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

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?

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

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.