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It is well known that the integral $\int_a^b e^{-n \cdot f(x)}dx$ can be approximate by $\sqrt{\frac{2\pi}{n|f''(x_0)|}}e^{-n\cdot f(x_0)}$ at $x_0$ the maximum of $f(x)$ in $(a,b)$ (for large $n$..).

What can be said about approximating $\int_a^b n^{g(x)}e^{-n \cdot f(x)}dx$?

Antonio Vargas
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nir
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  • This question asks about a case when $g(x) = -x$, $a=0$, and $b=\infty$. For that xample it's shown that the factor $n^{-x}$ doesn't shift the saddle point much and can just be replaced with $n^{-x_0}$ to get the correct first approximation. Higher-order terms in the asymptotic expansion are different from those for usual Laplace-type integrals, though. – Antonio Vargas Apr 23 '15 at 21:12

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