Laplace Transform of Bessel Function of the First Kind of Order One
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Theorem
Let $J_1$ denote the Bessel function of the first kind of order $1$.
Then the Laplace transform of $J_1$ is given as:
- $\laptrans {\map {J_1} t} = \dfrac {\sqrt {s^2 + 1} - s} {\sqrt {s^2 + 1} }$
Proof
| \(\ds \map {J_0'} t\) | \(=\) | \(\ds -\map {J_1} t\) | Derivative of $x^n$ by Bessel Function of the First Kind of Order $n$ of $x$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \laptrans {\map {J_1} t}\) | \(=\) | \(\ds -\laptrans {\map {J_0'} t}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {s \laptrans {\map {J_0} t} - \map {J_0} 0}\) | Laplace Transform of Derivative | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {s \laptrans {\map {J_0} t} - 1}\) | Bessel Function of the First Kind of Order Zero | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - s \dfrac 1 {\sqrt {s^2 + 1} }\) | Laplace Transform of Bessel Function of the First Kind of Order Zero | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\sqrt {s^2 + 1} - s} {\sqrt {s^2 + 1} }\) | rearranging |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Bessel Functions: $35$