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Math Help - integral

  1. #1
    Super Member Random Variable's Avatar
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    integral

     \int^{\pi}_{0} e^{\cos x} \cos (\sin x) \ dx
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Random Variable View Post
     \int^{\pi}_{0} e^{\cos x} \cos (\sin x) \ dx
    If f(x) =  e^{\cos x} \cos (\sin x) then f(2\pi-x) = f(x). So the integral from 0 to \pi is half the integral from 0 to 2\pi.

    Next, f(x) is the real part of  e^{\cos x} (\cos (\sin x) + i\sin(\sin x)) = e^{\cos x + i sin x} = e^{e^{ix}}. Put z = e^{ix}. Then the integral  \int^{\pi}_{0}\!\!\! f(x)\, dx becomes \text{Re}\frac12\oint e^z\,\frac{dz}{iz} (integral round the unit circle), which you can evaluate by the residue theorem. (I make the answer \pi.)
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  3. #3
    Super Member Random Variable's Avatar
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    It's not as nice as Opalg's solution, but how about the following:


    Let  I(a) = \int_{0}^{\pi} e^{a \cos x} \cos(a \sin x) \ dx = Re \Big(\int_{0}^{\pi} e^{ae^{ix}} \ dx \Big)

    now assuming it's OK to differentiate inside of the integral with respect to a

     I'(a) = Re \Big(\int_{0}^{\pi} e^{ix} e^{ae^{ix}} \ dx \Big) = Re \Big( \frac{1}{ai} \int_{a}^{-a}e^{u} \ du \Big) = Re \Big( \frac{i}{a} (e^{a}-e^{-a}) \Big) =0

    so  I(a) = C

    which means that I is independent of the value of  a


    to find C let a=0

    then  I(0)= C = \int^{\pi}_{0} dx = \pi

    and  \int_{0}^{\pi} e^{\cos x} \cos(\sin x) \ dx  = I(1) = \pi
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Random Variable View Post
    It's not as nice as Opalg's solution, but how about the following:


    Let  I(a) = \int_{0}^{\pi} e^{a \cos x} \cos(a \sin x) \ dx = Re \Big(\int_{0}^{\pi} e^{ae^{ix}} \ dx \Big)

    now assuming it's OK to differentiate inside of the integral with respect to a

     I'(a) = Re \Big(\int_{0}^{\pi} e^{ix} e^{ae^{ix}} \ dx \Big) = Re \Big( \frac{1}{ai} \int_{a}^{-a}e^{u} \ du \Big) = Re \Big( \frac{i}{a} (e^{a}-e^{-a}) \Big) =0

    so  I(a) = C

    which means that I is independent of the value of  a


    to find C let a=0

    then  I(0)= C = \int^{\pi}_{0} dx = \pi

    and  \int_{0}^{\pi} e^{\cos x} \cos(\sin x) \ dx = I(1) = \pi
    That is cool.
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