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Thread: Integrate Modified Bessel Function

  1. #1
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    Integrate Modified Bessel Function

    I have been given a modified Bessel function $\displaystyle I_{n}(z) $ and it it defined by the integral

    $\displaystyle I_{n}(z) = \frac{1}{\pi}\int_{0}^{\pi}d\theta{e^{zCos\theta}} Cos(n\theta)$

    I have been asked to show that

    (a) $\displaystyle I_{n-1}(z) - I_{n+1}(z)=\frac{2n}{z}I_{n}(z)$

    (b) $\displaystyle I_{n}(z)=I'_{N+1}(z)+\frac{n}{z}I_{n}(z)$

    Firstly i have started by trying to integrate the integral. Using $\displaystyle u=e^{zCos\theta}$ and $\displaystyle dv=Cos(n\theta)d\theta $. Therefore $\displaystyle du=-Sin\theta{e^{zCos\theta}}$ and $\displaystyle v=\int_{}cosn\theta{d\theta}=\frac{1}{n}Sin\theta{ n\theta}$

    Using integration by parts $\displaystyle \int_{}udv=uv-\int_{}vdu$

    I ended up with

    $\displaystyle I_{n}(z) = \frac{1}{\pi}\int_{0}^{\pi}d\theta{e^{zCos\theta}} Cos(n\theta)=\frac{1}{n}Sin\theta{n\theta}e^{zCos\ theta} + \int_{}\frac{1}{n}Sin\theta{Sin\theta}{n\theta}{e^ {zCos\theta}}$

    knowing that $\displaystyle sin\theta{sin\theta} = cos(n\theta+\theta) + cos(n\theta-\theta) $

    Therefore

    $\displaystyle I_{n}(z) = \frac{1}{n}Sin\theta{n\theta}e^{zCos\theta} + \int_{}\frac{1}{n}{cos(n\theta+\theta) + cos(n\theta-\theta)}{n\theta}{e^{zCos\theta}}$

    I'm not sure if i'm on the right track to solving this problem. Any ideas of what i can do next? How do i then show that

    (a) $\displaystyle I_{n-1}(z) - I_{n+1}(z)=\frac{2n}{z}I_{n}(z)$

    (b) $\displaystyle I_{n}(z)=I'_{N+1}(z)+\frac{n}{z}I_{n}(z)$
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  2. #2
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    Start with $\displaystyle I_{n-1}(z)-I_{n+1}(z)=\frac1\pi\int_0^\pi\mathrm e^{z\cos\theta}[\cos(n-1)\theta-\cos(n+1)\theta]\mathrm d\theta$ $\displaystyle =\frac2\pi\int_0^\pi\mathrm e^{z\cos\theta}\sin n\theta\,\sin\theta\,\mathrm d\theta=-\frac2{\pi z}\int_0^\pi\mathrm \sin\,n\theta\,\mathrm d(e^{z\cos\theta})$.

    Integrate the last term by parts, so that

    $\displaystyle I_{n-1}(z)-I_{n+1}(z)=\left[-\frac2{\pi z}\sin n\theta\,e^{z\cos\theta}\right]_0^\pi+\frac{2n}{\pi z}\int_0^\pi e^{z\cos\theta}\cos n\theta\,\mathrm d\theta=\frac{2n}z I_n(z)$.

    For the second one, start with $\displaystyle I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos(n+1)\theta\,\cos\theta\,\mathr m d\theta$.

    Thus $\displaystyle I_n(z)-I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}[\cos n\theta-\cos(n+1)\theta\,\cos\theta]\mathrm d\theta$ $\displaystyle =\frac1\pi\int_0^\pi e^{z\cos\theta}\sin(n+1)\theta\,\sin\theta\,\mathr m d\theta=\frac{n+1}z I_{n+1}(z)$ (similar to the last part of the previous derivation).

    Warning: the last identity is not quite the one given.
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  3. #3
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    What does it mean by I' (I prime) i can't seem to find much information about it.
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  4. #4
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    Quote Originally Posted by Karl Harder View Post
    What does it mean by I' (I prime) i can't seem to find much information about it.
    It's the derivative of $\displaystyle I$. If $\displaystyle I_{n+1}(z) = \frac{1}{\pi}\int_0^{\pi} e^{z \cos \theta} \cos (n+1) \theta\, d \theta $ then $\displaystyle I'_{n+1}(z) = \frac{1}{\pi}\int_0^{\pi} e^{z \cos \theta} \cos \theta \cos (n+1) \theta\, d \theta $.
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  5. #5
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    Quote Originally Posted by halbard View Post
    Start with $\displaystyle I_{n-1}(z)-I_{n+1}(z)=\frac1\pi\int_0^\pi\mathrm e^{z\cos\theta}[\cos(n-1)\theta-\cos(n+1)\theta]\mathrm d\theta$ $\displaystyle =\frac2\pi\int_0^\pi\mathrm e^{z\cos\theta}\sin n\theta\,\sin\theta\,\mathrm d\theta=-\frac2{\pi z}\int_0^\pi\mathrm \sin\,n\theta\,\mathrm d(e^{z\cos\theta})$.

    Integrate the last term by parts, so that

    $\displaystyle I_{n-1}(z)-I_{n+1}(z)=\left[-\frac2{\pi z}\sin n\theta\,e^{z\cos\theta}\right]_0^\pi+\frac{2n}{\pi z}\int_0^\pi e^{z\cos\theta}\cos n\theta\,\mathrm d\theta=\frac{2n}z I_n(z)$.

    For the second one, start with $\displaystyle I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos(n+1)\theta\,\cos\theta\,\mathr m d\theta$.

    Thus $\displaystyle I_n(z)-I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}[\cos n\theta-\cos(n+1)\theta\,\cos\theta]\mathrm d\theta$ $\displaystyle =\frac1\pi\int_0^\pi e^{z\cos\theta}\sin(n+1)\theta\,\sin\theta\,\mathr m d\theta=\frac{n+1}z I_{n+1}(z)$ (similar to the last part of the previous derivation).

    Warning: the last identity is not quite the one given.
    Does that mean that if

    $\displaystyle I_{n}(z)=\frac1\pi\int_0^\pi e^{zcos\theta} \cos n\theta\, d\theta $

    then

    $\displaystyle I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\,\cos\theta\, d\theta $
    Last edited by Karl Harder; Aug 30th 2009 at 04:33 AM.
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  6. #6
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    Quote Originally Posted by Karl Harder View Post
    Does that mean that if

    $\displaystyle I_{n}(z)=\frac1\pi\int_0^\pi e^{zcos\theta} \cos n\theta\, d\theta $

    then

    $\displaystyle I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\,\cos\theta\, d\theta $
    Yep, it sure does.
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  7. #7
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    I seem to be missing the $\displaystyle \cos{\theta} $ from the result. I'm not sure where i might have gone wrong. I have checked it 3 times each different ways and i still keep getting the same answer.

    $\displaystyle I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\, d\theta $

    I am using
    $\displaystyle I'_{n}(z) =I_{n+1}+\frac{n}{z} I_{n}(z) $

    which can be rewritten as
    $\displaystyle I'_{n}(z) =I_{n+1}(z)+\frac{1}{2} I_{n-1}(z)-\frac{1}{2}I_{n+1}(z) $ which was prooven in the first part of this question.

    Thus
    $\displaystyle I'_{n}(z) =\frac{1}{2}(I_{n+1}(z)+I_{n-1}(z))$

    And then i put the values in for $\displaystyle I_{n+1}(z) $ and $\displaystyle I_{n-1}(z))$

    and somehow i end up with the answer $\displaystyle I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\, d\theta $

    Although it should be
    $\displaystyle I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\,$$\displaystyle \cos \theta\ d\theta $
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