How do you find the inverse Laplace transform of $\frac{1}{\sqrt{s}(s-a)} $

Question

When I use the convolution method, I can't avoid getting a divergent integral.

Answer 1

I will solve this using convolution but I would LOVE to see someone do it using contour integration.

Using the fact that:

$\frac{1}{\sqrt t}\rightarrow \sqrt \frac{\pi}{s}$ which means $\frac{1}{\sqrt s}\rightarrow \sqrt \frac{1}{\pi t}$

$\frac{1}{s-a}\rightarrow e^{at}u(t)$, where $u(t)$ is the unit step (which means the exponential is zero for $t<0$, which is probably where you went wrong)

We need to find:

$$\sqrt \frac{1}{\pi t} * e^{at}=\frac {1}{\sqrt \pi}\int ^t_0\frac{e^{(at-a\tau)}}{\sqrt \tau}d\tau$$

The reason the integral is from $0$ to $\infty$ is because the exponential $e^{-at}$ is zero for $t<0$. If you have trouble visualizing, here is a gif. The exponential we want to convolve looks like this.

$$=\frac {1}{\sqrt \pi}e^{at}\int ^t_0\frac{e^{-a\tau}}{\sqrt \tau}d\tau$$

substitute $a\tau=b^2$, we have:

$$=\frac {1}{\sqrt \pi}e^{at}\int ^{\sqrt{at}}_0 \frac{2}{\sqrt a}{e^{-b^2}}db$$

$$=\frac{1}{\sqrt a}e^{at}\frac {2}{\sqrt \pi}\int ^{\sqrt{at}}_0 {e^{-b^2}}db$$

$$=\frac{1}{\sqrt a}e^{at}erf(\sqrt{at})$$

Which matches wolfram's answer. Again, I'd love to see someone do this using contour integration.