# Steepest descent approximation

1. Apr 6, 2012

### Mesmerized

Hi all, I am reading now Zee's book "Quantum Field Theory in a Nutshell", there in Apendix 2 of Chapter I.2 the method of steepest descent is briefly described. The part where I have a question is almost self contained and half a page long, so I attached the screen shot of it (formula 19). Anyway, the question is the following:
The only prior information required is the Gaussian integral $$\int_{-\infty}^\infty dx e^{-1/2ax^ 2} =(\frac {2\pi}{a})^ {1/2}$$
To compute the integral $$I=\int_{-\infty}^\infty dq e^{-1/h f(q)}$$ in the limit of very small h, we say that the integral is dominated by the minimum of f, and expanding f near that point up to quadratic terms, $$f(q)=f(a)+1/2f''(a)(q-a)^{2}+O[(q-a)^{3}]$$ we obtain
$$I = e^{-(1/h)f(a)} (\frac {2\pi h}{f''(a)})^{1/2} e^{-O(h ^{1/2})}$$
So the first part of the right hand side is ok, it's just the Gaussian integral, but how he knows that the corrections are of the form $$e^{-O(h ^{1/2})}$$?? Please, at least tell the general idea how it can be shown. Simply keeping also the terms of the next (third) order, i.e. (q-a)^3, gives a very hard integral, at least for me, to evaluate.

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2. May 1, 2012

### Thibaut_Pro20

It does not seem obvious : see for exemple,