Find C for Convergent Improper Integral

  • Thread starter demersal
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In summary: But if that fails, then I guess you could try integration by parts. In the meantime, I think I'll read up on partial fractions and see if that helps.In summary, you are trying to find the value of C for which the following integral converges. The integral converges for C=1.
  • #1
demersal
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Homework Statement


Find the value of the constant C for which the following integral converges. Evaluate the integral for this value of C:

[tex]\int[/tex] [tex]\frac{x}{x^2+1}[/tex] - [tex]\frac{C}{3x+1}[/tex]dx from 0 to infinity

Homework Equations


The Attempt at a Solution



[tex]\stackrel{lim}{t->inf.}[/tex] [tex]\int[/tex] [tex]\frac{x}{x^2+1}[/tex] dx - [tex]\stackrel{lim}{t->inf.}[/tex] [tex]\int[/tex] [tex]\frac{C}{3x+1}[/tex] dx
for (x^2/(x^2+1):
u = x^2 + 1
du = 2xdx
[tex]\stackrel{lim}{t->inf.}[/tex] (1/2)ln(u) dx
[tex]\stackrel{lim}{t->inf.}[/tex] (1/2)ln(x^2+1) ] [tex]\stackrel{t}{0}[/tex]
[tex]\stackrel{lim}{t->inf.}[/tex] (1/2)ln(t^2+1)

Now I am unsure of what to do. How do I know the limit of this first half? How can I use it to help me find what value of C will make it convergent? Your time and effort is greatly appreciated in helping me understand this :smile:\***Please note (I don't know how to format limits haha) that I mean the limit as t approaches infinity! Thanks!***
 
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  • #2
Don't integrate them separately, they both diverge (they are ~1/x for large x). Combine them algebraically first and see if you can find a value of C that eliminates that divergence.
 
  • #3
Ok, I will try that, but I am still unsure of my objective in integrating. Should I be looking to eliminate factors from the top and bottom to simplify the integrand before I integrate and take the limit? And I am just looking to make it so that the limit exists?
 
  • #4
You are trying to get C to cancel a term in the numerator that's causing the divergence. For example, ((C-1)*x+1)/(x^2+1) diverges if C=2. Since it behaves like 1/x for large x. But if C=1 it converges, since it's just an arctan. You canceled the x part that's causing problems. That kind of cancellation.
 
Last edited:
  • #5
Ahh, ok, I see what you mean by cancellation now. So I am basically trying to cancel out the effect of a smaller power of x over a larger power of x. But I'm not seeing a way to get rid of the x^3. Unless I am supposed to make the whole numerator 1? (I'm sorry I'm being so dense today, calculus tends to do that to me :rofl:)
 
  • #6
I think canceling out the x^2 in the numerator would be a good idea. What did you get for the combined integrand anyway?
 
  • #7
I got:
[(3-c)x^2+x-c] / [3x^3+x^2+3x+1]

And then if I plug in for 3=c, I get

lim as t approaches infinity of [tex]\int[/tex][tex]^{t}_{0}[/tex] [tex]\frac{x-3}{3x^3+x^2+3x+1}[/tex]

And now, as my luck would have it, I am stuck integrating. I cannot factor the denominator so I think integration by parts is out of the question and I don't think factoring by grouping would yield any sort of cancellation with the numerator. Any tips? :uhh:
 
  • #8
x^2*(3x+1)+(3x+1) is another way of writing the denominator. Sure you can't factor it? I'd say try partial fractions.
 

1. What is a convergent improper integral?

A convergent improper integral is an integral where the limits of integration are infinite or the integrand has a singularity within the limits. It is called "improper" because it does not meet the criteria for a standard definite integral. A convergent improper integral is one that has a finite value when evaluated.

2. How do you find the value of C for a convergent improper integral?

To find the value of C for a convergent improper integral, you must first evaluate the integral without the constant C. Then, you can use any known information about the integral, such as its convergence or divergence, to solve for the value of C. This process may involve using techniques such as substitution or integration by parts.

3. Can a convergent improper integral have a negative value for C?

Yes, a convergent improper integral can have a negative value for C. The value of C is determined by the properties of the integral and may be positive, negative, or zero. It is important to note that the value of C does not affect the convergence or divergence of the integral, but rather the specific numerical value.

4. What happens if the value of C for a convergent improper integral is undefined?

If the value of C for a convergent improper integral is undefined, it means that the integral may not have a finite value. This can occur when the integral does not meet the criteria for convergence, such as having an infinite limit of integration or a singularity within the limits. In this case, the integral is considered to be divergent and does not have a well-defined value.

5. Can the value of C for a convergent improper integral change depending on the chosen method of evaluation?

Yes, the value of C for a convergent improper integral can change depending on the chosen method of evaluation. Different methods of integration may yield different values for C, but as long as the integral is convergent, the value of C will not affect the final result. However, it is important to choose a method of evaluation that is appropriate for the integral to ensure an accurate value for C.

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