crawfs3
- 2
- 0
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. I am 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 don't think this is a valid way of evaluating something. Does anyone have any advice?
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. I am 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 don't think this is a valid way of evaluating something. Does anyone have any advice?