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Math Help - Evaluating an improper integral

  1. #1
    Junior Member
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    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.
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  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
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    64

    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
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