integration of $\int_{0}^{\infty} e^{-ax}\Sigma_{0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}dx $ I need to compute the following integral for $a>1$
$\int_{0}^{\infty} e^{-ax}\Sigma_{0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}dx $ 
My attempt 
$\int_{0}^{\infty} e^{-ax}\Sigma_{0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}dx$  = $ \Sigma_{0}^{\infty}\frac{(-1)^n }{4^n (n!)^2}$$ \int_{0}^{\infty}  e^{-ax} x^{2n}dx$= $ \Sigma_{0}^{\infty}\frac{(-1)^n }{4^n (n!)^2} \times (2n)!$
I am stuck here 
 A: $$\int_{0}^{\infty} e^{-ax}\sum_{0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}dx=  \sum_{0}^{\infty}\frac{(-1)^n }{4^n (n!)^2}\frac{(2n)!}{a^{2n+1}}=\frac{1}{a}\sum_{0}^{\infty}{-\frac{1}{2} \choose n}\frac{1}{a^{2n}}=\frac{1/a}{\sqrt{1+1/a^2}}=\frac{1}{\sqrt{1+a^2}}.
$$
A: The series is just $J_0(x)$, and thus your integral is
$$
\int_0^{+\infty}e^{-a x}J_0(x)\,dx=\frac{1}{\sqrt{1+a^2}}.
$$
Update
Since this was the integral you actually wanted to compute, let me add how it can be done using the integral formula of $J_0$. You already have an answer using the power series. 
I assume that you have the integral formula
$$
J_0(x)=\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos\theta)\,d\theta
$$
and that you are aware of (integrate by parts two times)
$$
\int_0^{+\infty} e^{-ax}\cos b x\,dx=\frac{a}{a^2+b^2}.
$$
Then (please try to fill in the details in the steps below), by changing order of integration, and invoking the integrals above
$$
\begin{aligned}
\int_0^{+\infty}e^{-a x}J_0(x)\,dx & =  \frac{2}{\pi}\int_0^{+\infty}e^{-ax}\int_0^{\pi/2}\cos(x\cos\theta)\,d\theta\,dx\\
&=\frac{2}{\pi}\int_0^{\pi/2}\int_0^{+\infty}e^{-ax}\cos(x\cos\theta)\,dx\,d\theta\\
&=\frac{2}{\pi}\int_0^{\pi/2}\frac{a}{a^2+\cos^2\theta}\,d\theta\\
&=\frac{2}{\pi}\int_0^{\pi/2}\frac{a}{a^2\tan^2\theta+1+a^2}\frac{1}{\cos^2\theta}\,d\theta\\
&=\frac{2}{\pi}\biggl[\frac{1}{\sqrt{1+a^2}}\arctan\Bigl(\frac{a\tan\theta}{\sqrt{1+a^2}}\Bigr)\biggr]_0^{\pi/2}\\
&=\frac{1}{\sqrt{1+a^2}}.
\end{aligned}
$$
