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 $\int^{i\infty}_{-i\infty}dz\frac{1}{z}\mathcal{M}(z)\lambda^z$ in which "$\lambda$" is a number which approches zero. Is the following way correct or not?

First we write it as $\int^{i\infty}_{-i\infty}dz\frac{1}{z}\mathcal{M}(z)e^{z\log{\lambda}}$ where $\mathcal{M}$ 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 is stationary at $z\sim 0$ when $\lambda\rightarrow 0$, then we just approximate the integral with the limit of the integrand when $z\rightarrow 0$, which is $\mathcal{M}(0)\log{\lambda}$.

Is this way of doing steepest descent reasonable?


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