Problem - T. Amdberhan and V. Moll (USA)
Show that
\[\int_0^\infty\frac{\cos(x)\sin\left(\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}\,\mathrm dx=\frac{\pi}{4}.\]Solution
Call the integral in question $I$. Let $x=\sinh(u)$ so that $\mathrm dx=\cosh(u)\,\mathrm du$ and $\sqrt{1+x^2}=\cosh(u)$. Then we find that
\[\begin{align*} I &= \int_0^\infty\cos\left(\sinh(u)\right)\sin\left(\cosh(u)\right)\,\mathrm du \\ &= \frac{1}{2}\int_0^\infty\sin\left(\cosh(u)-\sinh(u)\right)+\sin\left(\sinh(u)+\cosh(u)\right)\,\mathrm du \\ &= \frac{1}{2}\int_0^\infty\sin\left(e^{-u}\right)\,\mathrm du+\frac{1}{2}\int_0^\infty\sin\left(e^u\right)\,\mathrm du.\end{align*}\]In the leftmost integral in last line above, let $v=e^{-u}$. And in the rightmost integral in the same line, let $v=e^u.$ So
\[\begin{align*} I &=-\frac{1}{2}\int_1^0\frac{\sin(v)}{v}\,\mathrm dv+\frac{1}{2}\int_1^\infty\frac{\sin(v)}{v}\,\mathrm dv \\ &= \frac{1}{2}\int_0^\infty\frac{\sin(v)}{v}\,\mathrm dv \\ &=\frac{\pi}{4}.\end{align*}\]Where we’ve made use of the famous Dirichlet integral
\[\int_0^\infty\frac{\sin(x)}{x}\,\mathrm dx=\frac{\pi}{2}.\]Dirichlet’s Integral
Method 1 (The Feynman trick)
Define the function
\[I(s)=\int_0^\infty\frac{\sin(t)}{t}e^{-st}\,\mathrm dt,\qquad s>0\]and note that $\lim_{s\to0^+}I(0)=I$ and $\lim_{s\to\infty}I(s)=0.$ If we take the derivative we get
\[\begin{align*} I'(s) &= \int_0^\infty\frac{\partial}{\partial s}\frac{\sin(t)}{t}e^{-st}\,\mathrm dt \\ &= -\int_0^\infty\sin(t)e^{-st}\,\mathrm dt \\ &= -\Im\int_0^\infty e^{(i-s)t}\,\mathrm dt \\ &= -\Im\left[\frac{e^{(i-s)t}}{i-s}\right]\bigg\vert_0^\infty \\ &= \Im\frac{1}{i-s} \\ &= -\Im\left(\frac{s}{1+s^2}+\frac{i}{1+s^2}\right) \\ &= -\frac{1}{1+s^2}.\end{align*}\]Then
\[\begin{align*} I &= \lim_{s\to0^+}I(s)-\lim_{s\to\infty}I(s) \\ &= -\int_0^\infty I'(s)\,\mathrm ds \\ &= \int_0^\infty\frac{1}{1+s^2}\,\mathrm ds \\ &= \left[\text{acrtan}(s)\right]|_0^\infty \\ &= \frac{\pi}{2}.\end{align*}\]Method 2
\[\begin{align*}\int_0^\infty\frac{\sin(x)}{x}\,\mathrm dx&=\frac{1}{2}\int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm dx \\&=\frac{1}{2}\sum_{n=-\infty}^\infty\int_{2\pi n}^{2\pi(n+1)}\frac{\sin(x)}{x}\,\mathrm dx \\ &=\frac{1}{2}\sum_{n=-\infty}^\infty\int_{0}^{2\pi}\frac{\sin(\theta+2\pi n)}{\theta+2\pi n}\,\mathrm d\theta,\,\text{u-sub with }x=\theta+2\pi n \\ &=\frac{1}{2}\sum_{n=-\infty}^\infty\int_{0}^{2\pi}\frac{\sin(\theta)}{\theta+2\pi n}\,\mathrm d\theta \\ &=\frac{1}{2}\int_{0}^{2\pi}\sin(\theta)\sum_{n=-\infty}^\infty\frac{1}{\theta+2\pi n}\,\mathrm d\theta \\ &=\frac{1}{4}\int_0^{2\pi}\sin(\theta)\cot\left(\frac{\theta}{2} \right )\,\mathrm d\theta,\,\text{Euler's parital fraction expansion for cotan} \\&=\frac{1}{2}\int_0^{2\pi}\cos^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta.\end{align*}\]And this last integral can be evaluated using some symmetry,
\[\begin{align*}\frac{1}{2}\int_0^{2\pi}\cos^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta &=\frac{1}{2}\int_0^{2\pi}\left (1-\sin^2\left(\frac{\theta}{2} \right ) \right )\,\mathrm d\theta \\ &=\pi-\frac{1}{2}\int_0^{2\pi}\sin^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta \\ &=\pi-\frac{1}{2}\int_0^{2\pi}\cos^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta,\,\text{u-sub }\theta\mapsto2\pi-\theta \\ &\Rightarrow \frac{1}{2}\int_0^{2\pi}\cos^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta=\pi-\frac{1}{2}\int_0^{2\pi}\cos^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta \\ &\Rightarrow \frac{1}{2}\int_0^{2\pi}\cos^2\left(\frac{\theta}{2} \right )\,\mathrm d\theta=\frac{\pi}{2}.\end{align*}\]