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Given the integral:

$I = \int_0^a{e^{-\lambda g(x)}f(x)dx}$ Where $g(x)$ and $f(x$) are both low order positive polynomials, and $\lambda \gg 1$, Laplace's method is commonly used to approximate the integral by using the first or second derivatives of $g(x)$.

Now assuming we know everything about $f(x)$, but do not know $\lambda$. Assume also for simplicity that $g(x)=x$. Is there a way of expressing the integral $I$ only in terms of: $G = \int_0^a{e^{-\lambda g(x)}dx}$ and derivatives of the function $f(x)$? I've tried expanding $f(x)$ as a Taylor series, but after the first term, I end up with a denominator containing $\lambda$ which is not known. Ideally I'd like to have something like: $I \approx f(0)G + f'(0)\times(...)$ With no $\lambda$ dependence outside of $G$.

Is this possible?

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    @Raskolnikov - I am interested in the large $\lambda$ behavior. That's why I specified $\lambda \gg 1$..2012-08-09

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You can see that this can't work by finding the exact result for a linear function:

$ \int_0^a\mathrm e^{-x}(m+nx)\,\mathrm dx=mG+n\lambda^{-2}\left(1-(\lambda a+1)\mathrm e^{-\lambda a}\right)\;. $

Of course $G$ is an invertible function of $\lambda$, so you can always express $\lambda$ in terms of $G$ everywhere if you want, but I presume that's not what you had in mind.

Perhaps I misunderstood the question; in that case perhaps you should explain more about the difference between dependence on $\lambda$ and dependence on $G$.

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    unfortunately I'm stuck with one measurement only, i.e. $I$ for only one $f$ (which is a non trivial function of $x$), but something similar would be great.2012-08-09