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.