1. Jul 13, 2013

### yungman

I worked out and verify these two formulas:

$$\int_0^\pi \cos(x sin(\theta)) d\theta \;=\;\ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} \pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)(2n!)}\;=\; \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} \pi}{(2^2)(4^2)(6^2)...(2n)^2}$$
$$\int_0^\pi \sin(x sin(\theta)) d\theta \;=\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{(2n+1)!} \frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))} \;=\;\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{3^2 \cdot 5^2\cdot 7^2\cdot \cdot \cdot \cdot (2n+1)^2}$$

I verified these two statements to be correct:
$$J_0(x)=\frac{1}{\pi}\int_0^{\pi}\cos(x\sin\theta)d\theta=\frac{1}{2\pi}\int_0^{2\pi}\cos(x\sin\theta)d\theta$$
$$\frac{1}{2\pi}\int_0^{2\pi}\sin(x\sin\theta)d\theta=0$$

But then it concluded
$$J_0(x)=\frac{1}{\pi}\int_0^{\pi}e^{jx\sin\theta}d\theta$$
Which implied $\int_0^\pi \sin(x sin(\theta)) d\theta=0$. I cannot verify this. I can agree with $\int_0^{2\pi} \sin(x sin(\theta)) d\theta=0$.

But as you can see $\int_0^\pi \sin(x sin(\theta)) d\theta \;=\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{3^2 \cdot 5^2\cdot 7^2\cdot \cdot \cdot \cdot (2n+1)^2}≠0$. You cannot just jump from integrating over $2\pi$ back to $\pi$.

I verified that $J_0(x)=\frac{1}{\pi}\int_0^{\pi}e^{jx\sin\theta}d\theta$ from other sources. So what am I missing? Please help.

Thanks

2. Jul 13, 2013

### SteamKing

Staff Emeritus
3. Jul 13, 2013

### yungman

4. Jul 15, 2013

### yungman

I have gone through and understand page 11 and 12. My problem is the interval of integration. From different articles, they use different integral for the same equation and I don't think that will work. For example, it this article, the interval is always from $-\pi$ to $\pi$.
$$J_0(x)=\int_{-\pi}^{\pi}e^{jz\sin\theta}d\theta$$
Part of the reasoning is integration of $e^{jk\theta}$ from $-\pi$ to $\pi$ always equal zero. But this is absolutely not true in integrate from 0 to $\pi$.

The original post I sited claimed $J_0(x)=\int_{0}^{\pi}e^{jz\sin\theta}d\theta$. This is going to be very different from integrating from $-\pi$ to $\pi$.

It almost seems like in all the articles, integrating from $-\pi$ to $\pi$ or 0 to $2\pi$ gives double the value of from 0 to $\pi$. I just can't get pass this, please help.

5. Jul 15, 2013

### yungman

Anyone?

6. Jul 16, 2013

### Avodyne

The paper you cite appears to have a typo; the upper limit of integration should be 2π.
According to Mathematica, $\int_0^\pi \sin(x\sin(\theta)) d\theta = \pi H_0(x),$ where $H_0(x)$ is the Struve function.

7. Jul 16, 2013

### yungman

But this is not my question. I know $\int_0^\pi \sin(x\sin(\theta)) d\theta$ is not the same as $\int_0^{2\pi} \sin(x\sin(\theta)) d\theta$

I have two text books showing
$$J_0(x)=\int_{0}^{\pi}e^{jz\sin\theta}d\theta$$
$$J_0(x)=\int_{-\pi}^{\pi}e^{jz\sin\theta}d\theta=\int_{0}^{2\pi}e^{jz\sin\theta}d\theta$$
But yet I gone through the derivation to show $J_0(x)=\int_{-\pi}^{\pi}e^{jz\sin\theta}d\theta$

8. Jul 16, 2013

### SteamKing

Staff Emeritus
G.N. Watson wrote a very large book on Bessel functions which was first published in 1922.
A copy of this work is available here: http://archive.org/details/ATreatiseOnTheTheoryOfBesselFunctions

If you go to p. 19 of the text, Section 2.2, "Bessel's integral for the Bessel coefficients", Watson discusses in detail the derivation of Jn(z) and how the limits of integration can be manipulated.

HTH.

9. Jul 16, 2013

### yungman

Thanks, you are of great help. Again, I have to spend some time to read, write it out to see whether I can understand it. This is a really hard subject!!!

10. Jul 16, 2013

### Avodyne

These textbooks are wrong. The imaginary part of the right-hand side is not zero.

11. Jul 16, 2013

### SteamKing

Staff Emeritus
Care to elaborate?

12. Jul 16, 2013

### yungman

I have been looking through the few articles and notes I have.

$$\int sin^m(\theta) d\theta=-\frac{1}{m}sin^{m-1}(\theta)cos(\theta)+\frac{m-1}{m} \int sin^{m-2}(\theta) d\theta$$.
I gone through integration over intevals [0,$2\pi$],[$-\pi,\pi$][0,$\pi$] and even [0,$\pi$/2],
$$\sin^{m-1}\theta\cos\theta=0\;\hbox{ in all the above intervals.}$$
$$\Rightarrow \int sin^m(x) dx=\frac{m-1}{m} \int sin^{m-2}(x) dx \;\hbox { for the intervals above.}$$(1)

$$e^{jx\sin\theta}=\sum_{m=0}^{\infty}\frac{(jx\sin\theta)^m}{m!}$$
For $\int_0^{2\pi}e^{jx\sin\theta}d\theta$, only m=even result in non zero, so let $m=2k$. (2)
$$\Rightarrow\;\int_0^{2\pi}e^{jx\sin\theta}d\theta=\int_0^{2\pi}\left[1-\frac{x^2\sin^2\theta}{2!}+\frac{x^4\sin^4\theta}{4!}-\frac{x^6\sin^6\theta}{6!}\cdot\cdot\cdot\right]d\theta$$(3)

$$\hbox {For m=2k, }\; \int_0^{2\pi} sin^{2k}(\theta) d\theta=\frac{2k-1}{2k} \int sin^{2k-2}(\theta) d\theta=\frac{3\cdot 5\cdot 7\cdot\cdot\cdot (2k-1)}{4\cdot 6\cdot 8 \cdot\cdot\cdot (2k)}\pi$$
$$\hbox{As }\; \int_0^{2\pi} sin^2(\theta) d\theta=\pi$$

Also if you use the result of this but instead integrate from [0,$\pi$]. You'll get
$$\hbox {For m=2k, }\; \int_0^{\pi} sin^{2k}(\theta) d\theta=\frac{2k-1}{2k} \int sin^{2k-2}(\theta) d\theta=\frac{3\cdot 5\cdot 7\cdot\cdot\cdot (2k-1)}{4\cdot 6\cdot 8 \cdot\cdot\cdot (2k)}\frac{\pi}{2}$$
$$\hbox{As }\; \int_0^{\pi} sin^2(\theta) d\theta=\frac{\pi}{2}$$

This make it looks like
$$J_0(x)=\frac{1}{2\pi}\int_0^{2\pi}e^{jx\sin\theta}d\theta=\frac{1}{\pi}\int_0^{\pi} e^{jx\sin\theta}d\theta=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}e^{jx\sin\theta}d\theta$$

On the first pass, it looks reasonable. BUT the fraud is it is using the original assumption that only m=2k result in non zero. This is ABSOLUTELY NOT TRUE in the interval of integration is [0,$\pi$] and [0,$\pi$/2]. If you go back to (1) and (2) and (3)

Then work your way down, it will not give you the same answer. This is because integration of odd power of sine function is no longer zero and cannot just take m=2k anymore. The function is not real, it's complex like Avodyne said.

This is as far as my study. This is the very point I got stuck for days and until I put the 4 articles and went through the derivations to come to this point. I don't dare to say the book is wrong. But you are going to have to answer this first before I can be convinced. I know this is not something you can just have a simple answer, I hope some of you are willing to spend some time on this. Something is fishy here.

Thanks

Last edited: Jul 16, 2013
13. Jul 17, 2013

### yungman

I forgot to put this in:
$$e^{jx\theta}=\sum_{m=0}^{\infty}\frac{(jx\sin\theta)^m}{m!}=\frac 1 {0!}+\frac{jx\sin\theta}{1!}-\frac{x^2\sin^\theta}{2!}-\frac{jx^3\sin^3\theta}{3!}+\frac{x^4\sin^4\theta}{4!}+\frac{jx^5\sin^5\theta}{5!}\cdot \cdot \cdot \cdot$$
$$\int_0^{\pi}\sin\theta d\theta=2\;\Rightarrow\;\int_0^{\pi}\sin^m\theta d\theta=\frac{2\cdot 3\cdot 4\cdot \cdot \cdot \cdot (m-1)}{3\cdot 4\cdot 5\cdot \cdot \cdot \cdot m}\int_0^{\pi}\sin\theta d\theta\;=\;\frac{2\cdot 3\cdot 4\cdot \cdot \cdot \cdot (m-1)}{3\cdot 4\cdot 5\cdot \cdot \cdot \cdot m}2$$

The odd terms are all imaginary.

To be even clearer:
$$e^{jx\sin \theta}=\cos(x\sin\theta)+j\sin(x\sin\theta)=\sum_0^{\infty}\frac{(-1)^m x^{2m}\sin^{2m}\theta}{(2m)!}+j\sum_0^{\infty}\frac{(-1)^m x^{2m+1}sin^{2m+1}\theta}{(2m+1)!}=\sum_0^{\infty}\frac{(-1)^m x^{2m} sin^{2m}\theta}{(2m)!}\left[1+j\frac{x\sin\theta}{(2m+1)}\right]$$

You can see the imaginary terms appear if integration of sine is not zero. And
$$\int_0^{\pi}\sin\theta d\theta=2$$

Last edited: Jul 17, 2013
14. Jul 18, 2013

### yungman

Apparently there is another person here DeIdeal that agree with my assertion. https://www.physicsforums.com/showthread.php?t=701919

I have been searching a lot of articles and in my books, others do not represent it this way.

Don't tell me I wasted days on error of this.