Orthogonality of Legendre Polynomials from Jackson

[Demon117]

Hello all!

I am trying to work through and understand the derivation of the Legendre Polynomials from Jackson's Classical electrodynamics. I have reached a part that I cannot get through however. Jackson starts with the following orthogonality statement and jumps (as it seems) in his proof:

Equation 3.17 states:

[itex]\int P_{l'}[\frac{d}{dx} ([1-x^{2}]\frac{dP_{l}}{dx})+l(l+1)P_{l}(x)]dx=0[/itex]

He mentions integration by parts on the "first term" but I don't see how he gets to

Equation 3.18:

[itex]\int [(x^{2}-1)\frac{dP_{l}}{dx} \frac{dP_{l'}}{dx} +l(l+1)(P_{l'}(x)P_{l}(x))]dx=0[/itex]

I don't see this. Could someone please explain or give a hint to the intermediate step here? I'm afraid I just do not see it. Thanks.

 

[Born2bwire]

You understand integration by parts, yes? To wit,

[tex] \int udv = uv-\int vdu [/tex]

So by the chain rule we can also say

[tex] \int u \frac{dv}{dx}dx = uv-\int v \frac{du}{dx}dx [/tex]

So it is easy to see that [itex]u=P_{\ell'}[/itex] and [itex]v=\left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right][/itex]. In this manner we arrive at

[tex] \int_{-1}^1 P_{\ell'} \frac{d}{dx} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] dx = \left. P_{\ell'} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] \right|^{x=1}_{x=-1} - \int_{-1}^1 \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] \frac{d P_{\ell'}}{dx} dx [/tex]

The first term works out to be zero obviously and thus we say that,

[tex] \int_{-1}^1 P_{\ell'} \frac{d}{dx} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] dx = \int_{-1}^1 \left[ \left(x^2-1\right)\frac{dP_\ell}{dx}\right] \frac{d P_{\ell'}}{dx} dx [/tex]

 

[Demon117]

You understand integration by parts, yes? To wit,

[tex] \int udv = uv-\int vdu [/tex]

So by the chain rule we can also say

[tex] \int u \frac{dv}{dx}dx = uv-\int v \frac{du}{dx}dx [/tex]

So it is easy to see that [itex]u=P_{\ell'}[/itex] and [itex]v=\left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right][/itex]. In this manner we arrive at

[tex] \int_{-1}^1 P_{\ell'} \frac{d}{dx} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] dx = \left. P_{\ell'} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] \right|^{x=1}_{x=-1} - \int_{-1}^1 \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] \frac{d P_{\ell'}}{dx} dx [/tex]

The first term works out to be zero obviously and thus we say that,

[tex] \int_{-1}^1 P_{\ell'} \frac{d}{dx} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] dx = \int_{-1}^1 \left[ \left(x^2-1\right)\frac{dP_\ell}{dx}\right] \frac{d P_{\ell'}}{dx} dx [/tex]

Of course I understand Integration by parts, and in fact I just barely started it myself so this whole conversation is null. Thank you for your time, but I just had an aha! moment :) Sorry about that.