(This is problem 31e, Section 2.3, Folland's Real Analysis)
Obviously I know that if $a \in \mathbb{R}$ is such that $a>1$, then $$\int_0^{\infty}e^{-ax}J_0(x)dx=\int_0^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}e^{-ax}}{4^n(n!)^2},$$ but I'm not sure how to proceed. I tried expanding $e^{-ax}$, commuting the sum and the integral (but how would I justify doing that?), but am still stuck.
Any help is appreciated, thanks in advance.
Here's an alternate to the two previous good answers.
Since the integral depends on $a$ define $I(a) = \displaystyle \int_0^{\infty}e^{-ax}J_0(x)dx$. Differentiate this with respect to $a$ to get $I'(a) = \displaystyle \int_0^{\infty} -e^{-ax}xJ_0(x)dx$. Multiply this by $a^2$ and integrate by parts twice, at which point we have
$$a^2 I'(a) = -aI(a) - \int_0^{\infty}e^{-ax}[J'_0(x)+x J''_0(x)]dx$$
which can be rearranged to
$$a^2 I'(a) + aI(a)= - \int_0^{\infty}e^{-ax}[J'_0(x)+x J''_0(x)]dx$$
Add $I'(a)$ to each side, which gives
$$a^2 I'(a) + I'(a) + aI(a)= - \int_0^{\infty}e^{-ax}[x J_0(x) + J'_0(x)+x J''_0(x)]dx = 0$$
since $J_0(x)$ satisfies Bessel's equation, $x J_0(x) + J'_0(x)+x J''_0(x)=0$. You can now solve $a^2 I'(a) + I'(a) + aI(a)=0$, which gives
$$I(a) = \dfrac{c}{\sqrt{1 + a^2}}$$
Using $I(0) =\displaystyle \int_0^{\infty}J_0(x)dx = 1$ gives the desired result.
$$\int_0^\infty x^{2n} e^{-ax}\; dx = a^{-1-2n} (2n)! $$ (look up the Gamma function).
Compare the resulting sum to the binomial series for $(a^2+1)^{-1/2} = a^{-1} (1+a^{-2})^{-1/2}$.
Using the inequality $\binom{2n}{n} \leq (1+1)^{2n} = 4^n$, we have
$$ \left| \frac{(-1)^n}{n!^2} \left( \frac{x}{2} \right)^{2n} \right| \leq \frac{x^{2n}}{(2n)!}. $$
This shows that $\sum_{n=0}^{\infty} \left| \frac{(-1)^n}{n!^2} \left( \frac{x}{2} \right)^{2n} e^{-ax} \right| \leq e^{-(a-1)x}$ and hence by the Fubini's theorem you can interchange the infinite summation and the integral if $a > 1$:
$$ \int_{0}^{\infty} J_0(x)e^{-ax} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!^2} \frac{1}{4^n} \int_{0}^{\infty} x^{2n} e^{-ax} \, dx. \tag{*}$$
Once you have $\text{(*)}$ at your hands, you can safely follow the answer by @Robert Israel. That is, you can utilize the identity
$$ \int_{0}^{\infty} x^{2n}e^{-ax} \, dx = \frac{(2n)!}{a^{2n+1}} $$
together with
$$ \binom{-1/2}{n} = (-1)^n \binom{2n}{n} \frac{1}{4^n} $$
to simplify the RHS of $\text{(*)}$.