# Steepest Descent Approximation applied to Integral Form

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 $f(a)$?

• The quote is apparently footnote 38 on p. 61 in Ivancevic & Ivancevic: New Trends in Control Theory. The original quote contains several typos: An imaginary unit $i$ in the beginning disappears and a square root is missing in the final formula. – Qmechanic Feb 17 '18 at 20:06
• No this comes from an appendix in zee's qft in a nutshell – Permian Feb 17 '18 at 20:12
• $\uparrow$ Which page? – Qmechanic Feb 17 '18 at 20:15
• page 16......... – Permian Feb 17 '18 at 20:26

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