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:(adsbygoogle = window.adsbygoogle || []).push({});

The only prior information required is the Gaussian integral [tex]\int_{-\infty}^\infty dx e^{-1/2ax^ 2} =(\frac {2\pi}{a})^ {1/2}[/tex]

To compute the integral [tex]I=\int_{-\infty}^\infty dq e^{-1/h f(q)}[/tex] 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, [tex]f(q)=f(a)+1/2f''(a)(q-a)^{2}+O[(q-a)^{3}][/tex] we obtain

[tex]I = e^{-(1/h)f(a)} (\frac {2\pi h}{f''(a)})^{1/2} e^{-O(h ^{1/2})}[/tex]

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 [tex]e^{-O(h ^{1/2})}[/tex]?? 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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# Steepest descent approximation

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