An integral identity. 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?
 A: 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 
$$
\int_{0}^\infty e^{-y\sqrt{1+t^2}}\cos(xt)\ \mathrm dt = \frac{y}{\sqrt{x^2+y^2}}\int_{0}^\infty e^{-\sqrt{x^2+y^2}\sqrt{1+t^2}}\ \mathrm dt.
$$
