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}$$ \int_{0}^{\infty} e^{-ax} x^{2n}dx$ = $ \Sigma_{0}^{\infty}\frac{(-1)^n }{4^n (n!)^2} \times (2n)!$
I am stuck here. Any help is appreciated.
$$\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}}. $$
The series is just $J_0(x)$, and thus your integral is $$ \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} $$
