Steepest Descent Approximation applied to Integral Form

Question

To do an exponential integral of the form $I = \int_{-\infty}^{\infty}dq ~e^{-(1/\hbar)f(q)}$ we often have to resort to the steepest descent approximation. In the limit of $\hbar$ small, the integral is dominated by the minimum of $f(q)$. Expanding $f(q) = f(a)+\frac{1}{2}f''(a)(q-a)^2+O[(q-a)^3]$ and using

$$ \int_{-\infty}^{\infty} dx~ e^{-\frac{1}{2}{\rm a}x^2}=\sqrt{\frac{2\pi}{\rm a}},\tag{17}$$

we obtain

$$ I = e^{-(1/\hbar) f(a)}\sqrt{\frac{2\pi\hbar}{f''(a)}}e^{-O(\hbar^{\frac{1}{2}})}. \tag{27}$$

When $\hbar$ is small we get $(1/\hbar)$ goes to infinity so I don't see how this is dominated by $f(a)$?

Answer 1

Hints:

1. Rather than studying the complex method of steepest descent, I recommend to first study the real Laplace's method.

2. Perform a substitution $q=\sqrt{\hbar}x+a$ with $x$ as new integration variable.

References:

1. A. Zee. QFT in a nutshell, 2nd ed., 2010; p.16.