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

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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:

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

$\blacksquare$