# Laplace Transform of Bessel Function of the First Kind/Proof 2

## Theorem

Let $J_n$ denote the Bessel function of the first kind of order $n$.

Then the Laplace transform of $J_n$ is given as:

$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$

## Proof

 $\ds \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + (t^2-\alpha^2)x} s$ $=$ $\ds 0$ Laplace Transform of Bessel's Equation $\ds \frac {\d^2} {\d s^2}\mathcal{L}[x''] - \frac \d {\d s} \mathcal{L}[x'] + \frac {\d^2} {\d s^2} \mathcal{L}[x] - \alpha^2 \mathcal{L} [x]$ $=$ $\ds 0$ setting the initial conditions as $x(0)=1,\ x'(0)=0$ $\ds (s^2+1) \mathcal{L}''[x] + 3 s \mathcal {L}'[x] + (1 - \alpha^2) \mathcal{L}[x]$ $=$ $\ds 0$ Make the following change of variable $u = \sqrt {s^2 + 1} \mathcal {L} [x]$ $\ds u'' \sqrt {s^2 + 1} + \dfrac s {\sqrt {s^2 + 1} } u'$ $=$ $\ds \frac {\alpha^2 u} {\sqrt {s^2 + 1} }$ $\ds (u' \sqrt {s^2 + 1} )'$ $=$ $\ds \frac {\alpha^2 u} {\sqrt {s^2 + 1} }$ multiplying both sides by $u' \sqrt {s^2 + 1}$ $\ds \frac 1 2 ( (u' \sqrt {s^2 + 1} )^2)'$ $=$ $\ds \frac 1 2 \alpha^2 (u^2)'$ $\ds u' \sqrt {s^2 + 1}$ $=$ $\ds -\alpha u$ Constant of Integration removed by the Final Value Theorem of Laplace Transform $\ds \int \frac 1 u \rd u$ $=$ $\ds \int -\frac \alpha {\sqrt {s^2 + 1} } \rd s$ $\ds \map \ln u$ $=$ $\ds \alpha \ln (\sqrt {s^2 + 1} - s)$ $\ds u$ $=$ $\ds (\sqrt {s^2 + 1} - s)^\alpha$ reverting the substitution $\ds \mathcal{L} [x]$ $=$ $\ds \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }$

Hence:

 $\ds \mathcal{L}[J_\alpha (t)](s)$ $=$ $\ds \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }$

$\blacksquare$