Does the following indefininite integral converge?

May 2010
23
2
The integral from two to infinity

of (x-1)/(x^2-2)

So the first step I did was I said that the integral of
(x-1)/(x^2-2) was less than that of (x)/(x^2-2)
So if the integral of the second function converged, then the first did as well. I couldn't figure out the integral of this one though :(


The integral of
(x-1)/(x^2-2) was greater than that of (x-1)/(x^2)
So if the integral of the second function diverged, then the second did as well
Then I reasoned that the limit as N approaches infinity
of the integral of (x-1)/(x^2) from two to infinity was
-1+lnN-ln0

But the ln of 0 has no value, it doesn't exist so I'm confused as to whether the integral converges or diverges. Welp?
 

Opalg

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The integral from two to infinity

of (x-1)/(x^2-2)

So the first step I did was I said that the integral of
(x-1)/(x^2-2) was less than that of (x)/(x^2-2)
So if the integral of the second function converged, then the first did as well. I couldn't figure out the integral of this one though :(


The integral of
(x-1)/(x^2-2) was greater than that of (x-1)/(x^2)
So if the integral of the second function diverged, then the second did as well
Then I reasoned that the limit as N approaches infinity
of the integral of (x-1)/(x^2) from two to infinity was
-1+lnN-ln0

But the ln of 0 has no value, it doesn't exist so I'm confused as to whether the integral converges or diverges. Welp?
You are going about this in exactly the right way, and the second approach is the one that works. The only thing wrong is that the integral goes from 2 to infinity, not 0 to infinity, so you should not be getting the term ln0. If you replace that by ln2, you'll get the correct answer.
 
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May 2010
7
1
terminology

It is not an indefinite integral - for which the concept of convergence
is inappropriate. It is an improper (because of infinity) definite integral.
 
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