I want to find the inverse Laplace transform of $$f(s)={\frac{1}{s^{3/2}}}$$ Refer to the Laplace transform table, and I found that the result is $$F(t)=2\sqrt{\frac{t}{\pi}}$$ But I do not know how to get this result. I tried to use the Bromwich integral $$F(t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{1}{s^{3/2}}e^{st}\,ds$$ My progress so far has been stunted by the fact that we have a branch point at s=0. The contour should be like this, but I don't know how to perform the integration.

Any help is appreciated.

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  • $\begingroup$ $s=0$ is a branch point of $f(s)$ not an isolated singularity (thus not an essential singularity). For $t > 0$ use the change of variable $u = st$ to express $F(t)$ in term of $F(1)$ $\endgroup$
    – reuns
    Commented May 16, 2019 at 14:21
  • $\begingroup$ thanks for the comment. I revised my question. $\endgroup$ Commented May 17, 2019 at 9:04
  • $\begingroup$ See this. math.stackexchange.com/questions/1343764/laplace-inverse/… $\endgroup$
    – Ron Gordon
    Commented Jul 5, 2019 at 16:10

2 Answers 2


$s=0$ is not an essential singularity. It is a branch point. Choose a branch to calculate your integral, for example, choose branch $-\pi<\arg z<\pi$ and integrate along the contour that made up of:

  1. straight line from $c-iR$ to $c+iR$
  2. along large quarter circle to approximately $(c-R)+\varepsilon i$
  3. straight line to $\varepsilon i$
  4. right half of circle $\lvert z\rvert=\varepsilon$, to $-\varepsilon i$
  5. straight line to approximately $(c-R)-\varepsilon i$
  6. another large quarter circle to $c-iR$

Can you finish from here? Be careful when you take square-root.

  • $\begingroup$ I don't see what you are doing. With the principal branch of $s^{-3/2}$ and $c > 0$ for $t > 0, \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}s^{-3/2}e^{st}\,ds=\frac{1}{2\pi i}\int_{ct-i\infty}^{ct+i\infty}(u/t)^{-3/2}e^{u}\,d(u/t)= t^{1/2} F(1)$. To evaluate $F(1)$ shift the contour to the one enclosing $(-\infty,0]$ $\endgroup$
    – reuns
    Commented May 16, 2019 at 16:10
  • $\begingroup$ This is designed to change from the Bromwich integral to one encircling the nonpositive real axis, $\endgroup$ Commented May 16, 2019 at 16:28
  • $\begingroup$ Thanks for the reply. I revised my question. So the contour should be like this, but I don't know how to perform the integration. Would you like to help me further? $\endgroup$ Commented May 17, 2019 at 9:09

Haven't studied about the Bromwich integral yet but wouldn't it be sufficient to use the identity:

$L(t^n) = \dfrac{\Gamma(n+1)}{s^{n+1}}$

with $n = \dfrac{1}{2}$ and deduce hence?


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