How can I simplify this Bessel function integral?

[yungman]

Please help, I am running out of my wits trying to solving this equation:
\int_{0}^{2\pi}e^{jz\cos \theta}\sin(m\theta)d\theta
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 \theta=\theta-\pi
\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
\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
I am not seeing it get simpler.
I tried \;e^{jz\cos \theta}=\cos (z\cos \theta) +j\sin(z\cos \theta)\; Where you need to solve \;\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
First integral is zero, the second integral is not zero only
z\cos\theta=m\theta
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 m\theta no matter what.I tried \;e^{jz\cos \theta}=\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}\; Where you need to solve \;\int_0^{2\pi}\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}\sin(m\theta)d\theta.
then substitute \sin(m\theta)=\frac{e^{jm\theta}-e^{-jm\theta}}{2j} then using binomial expansion of \cos^p \theta=\left[\frac{e^{j\theta}+e^{-j\theta}}{2}\right]^p and use the fact \int_0^{2\pi} e^{jm\theta}d\theta=0. Still no luck.Thanks

 

[hilbert2]

The very reason why we define the Bessel functions is that integrals such as that can not be represented in terms of elementary functions.

Imagine a situation that the sine and cosine functions were unknown to us. In that case the differential equation y''(x)+y(x)=0 would seem intractable, no matter how we tried, we could not represent its solutions in terms of functions known to us. We could, however, form a power series representation for the solutions or use numerical integration. Then the logical next step would be to define two new functions sin x and cos x to be solutions of that DE with certain initial conditions.

 

[yungman]

The very reason why we define the Bessel functions is that integrals such as that can not be represented in terms of elementary functions.

Imagine a situation that the sine and cosine functions were unknown to us. In that case the differential equation y''(x)+y(x)=0 would seem intractable, no matter how we tried, we could not represent its solutions in terms of functions known to us. We could, however, form a power series representation for the solutions or use numerical integration. Then the logical next step would be to define two new functions sin x and cos x to be solutions of that DE.

This is part of the equation the the result is supposed to be zero! I am just trying to break it up into smaller pieces and hopefully someone can help.

The original equation I want to proof is

J_m(z)=\frac{1}{\pi j^m}\int_0^{\pi} e^{jz\cos \theta}\cos (m\theta)d\theta
From
J_m(z)=\frac{1}{2\pi }\int_0^{2\pi} e^{j(z\cos \theta-m\theta)}d\theta

So far I have:

Let $\theta=\frac{\pi}{2}-\theta$
\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 \theta-m\frac{\pi}{2}+m\theta)} d\theta=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos \theta+m\theta)}e^{-j(m\frac{\pi}{2})} \;d\theta
J_m(z)=\frac{1}{2\pi j^m}\int_0^{2\pi}e^{jz\cos \theta}[\cos(m\theta)+j\sin(m\theta)]d\theta
e^{-j(\frac{m\pi}{2})}=j^m=j^{-m}

So I have to get rid of the sine integral first! I don't know how to solve this and I am stuck for over a week. Please don't bump this post to the homework. I have been there, been to ALL the math forums I can find already. This is not an easy problem. This is about the last resort already.

Please help.

Thanks

 

[hilbert2]

Sorry, I wasn't paying enough attention when I wrote that reply.

You can see that the integral is zero by considering the symmetry properties of the integrand. Sin(mθ) is an odd function with respect to the midpoint (θ=pi) of the interval over which you integrate. Cos(θ) is an even function with respect to that midpoint. Therefore the integrand is a product of an odd and an even part and is odd in total. This implies that the value of the integral is zero.

 

[yungman]

Sorry, I wasn't paying enough attention when I wrote that reply.

You can see that the integral is zero by considering the symmetry properties of the integrand. Sin(mθ) is an odd function with respect to the midpoint (θ=pi) of the interval over which you integrate. Cos(θ) is an even function with respect to that midpoint. Therefore the integrand is a product of an odd and an even part and is odd in total. This implies that the value of the integral is zero.

Do you mean this:
\;e^{jz\cos \theta}=\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}\;
\int_0^{2\pi}\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}\sin(m\theta)d\theta=\int_0^{2\pi}\sum_0^{\infty}\frac {(jz)^k}{k!}\cos^k(\theta)\sin(m\theta)d\theta

So

\int_0^{2\pi}\cos^k\theta \sin\theta d\theta
Let $u=\cos\theta\Rightarrow\;du=-\sin\theta d\theta$
\Rightarrow\; \int_0^{2\pi}\cos^k\theta \sin\theta d\theta=\int_0^{\2\pi} \frac{\cos^k u}{\sin \theta}\sin (m\theta) d u

How does this become zero? The sine doesn't cancel out!

Thanks

 

[hilbert2]

What I meant, was that you can show that \int^{\pi}_{0}e^{jzcos(\theta)}sin(m\theta)d\theta=-\int^{2\pi}_{\pi}e^{jzcos(\theta)}sin(m\theta)d\theta and the two "halves" of the integral cancel each other. I'm not sure what you mean with that sum formula.

Because cos(x) is an even function, also f(cos(x)) is even, no matter what the function f is.

 

[yungman]

What I meant, was that you can show that \int^{\pi}_{0}e^{jzcos(\theta)}sin(m\theta)d\theta=-\int^{2\pi}_{\pi}e^{jzcos(\theta)}sin(m\theta)d\theta and the two "halves" of the integral cancel each other. I'm not sure what you mean with that sum formula.

Because cos(x) is an even function, also f(cos(x)) is even, no matter what the function f is.

Actually it's integrate over [0,2$\pi$].
\Rightarrow\;\int^{\pi}_{0}e^{jzcos(\theta)}sin(m\theta)d\theta

How do I know $e^{jz\cos\theta}$ an even function? That would do it!

Thanks for all your help.

 

[hilbert2]

How do I know $e^{jz\cos\theta}$ an even function? That would do it!

Let's write $f(θ)=e^{jz\cos\theta}$. We want to prove that this is even with respect to point $\theta=\pi$, in other words $f(\pi + \theta) = f(\pi -\theta)$. We already know that $cos(\pi + \theta) = cos(\pi - \theta)$. Now we get

$f(\pi+\theta)=e^{jz\cos(\pi+\theta)}=e^{jz\cos(\pi-\theta)}=f(\pi-\theta)$ and the claim is proved.

 

[yungman]

Let's write $f(θ)=e^{jz\cos\theta}$. We want to prove that this is even with respect to point $\theta=\pi$, in other words $f(\pi + \theta) = f(\pi -\theta)$. We already know that $cos(\pi + \theta) = cos(\pi - \theta)$. Now we get

$f(\pi+\theta)=e^{jz\cos(\pi+\theta)}=e^{jz\cos(\pi-\theta)}=f(\pi-\theta)$ and the claim is proved.

So
\int_0^{2\pi}\;\hbox{ (even function) X (odd function)}\; d\theta=0
As it is integrate [0,2$\pi$].

Do you happen to be able to help me how

\frac{1}{2\pi}\int_0^{2\pi}e^{jz\cos\theta}\cos(m\theta)d\theta=\frac{1}{\pi}\int_0^{\pi}e^{jz\cos\theta}\cos(m\theta)d\theta

Thanks for all your help, you really make my day.