# Laplace Inverse of the problem.

What is the Laplace Inverse of the given question? $$\frac{\sqrt{4+s^3}}{s^3}$$ I tried solving it by expanding $$\sqrt{1+\frac{s^3}{4}}$$ but the terms will not have Laplace Inverse. If I expand it like $$\sqrt{1+\frac{4}{s^3}}$$i.e.$$s^{-3/2}* \sqrt{1+\frac{4}{s^3}}$$ It will still have many terms of $$t^n$$ . Are there other methods?

• How do you define the square root? If $s^a$ is the principal branch and you take $s^{-3} \sqrt {s^3 + 4}$, the function doesn't have an ILT because of the branch cuts in the right half-plane. If you take $s^{-3/2} \sqrt {1 + 4 s^{-3}}$, that function does have an ILT. – Maxim Oct 20 '19 at 13:03
• I have edited the question to indicate it. – BJKShah Oct 20 '19 at 13:48

With CAS help I have:

$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{4+s^3}}{s^3}\right](t)=\frac{2 \sqrt{t} \, _1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{4 t^3}{27}\right)}{\sqrt{\pi }}$$

For general a:

$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{a+s^3}}{s^3}\right](t)=\\\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } a^n \left(s^3\right)^{-\frac{1}{2}-n} \binom{\frac{1}{2}}{n}\right](t)=\\\sum _{n=0}^{\infty } \mathcal{L}_s^{-1}\left[a^n \left(s^3\right)^{-\frac{1}{2}-n} \binom{\frac{1}{2}}{n}\right](t)=\sum _{n=0}^{\infty } \frac{a^n \sqrt{\pi } t^{\frac{1}{2}+3 n}}{2 \Gamma \left(\frac{3}{2}-n\right) \Gamma (1+n) \Gamma \left(\frac{3}{2}+3 n\right)}=\\\frac{2 \sqrt{t} \, _1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{1}{27} \left(a t^3\right)\right)}{\sqrt{\pi }}$$

• What type of function is $F3$ ? And what is CAS? – BJKShah Oct 20 '19 at 10:54
• $_1F_3$ is generalized hypergeometric function. CAS like Mathematica: computer algebra systems . – Mariusz Iwaniuk Oct 20 '19 at 10:59