# Thread: Evaluating an improper integral

1. ## Evaluating an improper integral

The integral is: integral(2/(t^2-1),t,2,infinity)

The integral of two over the quantity t squared minus one, from 2 to infinity.

I used trig substitution and ended up in the end with:

2*[sqrt(t^2-1)]+c for the integral, which is not what my calculator gets.

2 times the limit as B goes to infinity of [ln((b^2-1)^(1/2)-ln((3)^(1/2))]

The answer should be ln(3), which is 2*ln((3)^(1/2)), so I thought maybe I'm on the right track, but isn't the limit of as b approaches infinity of
ln((b^2-1)^(1/2), inifinity?

Here are my steps.

t=secu
dt=secutanudu
t^2-1=tan^2u

2*integral((cscdu)/tan^2u)
=2*integral(ln(abs(sinu)/abs(cosu+1)))

sinu= sqrt(t^2-1)/t

cosu=1/t

...

Is there something I am doing wrong.

2. ## Re: Evaluating an improper integral

$I=\displaystyle \lim_{a \to \infty} (\ln(1-a)-\ln(1+a)-(\ln(1-2)-\ln(1+2)))$

$I=\displaystyle \lim_{a \to \infty} \left(\ln \frac{1-a}{1+a} - (\ln(-1)-\ln 3)\right)$

$I=\ln\left(\displaystyle \lim_{a \to \infty} \frac{1-a}{1+a}\right) - \ln(-1)+\ln 3=\ln(-1)-\ln(-1)+\ln 3= \ln 3$