Find C for Convergent Improper Integral

[demersal]

Homework Statement

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

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

Homework Equations

The Attempt at a Solution

\stackrel{lim}{t->inf.} \int \frac{x}{x^2+1} dx - \stackrel{lim}{t->inf.} \int \frac{C}{3x+1} dx
for (x^2/(x^2+1):
u = x^2 + 1
du = 2xdx
\stackrel{lim}{t->inf.} (1/2)ln(u) dx
\stackrel{lim}{t->inf.} (1/2)ln(x^2+1) ] \stackrel{t}{0}
\stackrel{lim}{t->inf.} (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 \***Please note (I don't know how to format limits haha) that I mean the limit as t approaches infinity! Thanks!***

 

[Dick]

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.

 

[demersal]

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?

 

[Dick]

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.

 

[demersal]

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 )

 

[Dick]

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

 

[demersal]

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 \int^{t}_{0} \frac{x-3}{3x^3+x^2+3x+1}

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?

 

[Dick]

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.