Theorem: Laplace Transform of Bessel Function of the First Kind

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:
By definition of Bessel function of the first kind, $\map {J_0} t$ satisfies Bessel's equation:
{{begin-eqn}}
{{eqn | l = t^2 \, \map {\dfrac {\d^2} {\d t^2} } {\map {J_0} t} + t \, \map {\dfrac \d {\d t} } {\map {J_0} t} + \paren {t^2 - 0^2} {\map {J_0} t}
      | r = 0
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = t \, {\map {J_0' '} t} + \map {J_0'} t + t \, \map {J_0} t
      | r = 0
      | c = 
}}
{{end-eqn}}
From Laplace Transform of Derivative:
:$\laptrans {\map {J_0'} t} = s \laptrans {\map {J_0} t} - \map {J_0} 0$
and from Laplace Transform of Second Derivative:
:$\laptrans {\map {J_0' '} t} = s^2 \laptrans {\map {J_0} t} - s \, \map {J_0} 0 - \map {J_0'} 0$
From Bessel Function of the First Kind of Order $0$:
:$\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$
from which it follows immediately that:
:$\map {J_0} 0 = 1$
From Derivative of Bessel Function of the First Kind of Order $0$ we have:
{{ProofWanted|Derivative of Bessel Function of the First Kind of Order 0}}
from which it follows that:
:$\map {J'_0} 0 = 0$
Then from Derivative of Laplace Transform:
{{begin-eqn}}
{{eqn | l = \laptrans {t \, \map {J_0' '} t}
      | r = -\map {\dfrac \d {\d s} } {\map {J_0' '} t}
      | c = 
}}
{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s \, \map {J_0} 0 - \map {J_0'} 0}
      | c = 
}}
{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s}
      | c = $\map {J_0} 0 = 1$ and $\map {J'_0} 0 = 0$
}}
{{end-eqn}}
and:
:$\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$
Thus by taking the Laplace transform of $(1)$ we have:
Let $y = \laptrans {\map {J_0} t}$.
{{begin-eqn}}
{{eqn | l = \laptrans {t \, {\map {J_0' '} t} + \map {J_0'} t + t \, \map {J_0} t}
      | r = \laptrans 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\map {\dfrac \d {\d s} } {s^2 \, \laptrans {\map {J_0} t} - s} + \paren {s \laptrans {\map {J_0} t} - \map {J_0} 0} - \map {\dfrac \d {\d s} } {\map {J_0} t}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\map {\dfrac \d {\d s} } {s^2 y - s} + \paren {s y - 1} - \dfrac {\d y} {\d s}
      | r = 0
      | c = setting $y = \laptrans {\map {J_0} t}$ and noting that $\map {J_0} 0 = 1$ from above
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d y} {\d s}
      | r = -\dfrac {s y} {s^2 + 1}
      | c = rearranging and simplifying
}}
{{eqn | ll= \leadsto
      | l = \int \dfrac {\d y} y
      | r = -\int \dfrac {s \rd s} {s^2 + 1}
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac c {\sqrt {s^2 + 1} }
      | c = for some constant $c$
}}
{{end-eqn}}
Now we have that:
{{begin-eqn}}
{{eqn | l = \lim_{s \mathop \to \infty} s \, \map y s
      | r = \lim_{s \mathop \to \infty} \dfrac {c s} {\sqrt {s^2 + 1} }
      | c = 
}}
{{eqn | r = c
      | c = 
}}
{{end-eqn}}
and:
:$\ds \lim_{t \mathop \to 0} \map {J_0} t = 1$
Thus by the Initial Value Theorem of Laplace Transform:
:$c = 1$
and so:
:$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
{{qed}}