Here I would like to see the behavior of a function as an integral when its argument (which is a parameter in the integral) goes to zero. If I try to evaluate an integral $\int^{i\infty}_{-i\infty}dz\frac{\mathcal{M}(z)}{z}\lambda^z$ in which "λ" is a number which approaches zero. Is the following way correct or not?

First we write it as $\int^{i\infty}_{-i\infty}dz\frac{\mathcal{M}(z)}{z}e^{z\log{\lambda}}$ where $\lambda$ is some meromorphic function, but on the exponential the first derivative of the exponent doesn't have any zero, therefore I pull the $1/z$ factor onto the exponent: $\int^{i\infty}_{-i\infty}dz\mathcal{M}(z)e^{z\log{\lambda}-\log{z}}$, then the exponent $z\log{\lambda}-\log{z}$ is stationary at z∼0 when λ→0, then we just approximate the integral with the limit of the integrand when z→0, which is $\mathcal{M}(0)\log{\lambda}$.

Is this way of doing steepest descent reasonable?

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user user106592