Laplace Transform of Bessel Function of the First Kind of Order Zero/Proof 1

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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:

\(\ds \laptrans {\map {J_0} t}\) \(=\) \(\ds \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \laptrans {t^{2 k} }\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \dfrac {\paren {2 k}!} {s^{2 k + 1} }\) Laplace Transform of Positive Integer Power
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }\) Definition of Binomial Coefficient: $\dbinom {2 k} k = \dfrac {\paren {2 k}!} {\paren {k!}^2}$


Then:

\(\ds \dfrac 1 {\sqrt {s^2 + 1} }\) \(=\) \(\ds \dfrac 1 {\sqrt {s^2} \sqrt {1 + \paren {1 / s}^2} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 s \paren {1 + \paren {\dfrac 1 s}^2}^{-1/2}\) Definition of Rational Power
\(\ds \) \(=\) \(\ds \dfrac 1 s \sum_{k \mathop = 0}^\infty \dbinom {-1/2} k \paren {\dfrac 1 s}^{2 k}\) General Binomial Theorem
\(\ds \) \(=\) \(\ds \dfrac 1 s \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {4^k} \dbinom {2 k} k \paren {\dfrac 1 s}^{2 k}\) Binomial Coefficient of Minus Half
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \dfrac 1 {2^{2 k} } \dbinom {2 k} k \paren {\dfrac 1 {s^{2 k + 1} } }\) rearranging, and bringing $\dfrac 1 s$ inside the summation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }\) further rearrangement

The two expressions match, and the result follows.

$\blacksquare$