Find C for Convergent Improper Integral

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Homework Help Overview

The discussion revolves around finding the constant C that ensures the convergence of a specific improper integral involving rational functions. The integral in question is from 0 to infinity and includes terms that suggest potential divergence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the idea of combining the integrands before integrating to address divergence issues. Questions arise about the objectives of integration and how to simplify the integrand effectively. There is a focus on the cancellation of terms that contribute to divergence.

Discussion Status

Some participants have suggested combining the integrands and checking for terms that can be canceled to eliminate divergence. Others are attempting to clarify their understanding of the integration process and the implications of their manipulations on convergence.

Contextual Notes

There is mention of the behavior of the integrand as x approaches infinity, with specific attention to how the powers of x in the numerator and denominator affect convergence. Participants express uncertainty about the integration techniques applicable to the combined integrand.

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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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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.
 
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?
 
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:
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 :smile:)
 
I think canceling out the x^2 in the numerator would be a good idea. What did you get for the combined integrand anyway?
 
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? :rolleyes:
 
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.
 

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