How to evaluate the Laplace transform of the square root using Residue theory?

Question

My lecturer mentioned that it is possible to evaluate the Laplace integral transform (definition below) of $\sqrt{t}$ using complex analysis. How is that possible?

$$\hat f (s)=\int^{\infty} _0 {\sqrt{t} e^{-st}dt} $$

Answer 1

For $s > 0$ $$\int_0^\infty t^{1/2} e^{-st}dt = \int_0^\infty (u/s)^{1/2} e^{-s(u/s)}d(u/s) = s^{-3/2} \int_0^\infty u^{1/2} e^{-u}du \\= s^{-3/2} \Gamma(3/2)=s^{-3/2} \Gamma(1/2)1/2 = s^{-3/2} \pi^{1/2}/2 $$ (the last step is with the reflection formula)

Since $$\int_0^\infty t^{1/2} e^{-st}dt-s^{-3/2} \pi^{1/2} /2$$ is complex analytic for $\Re(s) > 0$ and it vanishes for $ s > 0$, it vanishes for $\Re(s) > 0$ and you got your Laplace transform.