# 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

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

Hence:

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

$\blacksquare$