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?

  • $\begingroup$ 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. $\endgroup$
    – Maxim
    Oct 20, 2019 at 13:03
  • $\begingroup$ I have edited the question to indicate it. $\endgroup$
    – BJKShah
    Oct 20, 2019 at 13:48

1 Answer 1


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:

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

  • $\begingroup$ What type of function is $F3$ ? And what is CAS? $\endgroup$
    – BJKShah
    Oct 20, 2019 at 10:54
  • 1
    $\begingroup$ $ _1F_3$ is generalized hypergeometric function. CAS like Mathematica: computer algebra systems . $\endgroup$ Oct 20, 2019 at 10:59

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