Pade integration

1. Aug 29, 2007

Sangoku

can we find for a well-behaved f(x) a Rational (Padé approximation) so

$$\int_{0}^{\infty}dx f(x) - \int_{0}^{\infty}dx Q(x) \approx 0$$ ??

Where Q(x) is a rational function, the main idea is that the integral for Q(x) can be performed exactly , whereas the initial integral of f(x) not.

2. Aug 29, 2007

HallsofIvy

Staff Emeritus
What do you mean by "well behaved"?

3. Aug 31, 2007

Sangoku

smooth and differentiable function, for example sin(x) or exp(x^3) ... so it is many times differentiable.

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