# Steepest descent for Mellin-type integration

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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
asked Nov 9, 2013

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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.

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli
answered Nov 9, 2013 by (40 points)
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.

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user user106592
And why do you assume that the fraction is not singular around zero? What if $M=z$?

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli
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 $\frac{\mathcal{M}'}{\mathcal{M}}$ :)

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user user106592
OK, this assumption does not look reasonable to me, but you know better what you need...

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli
@user106592: You may wish to look at en.wikipedia.org/wiki/Stationary_phase_approximation

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

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