I am considering the method of steepest descent from my notes. I have written that

$$\int_a^b dx e^{g(x)} \sim e^{g(x_0)} \int_{\infty}^{\infty}dx \exp \left[-\frac{1}{2}(x-x_0)^2|g^"(x_0)|\right] dx=e^{g(x_0)}\sqrt{\frac{2\pi}{|g^"(x_0)|}} $$ where we have used the fact that $g^"(x_0)<0$ and that the integrand is dominated by the region close to the maximum to extend limits to the infinities.

I cannot see how the fact $g^"(x_0)<0$ has been used to do the integration. I can see somehow it comes from the taylor expansion but sure how.

  • $\begingroup$ Im not sure I quite have the appropriate tags $\endgroup$ – Permian Apr 18 '15 at 11:14

You did a taylor expansion for $g$ around $x_0$, neglected all terms of order $3$ or above. Since $x_0$ is a maximum, you have that $g'(x_0)=0$ and $g''(x_0)\leq 0$, so

$$g(x)\approx g(x_0)+\frac{1}{2}(x-x_0)^2g''(x_0)=g(x_0)-\frac{1}{2}(x-x_0)^2|g''(x_0)|.$$

You used that $g''(x_0)\leq0$ to write $g''(x_0)=-|g''(x_0)|$.


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