Laplacian transform of division by square root of t?

Question

In this formula: $$f(t)=e^{-3t}t^{\frac{-1}2}$$ I saw examples on $t^n$ where $n>0$. But in above example $n<0$.

I don't know how to deal with the $t^{\frac{-1}2}$.

I know that $e^{-3t}=\frac1{s-3}$.

Thanks for helping. If I wasn't clear, please tell me.

Answer 1

I am just going to evaluate $L(\sqrt t)$.

Just substitute $\sqrt{st} = x$.

So, $L(\sqrt t) = \int_0^\infty e^{-st} \cdot t^{-1/2} dt$

$= \frac{2}{\sqrt s}\cdot \int_0^\infty e^{-x^2} dx = \frac{2}{\sqrt s} \cdot \frac{\sqrt \pi }{2} = \sqrt {\frac{\pi}{s}}$