Application of Fubini-Tonelli's Theorem on function $\frac{2}{\pi}e^{-ax}\cos(x\cos{\theta})$ The question asks me to prove that $$\int_0^\infty J(x)e^{-ax}dx=\frac{1}{\sqrt{1+a^2}},$$ where $a>0$ and $J(x)=\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos{\theta})d\theta.$
I started off by considering the function
$$f(x,\theta)=\frac{2}{\pi}e^{-ax}\cos(x\cos{\theta})\boldsymbol{1}_{\theta\in(0,\frac{\pi}{2})}$$ in the region $(x,\theta)\in(0,\infty)\times (0,\infty).$
Now $\vert f(x,\theta)\vert\leq \frac{2}{\pi}e^{-ax},$ so $$\int_0^\infty\left(\int_0^\infty\vert f(x,\theta)\vert dx\right)d\theta<\infty,$$ which means that by Fubini's we can do the integration in any order.
I don't think $\cos(x\cos{\theta})$ is integrable in terms of $\theta,$ so I'm thinking I should integrate $e^{-ax}\cos(x\cos{\theta})$ in terms of $x.$ But the integration of this looks quite horrible, so I was wondering (a) if my method is correct, and (b) is there a neater, more efficient way I could solve this problem?
 A: Let be 
$$I=\int_0^\infty J(x)e^{-ax}\mathrm dx$$
and 
$$
J(x)=\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos{\theta})\mathrm d\theta
$$
then we have
$$
I=\int_0^\infty \underbrace{\left(\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos{\theta})\mathrm d\theta\right)}_{J(x)}e^{-ax}\mathrm dx=\frac{2}{\pi}\int_0^{\pi/2}\underbrace{\left(\int_0^{\infty}\cos(x\cos{\theta})e^{-ax}\mathrm d x\right)}_{A(a,\cos\theta)}\mathrm d\theta
$$
as you stated.
Observe that
$$
A(a,b)=\int_0^\infty e^{-ax}\cos(b x)\mathrm d x=-\frac{e^{-ax}}{a^2+b^2}\left(a\cos(b x)-b\sin(bx)\right)\Big|_0^\infty=\frac{a}{a^2+b^2}\qquad \text{for }a>0
$$
that is $A(a,b)=\mathcal L\{\cos(bx)\}(a)$ is the Laplace transform of $\cos(bx)$.
So for $b=\cos\theta$
we have
$$\begin{align}
I&=\frac{2}{\pi}\int_0^{\pi/2}A(a,\cos\theta)\mathrm d\theta
=\frac{2}{\pi}\int_0^{\pi/2}\frac{a}{a^2+\cos^2\theta}\mathrm d\theta\\
&=\frac{2}{\pi}\left[{\frac{1}{\sqrt{1+a^2}}\tan^{-1}\left(\frac{a \tan x}{\sqrt{1+a^2}}\right)}\right]_0^{\pi/2}=\frac{2}{\pi}\frac{\frac{\pi}{2}}{\sqrt{1+a^2}}=\frac{1}{\sqrt{1+a^2}}
\end{align}
$$
