Inverse Laplace transorm of $\sqrt s/(s^2+1)$

Question

Can anyone calculate inverse Laplace of $$F(s) = \frac{\sqrt{s}}{{s^2+1}} $$ ?

Answer 1

Hint.

$$ \frac{\sqrt{s}}{{s^2+1}} = \frac{\sqrt{s}}{s}\frac{s}{s^2+1} $$

and

$$ \mathcal{L}^{-1}\left[\frac{1}{\sqrt{s}}\right] = \frac{\phi(t)}{\sqrt{\pi t}},\ \ \ \mathcal{L}^{-1}\left[\frac{s}{s^2+1}\right] =\phi(t)\cos t $$

with $\phi(t)$ the Heaviside unit step function.

Answer 2

With CAS help aka. Mathematica:

$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{s}}{s^2+1}\right](t)=\frac{2 \sqrt{t} \, _1F_2\left(1;\frac{3}{4},\frac{5}{4};-\frac{t^2}{4}\right)}{\sqrt{\pi }}$$ where:$\, _1F_2\left(1;\frac{3}{4},\frac{5}{4};-\frac{t^2}{4}\right)$ is the generalized hypergeometric function.

MMA code:

HoldForm[InverseLaplaceTransform[Sqrt[s]/(s^2 + 1), s, t] ==
(2 Sqrt[t]
HypergeometricPFQ[{1}, {3/4, 5/4}, -((a t^2)/4)])/Sqrt[\[Pi]] // TeXForm

For general:

$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{c s}}{a s^2+b}\right](t)=\frac{2 \sqrt{c} \sqrt{t} \, _1F_2\left(1;\frac{3}{4},\frac{5}{4};-\frac{b t^2}{4 a}\right)}{a \sqrt{\pi }}$$