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

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Theorem

Let $J_0$ denote the Bessel function of the first kind of order $0$.


Then the Laplace transform of $J_0$ is given as:

$\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$


Proof

From Bessel Function of the First Kind of Order Zero:

\(\ds \map {J_0} t\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}\)
\(\ds \) \(=\) \(\ds 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb\)


Hence:

\(\ds \laptrans {\map {J_0} t}\) \(=\) \(\ds \dfrac 1 s - \dfrac 1 {2^2} \dfrac {2!} {s^3} + \dfrac 1 {2^2 4^2} \dfrac {4!} {s^5} - \dfrac 1 {2^2 4^2 6^2} \dfrac {6!} {s^7} + \dotsb\) Laplace Transform of Positive Integer Power
\(\ds \) \(=\) \(\ds \dfrac 1 s \paren {1 - \dfrac 1 2 \paren {\dfrac 1 {s^2} } + \dfrac {1 \times 3} {2 \times 4} \paren {\dfrac 1 {s^4} } - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {\dfrac 1 {s^6} } + \dotsb}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 s \paren {\paren {1 - \dfrac 1 {s^2} } ^{-1/2} }\) General Binomial Theorem
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {s^2 + 1} }\) simplifying

$\blacksquare$


Sources

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