# Evaluation of Non-Elementary functions

1. Apr 6, 2009

### crawfs3

Hi everyone,

This should be a simple problem but l'Hospital Rule cannot be used in this case.

The Problem: INTEGRAL[f(v)dv] = INTEGRAL[(v^2)*exp(-c*(v^2))]. I cannot evaluate this integral at infinity because of the indeterminate form that appears. My partial solution is below.

Solution: When the function above is integrated we get:

(Sqrt(Pi)*erf(Sqrt(c)*v)/(4*c^1.5)) -(v/2c)*exp(-c*v^2) evaluated from x to infinity.

The problem is when infinity is plugged into the term with the exponent, an indeterminate form results. The exp(-infinity) is zero but the v multiplied in front wants to go to infinity. L'Hospital Rule cannot be used because this is a product and the rule can only be used for a quotient. Im thinking of just saying it goes to zero because the squared infinity is inside the exponential which would got zero much faster than the linear v term out in front of the exponent goes to infinity. But I dont think this is a valid way of evaluating something. Does anyone have any advice?

2. Apr 6, 2009

### Ben Niehoff

Try this:

$$xe^{-x} = \frac{x}{e^x}$$

3. Apr 11, 2009

### coomast

Indeed that is the way to evaluate the limit. You're reasoning was however not that bad. The exponential function with negative argument is a very strong one. There are not that many functions which multiplied by this one give rise to a non-zero value for large arguments. There are however some....

I evaluated the integral and got erfc (complementary) instead of the erf function. Is this correct?

coomast