Mellin's inversion integral, branch cut problem

[nearlynothing]

I recently had to solve a problem in which i had to find the inverse laplace transform of some function with a branch cut from - ∞ to 0, so i used a contour avoiding that branch cut like this

http://www.solitaryroad.com/c916/ole19.gif

my problem is as follows: i know the contributions from the integrals BDE and LNA are both zero, then the problem is to calculate the two integrals close to the real axis and the one along the small "circle", i calculated the ones next to the real axis already.

But i was wondering if for the one along the circle i can use the residue theorem and consider s = 0 as a pole, and thinking about the circle as if it was closed, as the lines next to the real axis are infinitely close to it, therefore "closing" the circle path, and also as the radius of the circle becomes infinitely small it would only contain the point s = 0 inside, in a polelike fashion.
I know I'm far from being mathematically rigid, but i did the calculations for two problems of the kind considering s = 0 as a pole for that particular integral and i got the right result on both.

 

[RedSonja]

I'm just learning this stuff myself, but I think residues are wrong, since the closed contour is the full contour and the small circle is open, and coming from above and below does not necessarily lead you to the same point, since the function may be multivalued. I believe you need to do the integral as

\begin{equation}
\int_{-\pi+\eta}^{\pi-\eta} f(\theta) d\theta
\end{equation}

with z=\epsilon e^{i\theta} and \eta, \epsilon → 0. Probably it is possible from your problem to estimate the upper limit for the size of this integral, and I think that for Mellin transform problems this integral is always zero?