Please help verifying Bessel function of zero order

[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}But if you look at page 7 of this article:http://www.math.kau.se/mirevine/mf2bess.pdf

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

 

[SteamKing]

Here is another derivation of J0(x) which might be clearer:

http://math.arizona.edu/~zakharov/BesselFunctions.pdf

See pp. 11-12

 

[yungman]

Here is another derivation of J0(x) which might be clearer:

http://math.arizona.edu/~zakharov/BesselFunctions.pdf

See pp. 11-12

Thanks, I need to read through this first.

 

[yungman]

Here is another derivation of J0(x) which might be clearer:

http://math.arizona.edu/~zakharov/BesselFunctions.pdf

See pp. 11-12

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.

 

[yungman]

Anyone?

 

[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.

 

[yungman]

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.

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 textbooks showing
J_0(x)=\int_{0}^{\pi}e^{jz\sin\theta}d\theta
Instead of
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 $J_0(x)=\int_{-\pi}^{\pi}e^{jz\sin\theta}d\theta$

 

[SteamKing]

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.

 

[yungman]

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.

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!

 

[Avodyne]

I have two textbooks showing
J_0(x)=\int_{0}^{\pi}e^{jz\sin\theta}d\theta

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

 

[SteamKing]

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

Care to elaborate?

 

[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\thetaOn 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

 

[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

 

[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.