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 $f$.

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 $

$=$

$\ds s \map F s - \map f 0$

$\ds $

$=$

$\ds \dfrac {\sqrt \pi} {2 s^{1/2} } \map \exp {-\dfrac 1 {4 s} }$

$\ds \leadsto \ \ $

$\ds \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }$

$=$

$\ds \sqrt {\dfrac \pi s} \map \exp {-\dfrac 1 {4 s} }$

$\blacksquare$

1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $5$