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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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Your problem is unrelated to Laplace's method, unless you are interested in large $\lambda$ behaviour, but that's not what you're asking it seems to me. –  Raskolnikov Aug 9 '12 at 5:24
    
@Raskolnikov - I am interested in the large $\lambda$ behavior. That's why I specified $\lambda \gg 1$.. –  nbubis Aug 9 '12 at 11:52

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up vote 2 down vote accepted

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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thank you. What I have is actually a physical measurement of $G$, that i'm trying to compare with, and I don't really know to much about the exponent other than that it falls off a lot faster than the smooth function $f(x)$. So what I take from here is that the best approximation I can get is simply $f(0)G$. –  nbubis Aug 9 '12 at 11:59
1  
@nbubis: If you can measure $I$ for given $f$, you could measure it for $x^n$ to get $I=\sum_nf^{(n)}I_n/n!$. –  joriki Aug 9 '12 at 12:07
    
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. –  nbubis Aug 9 '12 at 12:11

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