# 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

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

Hence:

 $\ds \map {\laptrans {\map {J_\alpha} t} } s$ $=$ $\ds \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }$

$\blacksquare$