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Math Help - Definite Integral

  1. #1
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    Definite Integral

    I am having trouble with this definite integral, especially since it involves infinity, and the way my professor explained the process has me completely baffled. Any help is very much appreciated, last time one of the members of this forum explained the process better than my professor! Thank you for any help !

    P.s. I haven't figured out how to display the definite integral with the math tag sorry about that...

    Definite Integral (a= 2 and b=oo infinity) of 1/(x^2-1) dx
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  2. #2
    Super Member PaulRS's Avatar
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    Note that: \frac{1}{x^2-1}=\frac{1}{2}\cdot\left(\frac{1}{x-1}-\frac{1}{x+1}\right)

    Thus: \int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\cdot \lim_{b\rightarrow{+\infty}}\left(\int_2^b\frac{dx  }{x-1}-\int_2^b\frac{dx}{x+1}\right)

    \int_2^b\frac{dx}{x-1}=\ln(b-1) and \int_2^b\frac{dx}{x+1}=\ln(b+1)-\ln(3)

    Thus: \lim_{b\rightarrow{+\infty}}\left(\int_2^b\frac{dx  }{x-1}-\int_2^b\frac{dx}{x+1}\right)=\ln(3)+\lim_{b\right  arrow{+\infty}}\ln\left(\frac{b-1}{b+1}\right)

    But: \lim_{b\rightarrow{+\infty}}\ln\left(\frac{b-1}{b+1}\right)=0 since \frac{b-1}{b+1}=\frac{1-\frac{1}{b}}{1+\frac{1}{b}}\rightarrow{1} and the logarithm is a continuous function.

    Thus: \int_2^{\infty}\frac{dx}{x^2-1}=\frac{\ln(3)}{2}
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bloomsgal8 View Post
    I am having trouble with this definite integral, especially since it involves infinity, and the way my professor explained the process has me completely baffled. Any help is very much appreciated, last time one of the members of this forum explained the process better than my professor! Thank you for any help !

    P.s. I haven't figured out how to display the definite integral with the math tag sorry about that...

    Definite Integral (a= 2 and b=oo infinity) of 1/(x^2-1) dx
    \int_2^{\infty}\frac{dx}{x^2-1}

    Using either the definition of arctanh(x)

    Or doing PFD

    to get

    \frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}

    1=A(x+1)+B(x-1)

    Letting x=1 we see that A=\frac{-1}{2}

    and letting x=-1 we see that B=\frac{-1}{2}


    So \frac{1}{x^2-1}=\frac{-1}{2}\bigg[\frac{1}{x-1}-\frac{1}{x+1}\bigg]

    So \int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\int_2^{\infty}\bigg[\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{2}\bigg[\ln(x+1)-\ln(x-1)\bigg]\bigg|_2^{\infty}=\frac{\ln(3)}{2}
    Last edited by Mathstud28; May 14th 2008 at 07:13 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int_2^{\infty}\frac{dx}{x^2-1}

    Using either the definition of arctanh(x)

    Or doing PFD

    to get

    \frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}

    1=A(x+1)+B(x-1)

    Letting x=1 we see that A=\frac{1}{2}

    and letting x=-1 we see that B=\frac{-1}{2}


    So \frac{1}{x^2-1}=\frac{1}{2}\bigg[\frac{1}{x-1}-\frac{1}{x+1}\bigg]

    So \int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\int_2^{\infty}\bigg[\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{2}\bigg[\ln(x+1)-\ln(x-1)\bigg]\bigg|_2^{\infty}=\infty=\frac{\ln(3)}{2}
    And as I said alternatively \int_2^{\infty}\frac{dx}{x^2-1}=arctanh(x)\bigg<br />
|_2^{\infty}=\frac{\ln(3)}{2}
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\int_2^{\infty}\bigg[\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{2}\bigg[\ln(x+1)-\ln(x-1)\bigg]\bigg|_2^{\infty}=\infty=\frac{\ln(3)}{2}
    Confuscious say: There something wrong in last line.

    -Dan
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  6. #6
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    Quote Originally Posted by bloomsgal8 View Post
    I am having trouble with this definite integral, especially since it involves infinity, and the way my professor explained the process has me completely baffled. Any help is very much appreciated, last time one of the members of this forum explained the process better than my professor! Thank you for any help !

    P.s. I haven't figured out how to display the definite integral with the math tag sorry about that...

    Definite Integral (a= 2 and b=oo infinity) of 1/(x^2-1) dx
    Well, as you might expect, we should first begin by finding the antiderivative of the function. So:

    F(x)=\int\frac{dx}{x^2-1}

    \frac{1}{x^2-1}=\frac{1}{2}(\frac{1}{x-1}-\frac{1}{x+1})

    F(x)=\frac{1}{2}\int\frac{dx}{x-1}-\frac{1}{2}\int\frac{dx}{x+1}

    F(x)=\frac{1}{2}ln \frac{x-1}{x+1}

    Now, recall the formula for definite integrals:

    \int_{a}^{b}f(x)dx=F(b)-F(a)

    Therefore:

    \int_{2}^{\infty}f(x)dx=F(\infty)-F(2)

    F(\infty)-F(2)=[\frac{1}{2}ln \frac{\infty-1}{\infty+1}]-[\frac{1}{2}ln \frac{2-1}{2+1}]

    F(\infty)-F(2)=ln \sqrt{3}
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