Inverse Laplace transform of $f(s)={\frac{1}{s^{3/2}}}$ using complex integration

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. • $s=0$ 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 $F(1)$ May 16 '19 at 14:21
• thanks for the comment. I revised my question. May 17 '19 at 9:04
• Jul 5 '19 at 16:10

$$s=0$$ 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.

• 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 $(-\infty,0]$ May 16 '19 at 16:10
• This is designed to change from the Bromwich integral to one encircling the nonpositive real axis, May 16 '19 at 16:28
• 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? May 17 '19 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?