Laplace Transform of Cosine of Root over Root

Theorem

$\laptrans {\dfrac {\cos \sqrt t} {\sqrt t} } = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }$

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

Proof 1

Let $\map f t = \sin \sqrt t$.

Then:

 $\ds \map {f'} t$ $=$ $\ds \dfrac {\cos \sqrt t} {2 \sqrt t}$ $\ds \map f 0$ $=$ $\ds 0$

So:

 $\ds \laptrans {\map {f'} t}$ $=$ $\ds \dfrac 1 2 \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }$ $\ds$ $=$ $\ds s \map F s - \map f 0$ Laplace Transform of Derivative $\ds$ $=$ $\ds \dfrac {\sqrt \pi} {2 s^{1/2} } \map \exp {-\dfrac 1 {4 s} }$ Laplace Transform of Sine of Root $\ds \leadsto \ \$ $\ds \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }$ $=$ $\ds \sqrt {\dfrac \pi s} \map \exp {-\dfrac 1 {4 s} }$

$\blacksquare$