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Math Help - evaluation of integral

  1. #1
    Super Member Random Variable's Avatar
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    evaluation of an integral

    \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+ a \cos x)}{\cos x} \ dx \ \ a>0

    I wrote the integral as  \int_{0}^{\frac{\pi}{2}} \int_{0}^{a} \frac{1}{1+t \cos x} \ dt \ dx .


    Then I switched the order of integration and used Weierstrass substitution to get  2 \int_{0}^{a} \frac{1}{\sqrt{1-t^{2}}} \arctan \sqrt{\frac{1-t}{1+t}} \ dt .


    Finally I let  t= \cos 2z and got  \frac{1}{2} \Big(\frac{\pi^{2}}{4} - \arccos^{2}(a) \Big) = \frac{1}{2} \arcsin (a) \Big(\pi - \arcsin (a) \Big)

    Sometimes Wolfram Alpha returns  \frac{1}{2a} \arcsin (a) \Big(\pi \sqrt{a^{2}} - a\arcsin (a) \Big) which reduces to my solution if  a is real and positive.


    But other times it returns  \frac{\pi \sqrt{a^{2}}}{2a} \arcsin(a) which reduces to  \frac{\pi}{2} \arcsin (a) if a is real and positive. On another forum this was stated as the solution. But this cannot be the solution for when  a >1 and it's not even the solution for when  a= 1 . Is it the solution for  0<a<1?


    EDIT: It's not the solution for  a= \frac{1}{2} either.
    Last edited by Random Variable; July 19th 2011 at 02:50 PM.
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  2. #2
    Grand Panjandrum
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    Re: evaluation of an integral

    Quote Originally Posted by Random Variable View Post
    \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+ a \cos x)}{\cos x} \ dx \ \ a>0

    I wrote the integral as  \int_{0}^{\frac{\pi}{2}} \int_{0}^{a} \frac{1}{1+t \cos x} \ dt \ dx .


    Then I switched the order of integration and used Weierstrass substitution to get  2 \int_{0}^{a} \frac{1}{\sqrt{1-t^{2}}} \arctan \sqrt{\frac{1-t}{1+t}} \ dt .


    Finally I let  t= \cos 2z and got  \frac{1}{2} \Big(\frac{\pi^{2}}{4} - \arccos^{2}(a) \Big) = \frac{1}{2} \arcsin (a) \Big(\pi - \arcsin (a) \Big)

    Sometimes Wolfram Alpha returns  \frac{1}{2a} \arcsin (a) \Big(\pi \sqrt{a^{2}} - a\arcsin (a) \Big) which reduces to my solution if  a is real and positive.


    But other times it returns  \frac{\pi \sqrt{a^{2}}}{2a} \arcsin(a) which reduces to  \frac{\pi}{2} \arcsin (a) if a is real and positive. On another forum this was stated as the solution. But this cannot be the solution for when  a >1 and it's not even the solution for when  a= 1 . Is it the solution for  0<a<1?


    EDIT: It's not the solution for  a= \frac{1}{2} either.
    If its any help here is a plot of the numerical integral
    Attached Thumbnails Attached Thumbnails evaluation of integral-integralplot.png  
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  3. #3
    Super Member Random Variable's Avatar
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    Re: evaluation of an integral

    I don't see how  \frac{\pi}{2} \arcsin (a) can ever be the solution (except for a=0). Perhaps Wolfram Alpha sometimes makes an incorrect simplification.
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  4. #4
    Grand Panjandrum
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    Re: evaluation of an integral

    Quote Originally Posted by Random Variable View Post
    I don't see how  \frac{\pi}{2} \arcsin (a) can ever be the solution (except for a=0). Perhaps Wolfram Alpha sometimes makes an incorrect simplification.
    When faced with difficulties of the kind we have with this integral I always ask myself why do I need to know? There may be ways of approximating this that meet your needs without having a closed form expression for the integral itself.

    For instance for small a we may expand the log as a power series and we end up with a simpler integrals.

    CB
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