Double check the derivation integral representation of Bessel Function

[yungman]

I am reading the article Mirela Vinerean:

http://www.math.kau.se/mirevine/mf2bess.pdf

On page 6, I have a question about
$$e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}]$$

I think there is a mistake at the last term. If you look at n=0, the equation will be:
$$J_0(x)[e^0+(-1)^0e^{-0}]\;=\;j_0(x)[1+1]\;=\;2J_0(x)$$
Which is not correct. The problem is n=0 is being repeated in both the n=+ve and n=-ve.

The equation should be:
$$e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^ne^{-jn\theta}]$$

With this, the first term covers the original n=+ve from 0 to ∞. The second term covers the original from -∞ to -1. Now you only have one term contain n=0. Am I correct?

Thanks

 

[yungman]

Also in the same page right below the equations of the first post:
$$\int_0^{\pi} \sin n\theta \sin m\theta d\theta\;=\;\frac{\pi}{2}\delta_{mn}$$

1) What is $\delta_{mn}$?
2) If m≠n, the result should be zero. But $\int_0^{\pi} \sin n\theta \sin m\theta d\theta\;≠0$ because the integration is from 0 to $\pi$, not from -$\pi$ to +$\pi$.

Thanks

 

[Infrared]

I can't help you with your first post. $\delta_{mn}$ is the Kronecker delta. It is 1 is m=n and 0 if m≠n. https://en.wikipedia.org/wiki/Kronecker_delta

That integral is zero if m≠n. One to show that is using complex exponentials $$\int_0^\pi sin(n\theta)sin(m\theta)d\theta = \int_0^\pi \frac{e^{in\theta}-e^{-in\theta}}{2i}\frac{e^{im\theta}-e^{-im\theta}}{2i} d\theta= -\frac{1}{4} \int_0^\pi e^{i(m+n)\theta}-e^{i(m-n)\theta}-e^{i(n-m)\theta}+e^{-i(m+n)\theta}d\theta$$. Since we know that this integral will be real, we can just consider the real parts of each of the expressions. $\mathrm{Re}(e^{ix})=cos(x)$ so we can write the integral as $$-\frac{1}{4} \int_0^\pi cos((m+n)\theta)-cos((n-m)\theta)-cos((m-n)\theta)+cos(-(m+n)\theta) d\theta$$. This is zero since cosine is odd about $\pi/2$ in sense that $cos(\pi/2+x)=-cos(\pi/2-x)$. Note that you need m≠n. Else, the $cos((n-m)\theta)$ is constantly one and does not integrate to zero.

 

[yungman]

I can't help you with your first post. $\delta_{mn}$ is the Kronecker delta. It is 1 is m=n and 0 if m≠n. https://en.wikipedia.org/wiki/Kronecker_delta

That integral is zero if m≠n. One to show that is using complex exponentials $$\int_0^\pi sin(n\theta)sin(m\theta)d\theta = \int_0^\pi \frac{e^{in\theta}-e^{-in\theta}}{2i}\frac{e^{im\theta}-e^{-im\theta}}{2i} d\theta= -\frac{1}{4} \int_0^\pi e^{i(m+n)\theta}-e^{i(m-n)\theta}-e^{i(n-m)\theta}+e^{-i(m+n)\theta}d\theta$$. Since we know that this integral will be real, we can just consider the real parts of each of the expressions. $\mathrm{Re}(e^{ix})=cos(x)$ so we can write the integral as $$-\frac{1}{4} \int_0^\pi cos((m+n)\theta)-cos((n-m)\theta)-cos((m-n)\theta)+cos(-(m+n)\theta) d\theta$$. This is zero since cosine is odd about $\pi/2$ in sense that $cos(\pi/2+x)=-cos(\pi/2-x)$. Note that you need m≠n. Else, the $cos((n-m)\theta)$ is constantly one and does not integrate to zero.

Thanks for the Kronecker delta, I forgot about this one.

I know the cosine part is fine. But if you look at page 6 of the provided link in the first post, the author claimed orthogonality properties with the sine function. That's the part I am challenging.

Thanks

 

[Infrared]

Thanks for the Kronecker delta, I forgot about this one.

I know the cosine part is fine. But if you look at page 6 of the provided link in the first post, the author claimed orthogonality properties with the sine function. That's the part I am challenging.

Thanks

I am confused. Didn't I just show that $\int_0^\pi sin(mx)sin(nx) dx =0$ where $m \neq n$, which is the orthogonality condition?

 

[yungman]

I can't help you with your first post. $\delta_{mn}$ is the Kronecker delta. It is 1 is m=n and 0 if m≠n. https://en.wikipedia.org/wiki/Kronecker_delta

That integral is zero if m≠n. One to show that is using complex exponentials $$\int_0^\pi sin(n\theta)sin(m\theta)d\theta = \int_0^\pi \frac{e^{in\theta}-e^{-in\theta}}{2i}\frac{e^{im\theta}-e^{-im\theta}}{2i} d\theta= -\frac{1}{4} \int_0^\pi e^{i(m+n)\theta}-e^{i(m-n)\theta}-e^{i(n-m)\theta}+e^{-i(m+n)\theta}d\theta$$. Since we know that this integral will be real, we can just consider the real parts of each of the expressions. $\mathrm{Re}(e^{ix})=cos(x)$ so we can write the integral as $$-\frac{1}{4} \int_0^\pi cos((m+n)\theta)-cos((n-m)\theta)-cos((m-n)\theta)+cos(-(m+n)\theta) d\theta$$. This is zero since cosine is odd about $\pi/2$ in sense that $cos(\pi/2+x)=-cos(\pi/2-x)$. Note that you need m≠n. Else, the $cos((n-m)\theta)$ is constantly one and does not integrate to zero.

An integral does not have to be real, I don't think you can assume it's real and get rid of the imaginary part.

A very simple test is not use exponential and just go with integration with n=0 and m=1

$$\int_0^{\pi} \sin \theta d\theta =-\cos\theta|_0^{\pi}=-[-1-1]=2$$

 

[Infrared]

An integral does not have to be real, I don't think you can assume it's real and get rid of the imaginary part.

It is real. If you wanted to verify this, you could expand all of the $e^{ik\theta}$ (where k is one of the linear combinations of m and n) terms as $cos(k\theta)+isin(k\theta)$ and find that all of the imaginary terms drop out. I skipped this step because the original integral with just sines must be real, and the integral with complex exponentials is only different by a factor of -1/4, so it too must be real.

Edit: I just saw your edit. If n=0, then the integrand is 0 and the integral is trivially zero.

 

[yungman]

It is real. If you wanted to verify this, you could expand all of the $e^{ik\theta}$ (where k is one of the linear combinations of m and n) terms as $cos(k\theta)+isin(k\theta)$ and find that all of the imaginary terms drop out. I skipped this step because the original integral with just sines must be real, and the integral with complex exponentials is only different by a factor of -1/4, so it too must be real.

The forum is so slow that we are cross posting. I just edited the last post, I don't even use exponential, just use n=0 and m=1. that will be just integrate a sine function and show it's not zero.

 

[Infrared]

The forum is so slow that we are cross posting. I just edited the last post, I don't even use exponential, just use n=0 and m=1. that will be just integrate a sine function and show it's not zero.

Yes, it is a bit annoying. See my last edit. Look more carefully at what happens to the integrand if one of m or n is zero.

 

[micromass]

An integral does not have to be real, I don't think you can assume it's real and get rid of the imaginary part.

You're taking the integral of a real function. That will always be real.

 

[yungman]

Yes, it is a bit annoying. See my last edit. Look more carefully at what happens to the integrand if one of m or n is zero.

Yes, I was wrong on the integration. I kept thinking if n=0, then it's an integration of a sine function and it's not zero. But if n=0, the whole thing is zero to start with!

Thanks


Can you help with my first post?

 

[Infrared]

Yes, I was wrong on the integration. I kept thinking if n=0, then it's an integration of a sine function and it's not zero. But if n=0, the whole thing is zero to start with!

Thanks


Can you help with my first post?

Sorry, I can't as I don't know anything about Bessel functions. I hope that you get the answer you need though.

 

[yungman]

Sorry, I can't as I don't know anything about Bessel functions. I hope that you get the answer you need though.

Actually the question has not much to do with Bessel Function. All you need to know is if n is an integer, $J_{-n}(x)=(-1)^n J_n(x)$, the rest is a series problem. It should be

$$\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}\;=\;\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^nJ_n(x)e^{-jn\theta}]$$

Not as the article that:
$$\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}]\;=\;\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=0}^{\infty}[(-1)^nJ_n(x)e^{-jn\theta}]$$

According to the article, n=0 are being repeated in both and result in twice the value for n=0.

 

[yungman]

Anyone please?