P. Dravek and G. Holubova, Elements of Partial Differential Equations, Section 3.4 Exercise 21 Part c):
Fix $x\in \mathbb{R}$ and $y>0$. Show that: $$\int_{0}^\infty e^{-y\sqrt{1+t^2}}\cos(xt)dt=\frac{y}{\sqrt{x^2+y^2}}\int_{0}^\infty e^{-\sqrt{1+t^2}\sqrt{x^2+y^2}}dt.$$
How to solve? Change of variable? Maybe use polar coordinates?
We may write $$\begin{align*} 2\int_{0}^\infty e^{-y\sqrt{1+t^2}}\cos(xt)\ \mathrm dt = \int_{-\infty}^\infty e^{-y\sqrt{1+t^2}+ixt}\mathrm dt &=\int_{-\infty}^\infty e^{-y\cosh u+ix\sinh u}\cosh u\ \mathrm du \end{align*}$$ by making substitution $t = \sinh u$. Let $\alpha\in (-\frac\pi{2},\frac\pi{2})$ be such that $$ \cos\alpha =\frac{y}r=\frac{y}{\sqrt{x^2+y^2}}, \quad \sin\alpha =\frac{-x}r= \frac{-x}{\sqrt{x^2+y^2}}. $$ Then, we have $$\begin{align*} \frac{y\cosh u-ix\sinh u}{r}&=\cos \alpha\cosh u +i\sin \alpha\sinh u\\&=\cosh(i\alpha)\cosh u +\sinh(i\alpha)\sinh u\\ &=\cosh(u+i\alpha). \end{align*}$$ It follows by Cauchy's integral formula $$\begin{align*} \int_{-\infty}^\infty e^{-y\cosh u+ix\sinh u}\cosh u\ \mathrm du&=\int_{-\infty}^\infty e^{-r\cosh(z+i\alpha)}\cosh z\ \mathrm dz\\ &=\int_{-\infty}^\infty e^{-r\cosh z}\cosh (z-i\alpha)\ \mathrm dz\\ &=\int_{-\infty}^\infty e^{-r\cosh u}\left(\cos\alpha \cosh u -i\sin\alpha\sinh u\right)\ \mathrm du\\ &=\int_{-\infty}^\infty e^{-r\sqrt{1+t^2}}\left(\cos\alpha -i\sin\alpha\frac{t}{\sqrt{1+t^2}}\right)\ \mathrm dt\\ &=\cos\alpha \int_{-\infty}^\infty e^{-r\sqrt{1+t^2}}\ \mathrm dt\\ &=\frac{2y}{\sqrt{x^2+y^2}}\int_{0}^\infty e^{-\sqrt{x^2+y^2}\sqrt{1+t^2}}\ \mathrm dt. \end{align*}$$ Hence $$ \int_{0}^\infty e^{-y\sqrt{1+t^2}}\cos(xt)\ \mathrm dt = \frac{y}{\sqrt{x^2+y^2}}\int_{0}^\infty e^{-\sqrt{x^2+y^2}\sqrt{1+t^2}}\ \mathrm dt. $$