I am trying to find the inverse Laplace transform of $f(z)$ using the residue theorem. Can you please check to see if what am doing below is correct? I am not really sure about what I am doing. Thanks.
$$f(z) = \frac{e^{-\sqrt{z}}}{z-1}$$
My attempt:
I calculated the residue at the pole/singularity at $z = 1$ as follows:
$$e^{zt}f(z)=\lim_{z\to 1} (z-1)\frac{e^{zt}e^{-\sqrt{z}}}{z-1} = e^{t-1}$$
Next I need to evaluate the integral at the branch cut, i.e, at $z = 0$:
$$z = re^{i\theta} = re^{\frac{i\pi}{2}},$$
I reckon since its a d-shaped contour from $R \to 0$ should be $\theta = \pi/2$,
$$\implies z = ir \therefore \sqrt{z} = \sqrt{ir},$$
so now along the branch cut have:
$$\int e^{zt}f(z)dz = \int \frac{e^{zt}e^{-\sqrt{z}}}{z-1}dz = \int \frac{e^{irt}e^{-\sqrt{ir}}}{ir-1}idr$$
and for $\theta = -\pi/2$ have the following:
$$z = re^{i\theta} = re^{\frac{-i\pi}{2}}.$$
I reckon we now along contour from $0 \to R$ should be $\theta = -\pi/2$:
$$\implies z = -ir \therefore \sqrt{z} = -\sqrt{ir}.$$
So now along the branch cut have:
$$\int e^{zt}f(z)dz = \int \frac{e^{zt}e^{-\sqrt{z}}}{z-1}dz = \int \frac{e^{irt}e^{\sqrt{ir}}}{-ir-1}(-idr)$$