TITLE: Lipshitz Integral for $a=0$
QUESTION [5 upvotes]: I knew that this, $$\displaystyle{\int_0^\infty e^{-ax}J_0(bx)dx=\frac{1}{\sqrt{a^2+b^2}}},$$ holds for $a>0$ but, in an exercise from Arfken, it said that this holds for $a\geq0$.
How can I prove that?

REPLY [2 votes]: Another way. We know that:
$$ J_0(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\exp\left(i x \sin(t)\right)\,dt \tag{1}$$
hence by assuming $a>0$ we may apply Fubini's theorem to get:
$$ \begin{eqnarray*}\int_{0}^{+\infty}e^{-ax}J_0(x)\,dx &=& \frac{1}{2\pi}\int_{-\pi}^{\pi}\int_{0}^{+\infty}\exp\left(i x \sin(t)-ax\right)\,dx\,dt\\&=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{dt}{a-i\sin(t)}\\&=&\frac{1}{\pi}\int_{0}^{\pi}\frac{a\,dt}{a^2+\sin^2(t)}\\&=&\frac{2a}{\pi}\int_{0}^{\pi/2}\frac{dt}{a^2+\sin^2(t)}\\&=&\frac{2a}{\pi}\int_{0}^{+\infty}\frac{du}{a^2(1+u^2)+u^2}\\&=&\color{red}{\frac{1}{\sqrt{1+a^2}}}\tag{2}\end{eqnarray*}$$
through the substitution $t=\arctan(u)$. Now $J_0(x)$ is not a Lebesgue-integrable function over $\mathbb{R}^+$, since $|J_0(x)|$ decays like $\frac{1}{\sqrt{x}}$, but it is improperly Riemann-integrable over $\mathbb{R}^+$. So assuming that the given identity is stated "in the Riemann way", $(2)$ holds also for $a=0$. The two-parameters identity of yours can be deduced from $(2)$ through a simple change of variable.