# Laplace Transform of Sine of Root/Proof 2

## Theorem

$\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$

where $\laptrans f$ denotes the Laplace transform of the function $f$.

## Proof

Let $\map y t := \sin \sqrt t$.

Differentiating twice with respect to $t$, we get:

$(1): \quad 4 t y'' + 2 y'' + y = 0$

Let $\map Y s = \laptrans {\map t y}$ be the Laplace transform of $y$.

Then taking the Laplace transform of $(1)$:

 $\ds -4 \map {\dfrac \d {\d s} } {\laptrans {\map {y''} t} } + 2 \laptrans {\map {y'} t} + \laptrans {\map y t}$ $=$ $\ds 0$ Derivative of Laplace Transform $\ds \leadsto \ \$ $\ds -4 \map {\dfrac \d {\d s} } {s^2 \, \map Y s - s \, \map y 0 - \map {y'} 0} + 2 \paren {s \, \map Y s - \map y 0} + \map Y s$ $=$ $\ds 0$ Laplace Transform of Derivative, Laplace Transform of Second Derivative $\ds \leadsto \ \$ $\ds 4 s^2 \, \map {Y'} s - \paren {6 s - 1} \map Y s$ $=$ $\ds 0$ simplifying $\ds \leadsto \ \$ $\ds \map Y s$ $=$ $\ds \dfrac c {s^{3/2} } \, \map \exp {\dfrac 1 {4 s} }$ solving the differential equation

For small $t$, we have:

$\sin \sqrt t \sim \sqrt t$

and:

$\laptrans {\sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} }$

For large $t$, we have:

$Y \sim \dfrac c {2 s^{3/2} }$

Hence by comparison:

$c = \dfrac {\sqrt \pi} 2$

Hence:

$\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$

$\blacksquare$