# Numerical integration - Gauss quadrature rule

How do we numerically integrate a rapidly decaying exponential function? A simple Gauss quadrature which is based on approximating the function by polynomial, I think will not work, since rapidly decaying exponential function will require more terms in the series. Is there a series approximation for the function?

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Did you try the variants of the basic Gauss quadrature like Gauss–Laguerre quadrature and Gauss–Hermite quadrature? –  user17762 Dec 24 '10 at 5:18
I haven't applied them yet but aren't they of different form? Will they work for a rapidly decaying function? I have the integral of the form $$\int f(x)e^{-kx}$$, where k lies in the range of [10,50] –  vijay Dec 24 '10 at 6:05
@vijay: is $f(x)$ smooth, or does it have singular behavior (e.g. at the left endpoint)? –  Ｊ. Ｍ. Dec 24 '10 at 8:08
@J.M. f(x) is smooth. Ok, I think I will be more clear here. The actual integration to be done goes like this,$$\int (\int f(v) e^{-kvx}dv \int g(v) e^{-kvx}dv)dx$$. Integrands with $f(v)$ & $g(v)$ can only be numerically evaluated. They have 10th order polynominal in $v$ in the denominator. –  vijay Dec 24 '10 at 8:58
...and how sure are you that the denominator polynomial has no zeroes within the interval of integration? –  Ｊ. Ｍ. Dec 24 '10 at 9:01

Variable substitution to make $k=1$, and use Gauss-Laguerre. For more advanced method, the double exponential transformation maybe relevant.

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I would say that vanilla DE is overkill here if the integrand already has a decaying exponential factor; however, Mori and Takahasi do have a modified version suitable for integrands with decaying exponentials; see their paper for details. Otherwise, Gauss-Laguerre ought to work well here, unless the $f(x)$ in $f(x)\exp(-kx)$ has wonky behavior on its own. The problem with Gauss-Laguerre is that it doesn't admit a natural error estimate; one has to usually compare the results for $n$ nodes and $n+1$ nodes. –  Ｊ. Ｍ. Dec 24 '10 at 8:07
I'll give a try and get back to you, if you don't mind.. thanks! –  vijay Dec 24 '10 at 8:56