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Help deriving this Bessel function formula

  1. Jul 19, 2013 #1
    I am studying Bessel Function in my antenna theory book, it said:
    [tex]\pi j^n J_n(z)=\int_0^{\pi} \cos(n\phi)e^{+jz\cos\phi}d\phi[/tex]


    I understand:
    [tex]J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta[/tex]
    Can you show me how do I get to
    [tex]\pi j^m J_m(z)=\int_0^{\pi} \cos(m\theta)e^{+jz\cos\theta}d\theta[/tex]

    I tried ##e^{jm\theta}=\cos m\theta +j\sin m \theta## but it is not easy. Please help.
     
    Last edited: Jul 19, 2013
  2. jcsd
  3. Jul 20, 2013 #2
    Anyone please?
     
  4. Jul 22, 2013 #3
    I since worked out a few steps: Let ##u=\frac{\pi}{2}-\theta##
    [tex]\Rightarrow\;J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta= \frac{1}{2\pi}\int_0^{2\pi} e^{j(z\cos u-m\frac{\pi}{2}+mu)} du=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u+mu)}e^{-j(m\frac{\pi}{2})} \;du [/tex]
    [tex]J_m(z)=\frac{e^{-(jm\frac{\pi}{2})}}{2\pi}\int_0^{2\pi}e^{jz\cos u}[\cos(m\theta)+j\sin(m\theta)]d\theta[/tex]
    [tex]e^{-j(\frac{m\pi}{2})}=\cos\frac{m\pi}{2}-j\sin\frac{m\pi}{2}[/tex]

    For m=odd, ##\;\cos\frac{m\pi}{2}=0## and ##\;-j\sin\frac{m\pi}{2}=j^{-m}##

    For m=even, ##\;j\sin\frac{m\pi}{2}=0##

    For m=0 ##\Rightarrow\;\cos\frac{m\pi}{2}=1,\;\;## m=2 ##\Rightarrow\;\cos\frac{m\pi}{2}=-1,\;\;## m=4 ##\Rightarrow\;\cos\frac{m\pi}{2}=-1,\;\;##.......Therefore ##\Rightarrow\;\cos\frac{m\pi}{2}=j^m=j^{-m};##

    [tex]J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}j\sin(m\theta)d\theta[/tex]

    Now the final step is to get rid of the second term. Please help.

    Thanks
     
    Last edited: Jul 22, 2013
  5. Jul 22, 2013 #4
    Anyone please?
     
  6. Jul 23, 2013 #5
    In the second formula which you have written in OP,break the integral from 0 to pi and from pi to 2pi.Then make a change of variables in second one to get both integrals as 0 to pi.Try to see what comes.
     
  7. Jul 23, 2013 #6
    Thanks for the reply, I have tried this before:

    Let [itex]\theta=\theta-\pi[/itex]
    [tex]\sin(m\theta-m\pi)=\sin(m\theta)\cos(m\pi)-\cos(m\theta)\sin(m\pi)=\sin(m\theta)\cos(m\pi)=\sin(m\theta)(-1)^m[/tex]
    [tex]\Rightarrow\;\int_{\pi}^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta=\int_0^{\pi} e^{jz\cos( \theta-\pi)} j\sin(m\theta-m\pi)d\theta=\int_{0}^{\pi}e^{-jz\cos \theta}j(-1)^m \sin(m\theta)d\theta[/tex]
    I am not seeing it get simpler. Please help.

    I tried [tex]\;e^{jz\cos \theta}=\cos (z\cos \theta) +j\sin(z\cos \theta)\;[/tex] Where you need to solve [tex]\;\int_0^{2\pi}[\cos (z\cos \theta) +j\sin(z\cos \theta)][\cos(m\theta)+j\sin(m\theta)] d\theta\;= \int_0^{2\pi}\cos (z\cos \theta)\cos(m\theta) d\theta - \int_0^{2\pi}\sin (z\cos \theta)\sin(m\theta) d\theta [/tex]
    Both integrals are not zero only if
    [tex]x\cos\theta=m\theta[/tex]
    But this really doesn't look right as x is the variable of a Bessel function and x is going to be limited to maximum value of [itex]m\theta[/itex] no matter what.


    I tried [tex]\;e^{jz\cos \theta}=\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}\;[/tex] Where you need to solve [tex]\;\int_0^{2\pi}\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}[\cos(m\theta)+j\sin(m\theta)]d\theta[/tex].

    Please help, I am getting very desperate after posting this question in all the forums I can find and get no result. I pretty much read all the articles on integral representation of Bessel function I can find on the web( believe me, there are not too many!!! you can count it with two hands!!!).

    Thanks
     
    Last edited: Jul 23, 2013
  8. Jul 24, 2013 #7
    All right,I think it has gone real serious.So time to solve the problem.Now from the integral representation you already know of bessel function(second eqn. in op),you can write
    eixSinθmeimθ.Jm(x)(This is actually the relation from which the integral representation is derived,also m runs from -∞ to ∞ over all integers).Now put θ→(∏/2+θ)(∏/2-θ will work as well) and you get
    eixCosθmimeimθ.Jm(x),now converting it into integral representation(by multiplying e-imθ on both sides and integrating)
    Jm(x)=(1/2∏im)∫02∏ ei(xCosθ-mθ) dθ.Half of the problem is solved now.Now write e(imθ)=Cosmθ+iSinmθ and also you will have to use the expansion of Cos(xCosθ) and Sin(xCosθ) which have expansion in terms of Cosnθ(BOTH OF THEM) multiplied by J0,J1 and so on.So on the basis of orthogonality of Sin and Cos over period of 2∏,You will be able to show that Sinmθ will not contribute and hence you will be able to obtain your relation as you want.One other thing is that the integral from 0 to 2∏ can be converted into 0 to ∏ by what I have said before in my earlier post taking into account the expansion of Cos(xCosθ) and Sin(xCosθ).Try to show it yourself now,it is almost done.(I hope it is not a complete solution,if it is then mentors can edit it)
     
  9. Jul 27, 2013 #8
    I have not check this post as I have worked out the problem
    [tex]J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta[/tex]

    [tex]\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta=\int_{-\pi}^{\pi}even\;\times\;odd \;d\theta\;=\;0[/tex]
    [tex]\Rightarrow\;J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta=\frac{j^{-m}}{2\pi}\left[\int_0^{\pi}e^{jz\cos u}\cos(m\theta)d\theta+\int_{-\pi}^{0}e^{jz\cos \theta}\cos(m\theta)d\theta\right][/tex]

    The rest is shown in:
    https://www.physicsforums.com/showthread.php?t=702751
     
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