Showing that $\int_0^{\infty} J_n(bx) dx = 1/b$

Question

I'm trying to prove that

$$ \int_0^{\infty}J_n(bx)dx=\dfrac{1}{b}$$

when $b > 0$ and $n$ is a nonnegative integer. From the recurrence relations I arrived at the identity

$$\int_0^{\infty}J_{n+1}(bx)dx=\int_0^{\infty}J_{n-1}(bx)dx,$$

but I do not see how to use this.

Answer 1

As proved in this other question, the $J$-Bessel function is normalized so that $$ \int_0^\infty J_n(x) dx = 1 $$ for $n \in \mathbb{N}_{\geq 0}$. This implies your claim after a short substitution.