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Math Help - Improper Integral

  1. #1
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    Improper Integral

    Hello!

    I am trying to evaluate the following integral:

    \int_{-\infty}^{0}(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}})\,dx

    The result is supposed to be ln2.

    Could somebody please help me with this?
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  2. #2
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    Re: Improper Integral

    Thanks from kareman
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  3. #3
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    Re: Improper Integral

    chiro thank you for your reply. I tried this way but I don't get the result ln2...
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  4. #4
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    Re: Improper Integral

    What result do you get?
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  5. #5
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    Re: Improper Integral

    I can't evaluate each integral that involves a fraction separately because I get limits that do not exist. Each integral that involves a fraction diverges.
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  6. #6
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    Re: Improper Integral

    That should not happen if a is non-zero.

    Show us what you have tried.
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  7. #7
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    Re: Improper Integral

    Since
    \frac{1}{\sqrt{x^2+a^2}}<\frac{1}{\sqrt{x^2}} the integral \int_{-\infty}^{0}\frac{1}{\sqrt{x^2+a^2}}\,dx diverges because the integral \int_{-\infty}^{0}\frac{1}{\sqrt{x^2}}\,dx diverges.

    The same applies to \int_{-\infty}^{0}\frac{1}{\sqrt{x^2+4a^2}}.

    Therefore it is not possible to use the formula \int\frac{1}{\sqrt{x^2+a^2}}\,dx=\ln(x+\sqrt{x^2+a  ^2}) for each fraction separately for this improper integral.

    Perhaps, I am missing something else here?
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  8. #8
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    Re: Improper Integral

    You've bought up a great point, but do remember that ln(x) - ln(y) = ln(x/y) (Try using that to get some cancellations).
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  9. #9
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    Re: Improper Integral

    Yes... if I proceed this way then for x=0:

    \ln(x+\sqrt{x^2+a^2})=\ln(a) for the first fraction and \ln(x+\sqrt{x^2+4a^2})=\ln(2a) for the second fraction which equals to -ln2, NOT ln2.

    However the result should be ln2, so what am I doing wrong?
    Last edited by kareman; February 7th 2013 at 10:56 AM.
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  10. #10
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    Re: Improper Integral

    Given arcsin(u) = ln(u + SQRT(u^2 + 1)) we get a cancellation for x = 0 but for x = -infinity we should get a non-zero term when everything is cancelled.

    We solve the limit of lim x->-infinity [(x/a) + SQRT((x/a)^2 + ]/[(x/4a) + SQRT((x/4a)^2 + 1))] and then take the log of that number.
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  11. #11
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    Re: Improper Integral

    The answer is ln2 .
    but I don't know how to type math symbols to show you the result if anyone wants..please let me know....
    Minoas
    P.S Careman are you a Greek?
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  12. #12
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    Re: Improper Integral

    OK, after struggling too much time with this, I tried again with a clear mind. Here is my calculation (I have checked the LaTeX code, but the symbols for the square root (in two cases) and for the fraction (in one case) don't show up normally):

    \int_{-\infty}^{0}(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}})\,dx=

    \lim_{b\rightarrow-\infty}[\ln{(x+\sqrt{x^2+a^2}})-\ln{(x+\sqrt{x^2+4a^2)}}]|_b^0=

    \lim_{b\rightarrow-\infty}[\ln{|a|}-\ln{|2a|}-ln{(b+\sqrt{b^2+a^2}})+\ln{(b+\sqrt{b^2+4a^2)}}]=

    -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{(b+\sqrt{b^2+4a^2)}}-ln{(b+\sqrt{b^2+a^2}})]=

    -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{(b+\sqrt{b^2+4a^2})}{(b+\sqrt{b^2+a^2})}  }]=

    -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{b+\sqrt{b^2(1+\frac{4a^2}{b^2})}}{b+\sqr  t{b^2(1+\frac{a^2}{b^2})}}]=

    -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{b+|b|\sqrt{(1+\frac{4a^2}{b^2})}}{b+|b|\  sqrt{(1+\frac{a^2}{b^2})}}]

    b goes to -\infty therefore |b|=-b, so

    -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{b-b\sqrt{(1+\frac{4a^2}{b^2})}}{b-b\sqrt{(1+\frac{a^2}{b^2})}})}=

    -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{1-\sqrt{(1+\frac{4a^2}{b^2})}}{1-\sqrt{(1+\frac{a^2}{b^2})}})}

    the limit is in the indeterminate form \frac{0}{0} so by using L'Hospital's rule I get

    -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{-\frac{1}{2\sqrt{(1+\frac{4a^2}{b^2})}}(1+\frac{4a^  2}{b^2})'}{-\frac{1}{2\sqrt{(1+\frac{a^2}{b^2})}}(1+\frac{a^2}  {b^2})'}=

    -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{\frac{1}{2\sqrt{(1+\frac{4a^2}{b^2})}  }(4a^2(-2b^{-1}))}{\frac{1}{2\sqrt{(1+\frac{a^2}{b^2})}}(a^2(-2b^{-1}))}=

    -\ln{2}+\ln{\lim_{b\rightarrow-\infty}4\frac{\sqrt{1+\frac{a^2}{b^2}}}{\sqrt{1+\f  rac{4a^2}{b^2}}}=

    -\ln{2}+\ln{4}=

    -\ln{2}+\ln{2^2}=

    -\ln{2}+2\ln{2}=

    \ln{2}


    Alternatively, it occured to me that the function \frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}} is an even function, therefore the integral from -\infty to 0 has the same value as the integral from 0 to \infty, which is easier to calculate (without using L'Hospital's rule).


    @MINOAMAN: Yes, I am Greek.
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