Alan0354
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I have seen different books claimed \[J_0(x)=\frac{1}{2\pi}\int_0^{2\pi}e^{jx\sin\theta}d\theta=\frac{1}{\pi}\int_0^{\pi} e^{jx\sin\theta}d\theta=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}e^{jx\sin\theta}d\theta\]. I have issue with this. Here is my reasoning:
\[\int \ sin^{m}(\theta) d\theta=-\frac{1}{m}sin^{m-1}(\theta)cos(\theta)+\frac{m-1}{m} \int\; sin^{m-2}(\theta) d\theta \].
I gone through integration over intevals \[[0,2\pi]\],\[[-\pi,\pi]\]\[[0,\pi]\].
\[\sin^{m-1}\theta\cos\theta=0\;\] in all the above intervals.
\[\Rightarrow\; \int \;sin ^m (x) dx=\frac{m-1}{m} \int\; sin^{m-2}(x) dx \;\] 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)
\[For \;m=2k, \;\; \int_0^{2\pi} sin^{2k}(\theta) d\theta=\frac{2k-1}{2k} \int \;sin^{2k-2}(\theta) d\theta=\frac{3\cdot 5\cdot 7\cdot\cdot\cdot (2k-1)}{4\cdot 6\cdot 8 \cdot\cdot\cdot (2k)}\pi\]
\[As \;\; \int_0^{2\pi} sin^2(\theta) d\theta=\pi\]
Also if you use the result of this but instead integrate from \[[0,2\pi]\]. You'll get
\[For \;m=2k\;\; \int_0^{\pi} sin^{2k}(\theta) d\theta=\frac{2k-1}{2k} \int\; sin^{2k-2}(\theta) d\theta=\frac{3\cdot 5\cdot 7\cdot\cdot\cdot (2k-1)}{4\cdot 6\cdot 8 \cdot\cdot\cdot (2k)}\frac{\pi}{2}\]
\[As \;\; \int_0^{\pi} sin^2(\theta) d\theta=\frac{\pi}{2}\]
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 \[[0,\pi]\]. For this interval:.
\[e^{jx\sin\theta}=\sum_{m=0}^{\infty}\frac{(jx\sin\theta)^m}{m!}=\frac {1}{0!}+\frac{jx\sin\theta}{1!}-\frac{x^2\sin^\theta}{2!}-\frac{jx^3\sin^3\theta}{3!}+\frac{x^4\sin^4\theta}{4!}+\frac{jx^5\sin^5\theta}{5!}\cdot \cdot \cdot \cdot\]
\[\int_0^{\pi}\sin\theta \;d\theta=2\;\Rightarrow\;\int_0^{\pi}\sin^m\theta \;d\theta=\frac{2\cdot 3\cdot 4\cdot \cdot \cdot \cdot (m-1)}{3\cdot 4\cdot 5\cdot \cdot \cdot \cdot\; m}\int_0^{\pi}\sin\theta\; d\theta\;=\;\frac{2\cdot 3\cdot 4\cdot \cdot \cdot \cdot \;(m-1)}{3\cdot 4\cdot 5\cdot \cdot \cdot \cdot \;m}2\]
The odd terms are all imaginary.
To be even clearer:
\[e^{jx\sin \theta}=\cos(x\sin\theta)+j\sin(x\sin\theta)=\sum_0^{\infty}\frac{(-1)^m x^{2m}\sin^{2m}\theta}{(2m)!}+j\sum_0^{\infty}\frac{(-1)^m x^{2m+1}sin^{2m+1}\theta}{(2m+1)!}=\sum_0^{\infty}\frac{(-1)^m x^{2m} sin^{2m}\theta}{(2m)!}\left[1+j\frac{x\sin\theta}{(2m+1)}\right]\]
Thank
\[\int \ sin^{m}(\theta) d\theta=-\frac{1}{m}sin^{m-1}(\theta)cos(\theta)+\frac{m-1}{m} \int\; sin^{m-2}(\theta) d\theta \].
I gone through integration over intevals \[[0,2\pi]\],\[[-\pi,\pi]\]\[[0,\pi]\].
\[\sin^{m-1}\theta\cos\theta=0\;\] in all the above intervals.
\[\Rightarrow\; \int \;sin ^m (x) dx=\frac{m-1}{m} \int\; sin^{m-2}(x) dx \;\] 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)
\[For \;m=2k, \;\; \int_0^{2\pi} sin^{2k}(\theta) d\theta=\frac{2k-1}{2k} \int \;sin^{2k-2}(\theta) d\theta=\frac{3\cdot 5\cdot 7\cdot\cdot\cdot (2k-1)}{4\cdot 6\cdot 8 \cdot\cdot\cdot (2k)}\pi\]
\[As \;\; \int_0^{2\pi} sin^2(\theta) d\theta=\pi\]
Also if you use the result of this but instead integrate from \[[0,2\pi]\]. You'll get
\[For \;m=2k\;\; \int_0^{\pi} sin^{2k}(\theta) d\theta=\frac{2k-1}{2k} \int\; sin^{2k-2}(\theta) d\theta=\frac{3\cdot 5\cdot 7\cdot\cdot\cdot (2k-1)}{4\cdot 6\cdot 8 \cdot\cdot\cdot (2k)}\frac{\pi}{2}\]
\[As \;\; \int_0^{\pi} sin^2(\theta) d\theta=\frac{\pi}{2}\]
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 \[[0,\pi]\]. For this interval:.
\[e^{jx\sin\theta}=\sum_{m=0}^{\infty}\frac{(jx\sin\theta)^m}{m!}=\frac {1}{0!}+\frac{jx\sin\theta}{1!}-\frac{x^2\sin^\theta}{2!}-\frac{jx^3\sin^3\theta}{3!}+\frac{x^4\sin^4\theta}{4!}+\frac{jx^5\sin^5\theta}{5!}\cdot \cdot \cdot \cdot\]
\[\int_0^{\pi}\sin\theta \;d\theta=2\;\Rightarrow\;\int_0^{\pi}\sin^m\theta \;d\theta=\frac{2\cdot 3\cdot 4\cdot \cdot \cdot \cdot (m-1)}{3\cdot 4\cdot 5\cdot \cdot \cdot \cdot\; m}\int_0^{\pi}\sin\theta\; d\theta\;=\;\frac{2\cdot 3\cdot 4\cdot \cdot \cdot \cdot \;(m-1)}{3\cdot 4\cdot 5\cdot \cdot \cdot \cdot \;m}2\]
The odd terms are all imaginary.
To be even clearer:
\[e^{jx\sin \theta}=\cos(x\sin\theta)+j\sin(x\sin\theta)=\sum_0^{\infty}\frac{(-1)^m x^{2m}\sin^{2m}\theta}{(2m)!}+j\sum_0^{\infty}\frac{(-1)^m x^{2m+1}sin^{2m+1}\theta}{(2m+1)!}=\sum_0^{\infty}\frac{(-1)^m x^{2m} sin^{2m}\theta}{(2m)!}\left[1+j\frac{x\sin\theta}{(2m+1)}\right]\]
Thank
