $e^{-xy}$ turns out to be too rough near $x=0$, because $1/x\rightarrow\infty$ but $\sin x/x\rightarrow1$ when $x\searrow0$. We therefore need to use a tighter bound near $x=0$ to make the integral finite. Observe that, since $\sin x/x\rightarrow1$, given $\varepsilon>0$ there exists $\delta>0$ such that $\left|\sin x/x-1\right|< \varepsilon$ for $0< x< \delta$. Hence, $$\begin{eqnarray*} \int_{0}^{a}\int_{0}^{\infty}e^{-xy}\sin x\,dy\,dx & = & \int_{0}^{\delta}\int_{0}^{\infty}e^{-xy}\sin x\,dy\,dx+\int_{\delta}^{a}\int_{0}^{\infty}e^{-xy}\sin x\,dy\,dx\\ & \le & \int_{0}^{\delta}\frac{\sin x}{x}dx+\int_{\delta}^{a}\frac{1}{x}dx\\ & \le & \left(1+\varepsilon\right)\delta+\ln a-\ln\delta < \infty. \end{eqnarray*}$$ Similarly, $$\int_{0}^{a}\int_{0}^{\infty}e^{-xy}\sin x\,dy\,dx \ge \left(1-\varepsilon\right)\delta-\ln a+\ln\delta > -\infty.$$ It follows that the integral indeed exists. Now we can perform the double integral in two orders to obtain $$\int_{0}^{a}\left(\int_{0}^{\infty}e^{-xy}\, dy \right)\sin x\, dx=\int_{0}^{a}\frac{\sin x}{x}dx$$ and $$\begin{eqnarray*} \int_{0}^{\infty}\left(\int_{0}^{a}e^{-xy}\sin x \, dx \right) dy & = & \int_{0}^{\infty}\left(\dfrac{1}{y^{2}+1}-\dfrac{\left( \cos a + y\,\sin a \right)e^{-ay}}{y^{2}+1}\right) dy \\ & = & \frac{\pi}{2}-\cos a\int_{0}^{\infty}\frac{e^{-ay}}{1+y^{2}}dy-\sin a\int_{0}^{\infty}\frac{y\,e^{-ay}}{1+y^{2}} dy \end{eqnarray*}$$ Hence, $$\begin{eqnarray*} \left|\int_{0}^{a}\frac{\sin x}{x}dx-\frac{\pi}{2}\right| & = & \cos a\int_{0}^{\infty}\frac{e^{-ay}}{1+y^{2}}dy+\sin a\int_{0}^{\infty}\frac{y\,e^{-ay}}{1+y^{2}}dy\\ & \le & \int_{0}^{\infty}\left(1+y\right)e^{-ay}\, dy\\ & = & a^{-2}+a^{-1}. \end{eqnarray*}$$ By taking $a\rightarrow\infty$ we obtain the desired result.

You can find different proofs of the same result here. Besides, this thread shows why $\sin{x}/x$ is Riemann integrable but not Lebesgue integrable on $(0,\infty)$.