I want to evaluate the integral $$\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$$
I tried using Feynman’s trick and introduced the parameter $e^{-\sqrt{1+t^2} x}$, then differentiating with respect to $x$ under the integral. However I don’t see a good way to evaluate the resulting integral. Is there a contour method?
This was an AMM problem ($12145$).
Best Answer
You can let $t = \sinh(x)$ to get \begin{align} \int \limits_{-\infty}^\infty \frac{\cos(t) \sin(\sqrt{1+t^2})}{\sqrt{1+t^2}} \, \mathrm{d} t &= \int \limits_{-\infty}^\infty \cos(\sinh(x)) \sin(\cosh(x)) \, \mathrm{d} x = \int \limits_{-\infty}^\infty \frac{1}{2} \left[\sin(\mathrm{e}^{x}) + \sin(\mathrm{e}^{-x})\right] \, \mathrm{d} x \\ &= \int \limits_{-\infty}^\infty \sin(\mathrm{e}^{x}) \, \mathrm{d} x \stackrel{\mathrm{e}^{x} = u}{=} \int \limits_0^\infty \frac{\sin(u)}{u} \, \mathrm{d} u = \frac{\pi}{2} \end{align} using the famous Dirichlet integral and $\sin(a) \cos(b) = \frac{1}{2} [\sin(a+b)+\sin(a-b)]$.