Infinite Integration of Bessel Function of the First Order and Zero Degree

Question

I want to prove that $$\int_0^\infty J_0(bx)dx=\frac1b \tag{1} $$ What I have found online is that you take the Laplace Transform of the Bessel Function and we get $$\int_0^\infty e^{-ax}J_0(bx)dx=\frac1{\sqrt{a^2+b^2}} \tag{2} $$ and we set $a=0$. However, for the derivation of the above result I have assumed that $a>0$, so that during the integration on the infinite end the integrant goes to zero. $$ I=\frac1{2\pi}\int_0^\infty dx\int_{-\pi}^\pi d\theta \cdot e^{(ib\sin \theta-a)x}=\frac1{2\pi}\int_{-\pi}^\pi\frac{d\theta}{a-ib\sin\theta}=\cdots$$ For the proof of $(1)$ we have to take $a=0$. How can we justify this substitution or even the $a\rightarrow ia$?