1

I want to derive equation 3.26 from jackson's book, classical electrodynamics.

$(2l+1)\int_{0}^{1}P_l(x)dx=(-\frac{1}{2})^{(l-1)/2}\dfrac{(2l+1)(l-2)!!}{2(\dfrac{l+1}{2})!}$

where l is odd, using the Rodrigues formula

$P_l(x)=\dfrac{l}{2^l l!}\dfrac{d^l}{dx^l}(x^2-1)^l$

I have tried a direct substitution

\begin{split} (2l+1)\int_{0}^{1}P_l(x)dx&=(2l+1)\int_{0}^{1}\dfrac{l}{2^l l!}\dfrac{d^l}{dx^l}(x^2-1)^ldx\\ &=\dfrac{(2l+1)l}{2^l l!}\int_o^1 \dfrac{d^l}{dx^l} \sum_{k=0}^{l}(-1)^k\binom{l}{k}x^{2(l-k)}dx \end{split}

But I don't know how to continue from here, can you give me some hints?

Otv
  • 21
  • I don't know to get to the final answer, but I do know how to get further. Taking the integral changes the derivative from order $l$ to $l-1$. The rest in the integrand remains the same. Notice that d$^a x^b /$d$x^a$$= (b!) x^{b-a}/[(b-a)!]$ for $b \geq a$ where $a$ and $b$ are both non-negative integers. Now, evaluating the integral should be straightforward. – Kevin Nov 27 '22 at 14:21
  • Since $l$ is odd, letting $l = 2n+1$ may help with further calculations. – Kevin Nov 27 '22 at 14:27

0 Answers0