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

amWhy
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peperasmenos
  • 13

2 Answers2

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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.

Cesareo
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  • the problem is that the integral which shows up in Convolutions type cant be solved – peperasmenos Feb 08 '20 at 16:02
  • It gives $$\sqrt{2} \left(C\left(\sqrt{\frac{2}{\pi }} \sqrt{t}\right) \cos (t)+S\left(\sqrt{\frac{2}{\pi }}\sqrt{t}\right) \sin (t)\right)$$ with $C,S$ the corresponding Fresnel functions. – Cesareo Feb 08 '20 at 16:20
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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:

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

Mariusz Iwaniuk
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