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Thread: integral

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

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

    Next, f(x) is the real part of$\displaystyle e^{\cos x} (\cos (\sin x) + i\sin(\sin x)) = e^{\cos x + i sin x} = e^{e^{ix}}$. Put $\displaystyle z = e^{ix}$. Then the integral $\displaystyle \int^{\pi}_{0}\!\!\! f(x)\, dx $ becomes $\displaystyle \text{Re}\frac12\oint e^z\,\frac{dz}{iz}$ (integral round the unit circle), which you can evaluate by the residue theorem. (I make the answer $\displaystyle \pi$.)
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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 $\displaystyle 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 $\displaystyle a$

    $\displaystyle 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 $\displaystyle I(a) = C$

    which means that $\displaystyle I$ is independent of the value of $\displaystyle a$


    to find $\displaystyle C $ let $\displaystyle a=0$

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

    and $\displaystyle \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 $\displaystyle 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 $\displaystyle a$

    $\displaystyle 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 $\displaystyle I(a) = C$

    which means that $\displaystyle I$ is independent of the value of $\displaystyle a$


    to find $\displaystyle C $ let $\displaystyle a=0$

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

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