How Do You Integrate \( C_\ell \) Using Integration by Parts?

[bjnartowt]

Homework Statement

consider the integral,
{C_\ell } = \frac{{2\ell + 1}}{{2{j_\ell }(kr)}}\frac{1}{{{2^\ell }\ell !}}\int_{ - 1}^{ + 1} {\frac{{{d^\ell }({{({x^2} - 1)}^\ell })}}{{d{x^\ell }}}{e^{{\bf{i}}krx}}dx}

how do you do it?

Homework Equations

l = integer, as in the oft-occurring l*(l+1)*hbar
j sub l = the l-th bessel function
x = cos(theta)

The Attempt at a Solution

this is related to Shankar 12.6.10

 

[vela]

Use the addition theorem for spherical harmonics.

http://en.wikipedia.org/wiki/Spherical_harmonics#Addition_theorem

 

[bjnartowt]

Hi Vela, thanks for your post. I looked up the addition theorem for spherical harmonics as it appeared in Jackson's electrodynamics text (rather than Wikipedia), and tried to use that to represent the Legendre polynomial I have to integrate. I wound up just writing a tautology (e.g., x = x; true, and totally useless). Based on what the attached .pdf says, could you elaborate on what it means to use the addition theorem? Is there some standard trick I should be doing that's written in a math-methods text (e.g., Boas or Arfken/Weber) that I should know of or could read up on?

 

[vela]

Sorry, I was thinking of a completely different problem. Just ignore what I said earlier.

I found the problem in Arfken on page 665, problem 12.4.7. You got to:
C_lj_l(kr) = \frac{2l+1}{2}\int_{-1}^1 P_l(x)e^{ikrx}\,dx
The hint for the problem says to then differentiate this l times with respect to kr, then set r=0, and, finally, evaluate the remaining integral.

 

[bjnartowt]

Ah, brilliant! Something new to try--I found the relation you're talking about, and Arfken/Weber provides enough instruction to get me started and keep me busy for awhile... Be back in a bit, maybe...

 

[sgd37]

it is early so this may be unfounded but could you not use integration by parts

to have a series of expressions like

\Sigma_{j=0}^{l-1} [ \frac{d^j (x^2 -1)^l}{dx^j} e^{ikrx}]_{-1}^{1} + (-1)^l i^l (kr)^l \int_{-1}^{+1} (x^2 - 1 )^l e^{ikrx}

since the series ends at l-1 the terms in the series will all have a factor of (x^2 -1) which is 0 at both +1 and -1 so you are left with the final integral does that help at all