Laplace Transform of Bessel Function of the First Kind of Order Zero/Proof 3

Proof
By definition of Bessel function of the first kind, $\map {J_0} t$ satisfies Bessel's equation:

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:

from which it follows that:
 * $\map {J'_0} 0 = 0$

Then from Derivative of Laplace Transform:

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 $y = \laptrans {\map {J_0} t}$.

Now we have that:

and:
 * $\displaystyle \lim_{t \mathop \to 0} \map {J_0} t = 1$

Thus by the Initial Value Theorem of Laplace Transform:
 * $c = 1$

and so:
 * $\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$