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Generally, the method of steepest descent describes the asymptotic behavior of integrals of the form $$\int_{-\infty}^\infty h_t(x)\exp(-tg(x)) \,dx$$ in terms of $t$. As long as $h_t(x)$ is controlled nicely as $t\to\infty$ and $g(x)$ has a global minimum at a finite $x_{min}$, we know that the leading order of the integral is $$h_t(x_{min})\exp(-tg(x_{min}))/\sqrt {t/2\pi}.$$ However, I currently have a problem for which I am interested in $$\int_{0}^\infty h_t(x)\exp(-tg(x)) \,dx$$ and g(x) is monotonically decreasing with $g(0)=\infty$ and $g(\infty)=1$. Are there any standard asymptotic results for the above integral? In my case $h_t(x)=h(x)$, i.e. independent of $t$ with $\int_0^\infty h(x) \,dx <\infty$. Assume whatever smoothness conditions you'd like about $h$.

Ivan
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It's a bit difficult to answer with the information that you provide, but this will probably work (and is a standard method): make the substitution $y=-g(x)$, so that your integral is $I(t)=\int_{-\infty}^{-1} f(y)\exp(ty)dy$ (for the appropriate $f$). Now use per partes $$I(t)=f(-1)\exp(-t)/t-\int_{-\infty}^{-1}f'(y)\exp(ty)/t\,dy,$$ then use per partes again and again, to get a series $\sum_{n\geq1}c_n\exp(-t)/t^n$.

user8268
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  • Thank you for your answer. I'd like to point out that $f(-1)=\lim_{x\to\infty} h(x)/g'(x)$. So, if I choose $h$ to be a rational function and $g$ exponential, this is zero. The same follows for all other operations of integration by parts. In other words, the series that you wrote does not converge because the remainder is large. Further, this misses the point of the steepest descent method, which is to derive the asymptotic behavior from derivatives of $h$ and $g$ at particular points. – Ivan Sep 11 '13 at 23:15
  • By "choose" I mean if I am "given". I'd like to know if there is a generalization of the steepest descent method to the type of functions specified above. – Ivan Sep 11 '13 at 23:22
  • Ah, and I meant to say if $h$ is exponential and $g$ is rational. Sorry about that... – Ivan Sep 11 '13 at 23:41
  • @Ivan I think your question would work better if you asked it for concrete $h$ and $g$ (not necessarily the ones you actually want, but for some simple ones having the right behaviour) – user8268 Sep 12 '13 at 08:57