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Convergence or divergence

by fk378
Tags: convergence, divergence
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fk378
#1
Nov18-07, 09:05 PM
P: 367
1. The problem statement, all variables and given/known data
Determine whether the following integral is convergent or divergent. If convergent, what does it converge to?

dx/(4x^2 + 4x + 5) [-infinity, infinity]

2. Relevant equations
comparison theorem?


3. The attempt at a solution
I think it is convergent, so I set the original integral less than or equal to dx/4x^2.
Solving the integral and setting up a limit, I got the limit as t-->infinity of (4x^-1)/-1 evaluated from [-infinity, t]. Now here is where I get lost. Evaluating it at t gives 0, but when I plug in the lower limit of 0, isn't it possible to say that the 0 either belongs in the numerator if there is a -1 exponent there, as well as saying that it doesn't exist if I move (4x^-1) into the denominator?
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dynamicsolo
#2
Nov18-07, 09:19 PM
HW Helper
P: 1,664
Quote Quote by fk378 View Post
I think it is convergent, so I set the original integral less than or equal to dx/4x^2.
Solving the integral and setting up a limit, I got the limit as t-->infinity of (4x^-1)/-1 evaluated from [-infinity, t]. Now here is where I get lost. Evaluating it at t gives 0, but when I plug in the lower limit of 0, isn't it possible to say that the 0 either belongs in the numerator if there is a -1 exponent there, as well as saying that it doesn't exist if I move (4x^-1) into the denominator?
I think you want to be careful about comparing your integral to the integral for (4x)^(-2), because that function is not continuous at x = 0 , so you can't use the Fundamental Theorem on it and its integral doesn't converge as x approaches 0.

For the purpose of proof, you may be better off evaluating the original integral itself, since that integrand is continuous. You are correct that it is convergent and can be evaluated using a "completion of squares" technique (since the quadratic is irreducible).
fk378
#3
Nov18-07, 09:34 PM
P: 367
So if I want to plug in 0 into (4x^-1) then I have to move the 4x to the bottom instead of keeping it at the top with the -1 exponent?

dynamicsolo
#4
Nov18-07, 09:49 PM
HW Helper
P: 1,664
Convergence or divergence

Quote Quote by fk378 View Post
So if I want to plug in 0 into (4x^-1) then I have to move the 4x to the bottom instead of keeping it at the top with the -1 exponent?
It's all the same: the point is that this is 1/(4x), so putting zero in for x will make this undefined. The type II improper integral about x = 0 for dx/(4x^2) diverges, so it's no help in making a comparison with your integral.

Fortunately, the type I improper integral you are given is not particularly difficult to evaluate.


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