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 $f(a)$?

  • $\begingroup$ 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. $\endgroup$ – Qmechanic Feb 17 '18 at 20:06
  • $\begingroup$ No this comes from an appendix in zee's qft in a nutshell $\endgroup$ – Permian Feb 17 '18 at 20:12
  • $\begingroup$ $\uparrow$ Which page? $\endgroup$ – Qmechanic Feb 17 '18 at 20:15
  • $\begingroup$ page 16......... $\endgroup$ – Permian Feb 17 '18 at 20:26


  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.


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

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