# Steepest descent for Mellin-type integration

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 $$I(\lambda) = \int^{i\infty}_{-i\infty}\mathrm{d}z\frac{\mathcal{M}(z)}{z}\lambda^z$$ in which $\lambda$ is a number which approaches zero. Is the following way correct or not?

First we write it as $$I(\lambda)=\int^{i\infty}_{-i\infty}\mathrm{d}z\frac{\mathcal{M}(z)}{z}e^{z\log{\lambda}}$$ 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: $$I(\lambda)=\int^{i\infty}_{-i\infty}\mathrm{d}z\,\mathcal{M}(z)e^{z\log{\lambda}-\log{z}},$$ then the exponent $z\log{\lambda}-\log{z}$ is stationary at $z\sim0$ when $\lambda\to0$, then we just approximate the integral with the limit of the integrand when $z\to0$, which is $\mathcal{M}(0)\log{\lambda}$.

Is this way of doing steepest descent reasonable?

This does not seem reasonable, at least not at first glance. It's hard to believe that the integral does not depend on the behavior of $M(z)$ near $z=0$. Maybe you should move $M(z)$ to the exponent as well.
• Yes, it is possible to include $\mathcal{M}(s)$ in the exponent, the result of the saddle point turns into $\log{\lambda}-\frac{1}{z}+\frac{\mathcal{M}'}{\mathcal{M}}=0$ and I assume $\frac{\mathcal{M}'}{\mathcal{M}}$ is not singular around zero. What I worry about is, the usual form of the exponent for steepest descent has an overall big coefficient like $\int dz A(z) e^{\lambda B(z)}$, but here the form of the exponential seems to be non-standard. I'm not sure if there are pitfalls. Nov 9 '13 at 21:32
• And why do you assume that the fraction is not singular around zero? What if $M=z$? Nov 9 '13 at 21:39
• A better way of saying is, I'm doing the integral in this way for those $\mathcal{M}$ that satisfies such regularity condition, not for a general $\mathcal{M}$, in the problem I'm facing with, actually I don't know what $\mathcal{M}$ is, so I have to modestly impose some condtions first so that I can somehow proceed. So let's first take a regular $\frac{\mathcal{M}'}{\mathcal{M}}$ :) Nov 9 '13 at 21:48