I want to find the laplace inverse of
$$s^{-3/2}$$
the steps given in the solution manual are as follows:
$$\frac{2}{\sqrt\pi}\frac{\sqrt\pi}{2s^{3/2}}=2\sqrt{\frac t\pi}$$
I know the first part $\frac2{\sqrt\pi}$ is obtained using the gamma function $\Gamma(\frac32)$, but not quite sure how the rest is obtained.
On
To find the inverse LT, we may use the definition, which requires integration in the complex plane:
$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{s t} $$
where $C$ is the following contour:
We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.
On $C_2$, the real part of the argument of the exponential is $R t \cos{\theta}$, where $\theta \in [\pi/2,\pi)$. Clearly, $\cos{\theta} < 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$ as:
$$R \int_{\pi/2}^{\pi} d\theta R^{-3/2} e^{R t \cos{\theta}} = R^{-1/2} \int_0^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \le \frac{\pi}{2 R^{3/2} t} $$
On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$. This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.
On $C_4$, The integral behaves as
$$i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \epsilon^{-3/2} e^{-i 3 \phi/2} e^{\epsilon e^{i \phi} t} = -i 4 \epsilon^{-1/2} + O \left ( \epsilon^{1/2} \right )$$
While this is singular as $\epsilon \to 0$, this divergence will cancel as follows. On $C_3$, we have $z=e^{i \pi} x$ and the integral there is
$$e^{i \pi} \int_R^{\epsilon} dx \, x^{-3/2} e^{-i 3 \pi/2} e^{-x t} $$
On $C_5$, we have $z=e^{-i \pi} x$ and the integral there is
$$e^{-i \pi} \int_R^{\epsilon} dx \, x^{-3/2} e^{i 3 \pi/2} e^{-x t} $$
Combining the two integrals, we get, as $R \to \infty$,
$$\begin{align}i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \, e^{-x t} &= i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \, \left ( e^{-x t}-1 \right ) + i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2}\\ &= i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \, \left ( e^{-x t}-1 \right ) + i 4 \epsilon^{-1/2} \end{align}$$
As stated, the divergences cancel and the contour integral is equal to, as $\epsilon \to 0$:
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{s t} -i 2 \int_{0}^{\infty} dx \, x^{-3/2} \, \left ( e^{-x t}-1 \right )$$
Note that the 2nd integral indeed converges. By Cauchy's theorem, the contour integral is zero. Thus, the ILT is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{s t} = \frac1{\pi} \int_{0}^{\infty} dx \, x^{-3/2} \, \left ( 1-e^{-x t} \right )$$
The integral on the right is not too bad to evaluate:
$$\begin{align} \int_{0}^{\infty} dx \, x^{-3/2} \, \left ( 1-e^{-x t} \right ) &= 2 \int_0^{\infty} dx \, \frac{1-e^{-t x^2}}{x^2} \\ &= \int_{-\infty}^{\infty} dx \, \int_0^t du \, e^{-u x^2} \\ &= t \int_0^1 dv \, \int_{-\infty}^{\infty} dx \, e^{-t v x^2} \\ &= \sqrt{\pi t} \int_0^1 dv \, v^{-1/2} \\ &= 2 \sqrt{\pi t}\end{align}$$
Finally, we have the ILT in closed form:
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{s t} = \frac{2}{\sqrt{\pi}} t^{1/2} $$
The laplace transform of $t^{\frac{1}{2}}$ is $\frac{\Gamma(\frac{3}{2})}{s^{\frac{3}{2}}}$, so the inverse transform of $s^{-\frac{3}{2}}$ is $\frac{t^{\frac{1}{2}}}{\Gamma(\frac{3}{2})}$.