1. Jul 23, 2013

### yungman

$$\int_{0}^{2\pi}e^{jz\cos \theta}\sin(m\theta)d\theta$$
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

Last edited: Jul 23, 2013
2. Jul 23, 2013

### 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.

Last edited: Jul 23, 2013
3. Jul 23, 2013

### yungman

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.

Thanks

Last edited: Jul 23, 2013
4. Jul 23, 2013

### 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.

Last edited: Jul 23, 2013
5. Jul 23, 2013

### yungman

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

Last edited: Jul 23, 2013
6. Jul 23, 2013

### 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.

7. Jul 23, 2013

### yungman

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!!

8. Jul 23, 2013

### hilbert2

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.

9. Jul 23, 2013

### yungman

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.