$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{1}{s^2}e^{s(t - \frac{1}{2}x^2)}ds$ - different answers depending on value of $t$?

Question

After taking an inverse Laplace transform I have the following -

$$y = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{1}{s^2}e^{s(t - \frac{1}{2}x^2)}ds$$

In my notes it says

1. if $t < \frac{1}{2}x^2$ the "contour can be closed to the right" and hence $y = 0$ for $t < \frac{1}{2}x^2$
2. if $t > \frac{1}{2}x^2$ the "contour must be be closed to the left" and hence $y = t - \frac{1}{2}x^2$ for $t > \frac{1}{2}x^2$

I can see that the result for 2. comes from evaluating the integral using the residue at the singularity $s = 0$. What I don't fully get is why we have two cases for different values of $t$. Can someone elaborate on what is actually happening here and why the solution for $y$ is different depending on $t$ being less than or greater than $\frac{1}{2}x^2$?

Answer 1

For inverse Laplace transforms, we require that $\gamma$ be larger than the real part of any pole of the Laplace transform, in this case, $1/s^2$. There is good reason for this. Recall that the Laplace transform is defined as

$$\int_0^{\infty} dt \, f(t) e^{-s t} $$

which always converges when $t \gt 0$, right? Wrong. One must consider any exponential behavior in $f$, e.g., $f(t) = g(t) e^{\sigma t}$. Such behavior shifts the $s$ coordinate to the right, i.e., the integral only converges when $s \gt \sigma $ when $t \gt 0$. Thus we guarantee convergence when we place the line of integration to the right of any pole of $F$. When $t \gt 0$, we close the contour to the left to enclose the poles and get a nonzero ILT. This leaves the ILT being identically zero when $t \lt 0$ because no poles are ever enclosed to the right.