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Please help, I am running out of my wits trying to solving this equation:

[tex]\int_{0}^{2\pi}e^{jz\cos \theta}\sin(m\theta)d\theta[/tex]

This is part of a bigger equation involving integral representation of Bessel function. I posted on all the math forums and nobody can help yet. I posted on the homework forum here about the Bessel function and still no luck. I read about all the articles on integral representation of Bessel function ( there are less than two handful of it!!!) and have no luck.

I tried letting [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}\sin(m\theta)d\theta=\int_0^{\pi} e^{jz\cos( \theta-\pi)} \sin(m\theta-m\pi)d\theta=\int_{0}^{\pi}e^{-jz\cos \theta}(-1)^m \sin(m\theta)d\theta[/tex]

I am not seeing it get simpler.

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)]\sin(m\theta) d\theta\;= \int_0^{2\pi}\cos (z\cos \theta)\sin(m\theta) d\theta +j \int_0^{2\pi}\sin (z\cos \theta)\sin(m\theta) d\theta [/tex]

First integral is zero, the second integral is not zero only

[tex]z\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!}\sin(m\theta)d\theta[/tex].

then substitute [itex]\sin(m\theta)=\frac{e^{jm\theta}-e^{-jm\theta}}{2j}[/itex] then using binomial expansion of [itex]\cos^p \theta=\left[\frac{e^{j\theta}+e^{-j\theta}}{2}\right]^p[/itex] and use the fact [itex]\int_0^{2\pi} e^{jm\theta}d\theta=0[/itex]. Still no luck.

Thanks

[tex]\int_{0}^{2\pi}e^{jz\cos \theta}\sin(m\theta)d\theta[/tex]

This is part of a bigger equation involving integral representation of Bessel function. I posted on all the math forums and nobody can help yet. I posted on the homework forum here about the Bessel function and still no luck. I read about all the articles on integral representation of Bessel function ( there are less than two handful of it!!!) and have no luck.

I tried letting [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}\sin(m\theta)d\theta=\int_0^{\pi} e^{jz\cos( \theta-\pi)} \sin(m\theta-m\pi)d\theta=\int_{0}^{\pi}e^{-jz\cos \theta}(-1)^m \sin(m\theta)d\theta[/tex]

I am not seeing it get simpler.

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)]\sin(m\theta) d\theta\;= \int_0^{2\pi}\cos (z\cos \theta)\sin(m\theta) d\theta +j \int_0^{2\pi}\sin (z\cos \theta)\sin(m\theta) d\theta [/tex]

First integral is zero, the second integral is not zero only

[tex]z\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!}\sin(m\theta)d\theta[/tex].

then substitute [itex]\sin(m\theta)=\frac{e^{jm\theta}-e^{-jm\theta}}{2j}[/itex] then using binomial expansion of [itex]\cos^p \theta=\left[\frac{e^{j\theta}+e^{-j\theta}}{2}\right]^p[/itex] and use the fact [itex]\int_0^{2\pi} e^{jm\theta}d\theta=0[/itex]. Still no luck.

Thanks

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