Laplace transform of the Bessel Function of order 1

Question

Guys can you help me prove that $$\mathcal{L}\left\{\frac{t}{\alpha}\,J_1(\alpha t)\right\} = \frac{1}{(s^2 + \alpha^2)^\frac{3}{2}}$$ ?

Answer 1

By the series definition of Bessel functions of the first kind: $$ z\cdot J_1(z) = \frac{1}{2}\sum_{n\geq 0}\frac{z^{2n+2}(-1)^n}{4^n n!(n+1)!} \tag{1} $$ and since $\mathcal{L}(z^{2n+2})=\frac{(2n+2)!}{s^{2n+3}}$ we have: $$ \mathcal{L}\left(z\cdot J_1(z)\right) = \frac{1}{2}\sum_{n\geq 0}\frac{(-1)^n (2n+2)!}{4^n n!(n+1)! s^{2n+3}}=\frac{1}{(1+s^2)^{3/2}}\tag{2} $$ by well known Taylor series. The claim follows by rescaling.

Answer 2

One way to do this is to use the Bessel differential equation. By definition, $J_n(\alpha t)$ solves $$ t^2f''(t) + t f'(t) + (\alpha^2t^2 - n^2)f(t) = 0 $$ The Laplace transform of $J_1(\alpha t)$ will then solve the Laplace transform of this equation, which we can obtain through identities about how derivatives transform: $$ (\alpha^2 + s^2)F''(s) +3F'(s) + (1-n^2)F(s) = 0 $$ (You should work out how to derive this from the identities of Laplace transforms.)

Now this equation isn't any simpler than the previous unless $n = 1$, which fortuitously is the exact scenario we're looking at. In this case, we have $$ (\alpha^2 + s^2)F''(s) +3F'(s) = 0 \Longrightarrow F(s) = C_2 - \frac{C_1 s}{\alpha\sqrt{s^2 + \alpha^2}} $$ So $\mathcal{L}[J_1(\alpha t)] = C_2 - C_1 s/(\alpha\sqrt{s^2 + \alpha^2})$, for some constants $C_1$ and $C_2$. However, that's not quite what we're looking for. We want $\mathcal{L}[(t/\alpha)J_1(\alpha t)]$. But we can once again use derivative identities to get $$ \mathcal{L}\left[\frac{t}{\alpha}J_1(\alpha t)\right] = -\frac{1}{\alpha}\frac{d}{ds}\mathcal{L}\left[J_1(\alpha t)\right] = \frac{1}{\alpha^2}\frac{d}{ds}\left(\frac{C_1s}{\sqrt{s^2+\alpha^2}}\right) = \frac{C_1}{(s^2+\alpha^2)^{3/2}} $$ I'll leave the proof that $C_1 = 1$ to you.