Bessel Function, Orthogonality and More

[Post]

Hello,
I'm trying to show that

Integral[x*J₀(a*x)*J₀(a*x), from 0 to 1] = 1/2 * J₁(a)^2

Here, (both) a's are the same and they are a root of J₀(x). I.e., J₀(a) = 0.

I have found and can do the case where you have two different roots, a and b, and the integral evaluates to zero (orthogonality). How do I go about showing this relationship? I can't find details anywhere.

 

[phyzguy]

Try expanding J0 in a power series, collect terms in like powers, and integrate. Then you can also expand the right side in a power series and show the two are equal.

 

[Post]

Hi,
Sorry for my ignorance, but if expanding into a power series don't we have two infinite sums multiplied together? I attempted it but wasn't able to get anywhere nicely (maybe it's beyond me)

I was thinking something more along the lines of this:
http://physics.ucsc.edu/~peter/116C/bess_orthog.pdf
but I don't see the proper modifications that will give me my identity.

Any further hints would be amazing!

 

[phyzguy]

Why isn't equation 15 of the link you sent exactly what you are looking for?

 

[blue_raver22]

Hello,

Thank you for your interest in Bessel functions and orthogonality. This is a very interesting and important topic in mathematics and physics.

To prove the relationship you have mentioned, we will start by using the definition of Bessel functions:

J0(x) = 1/pi * Integral[cos(x*cos(t))dt, from 0 to pi]

Now, let's use the identity cos(x) = (e^(ix) + e^(-ix))/2 and substitute it into the definition of J0(x):

J0(x) = 1/pi * Integral[(e^(ix*cos(t)) + e^(-ix*cos(t)))/2 dt, from 0 to pi]

Next, we will use the property of Bessel functions that J0(x) = J0(-x). This means that we can rewrite the integral as:

J0(x) = 1/pi * Integral[(e^(ix*cos(t)) + e^(-ix*cos(t)))/2 dt, from -pi/2 to pi/2]

Now, let's use the trigonometric identity cos(x)*cos(y) = (cos(x+y) + cos(x-y))/2 and substitute it into the integral:

J0(x) = 1/pi * Integral[(e^(ix*cos(t)) + e^(-ix*cos(t)))/2 dt, from -pi/2 to pi/2]

= 1/pi * Integral[(e^(ix*cos(t)+ix) + e^(ix*cos(t)-ix))/2 dt, from -pi/2 to pi/2]

= 1/pi * Integral[(e^(ix*cos(t)+ix) + e^(ix*cos(t)-ix))/2 dt, from 0 to pi]

= 1/pi * Integral[(e^(ix*cos(t)+ix) + e^(ix*cos(t)-ix))/2 * cos(t) dt, from 0 to pi]

= 1/2 * Integral[e^(ix*cos(t)+ix)*cos(t) dt, from 0 to pi] (using the property that J0(x) = J0(-x))

= 1/2 * Integral[e^(ix*cos(t)+ix)*cos(t) dt, from -pi/2 to pi/2]

= 1/2 * Integral[e^(ix*cos(t)+ix)*cos(t) dt, from -pi/2 to pi/2] +