find integral with limits 3 to infinit of 2dx/(x^2 - 1)

[hank]

find integral with limits 3 to infinity of 2dx/(x^2 - 1).


Ok, so the first thing I do is to split the integral into two parts utilizing integration of rational functions with partial fractions.

This gives me:

= int of 3 to infity of dx/(x - 1) minus int of 3 to infinity of dx/(x + 1).

Then I decide to attempt to find the limit of the first integral:

lim B->inf of the integral with limits 3 to b of dx/(x - 1).

I integrate and get lim B->inf of ln(x - 1), 3 to B.

Then when I do the plugging in, I get:

lim B->inf ( infinity - ln(2)) which should equal infinity, right?

This makes the whole problem divergent, but the book says the answer is ln(2).

What am I missing?

 

[royhaas]

Do not split after partial fractions. The antiderivative is ln{(x-1)/(x+1)}, obtained immediately by integrating the partial fraction decomposition. Now evaluate.

 

[hank]

Oh, I think I got it now.

You end up with ln (inf/inf) - ln (2/4).
Since ln (inf/inf) is an indeterminate form, you must apply L'Hopitale's rule which gives you ln (1) - ln(1/2).
Which, using properties of logs gives you ln(1/1/2) which equals ln(2).

Is this correct?

 

[pka]

CORRECT

 

[hank]

Cool.

I'm just wondering now why you don't seperate the two.

Is int (dx/(x - 1) - int (dx/(x + 1) the same as int [ (dx/(x - 1) - (dx/(x + 1)]

Assuming that the limits are 3 to infinity?

 

[pka]

Correct again.
The integral is an additive operator.

 

[hank]

Awesome, thanks.

 

[tkhunny]

ln (inf/inf)

Please don't ever write that again.

You could be on thin ice, here. If you split it apart, both integrals must exist. Do they?

I don't disagree with the validity in this case. I just want you to think about what you are doing. Adding up infinitely many things just isn't the same as adding up some finite number of things.

Good work. Way to go exploring! You will learn more than many others if you keep it up.