Normalization for final result

Define $A^k_{n,s} = \int^1_1 P^k_n(x)P^{k}_s(x)dx$ , then by separation of variables:

$$A^k_{n,s} = \int^1_1(1-x^2)^k\left(\frac{d^k}{dx^k} P_n(x)\right)\left(\frac{d^k}{dx^k}P_s(x)\right)dx$$

$$= \int^1_1(1-x^2)^k\left(\frac{d^k}{dx^k} P_n(x)\right)\left(\frac{d^{k-1}}{dx^{k-1}}P_s(x)\right)dx$$

$$-\int\left(\frac{d^{k-1}}{dx^{k-1}}P_s(x)\right)\frac{d}{dx}\left((1-x^2)^k\frac{d^k}{dx^k}P_n(x)\right)dx$$ (3.14)

Introduce $z'=\frac{d^k}{dx^k}P_n(x)$ . From our previous discussion, we find that,

$$\frac{d}{dx}\left((1-x^2)z'\right)=\left((k(k+1)-l(l+1)\right)z=0$$ (3.15)

Or just as easily,

$$\frac{d}{dx}\left((1-x^2)^{k+1}z'\right) = \left(k(k+1)-l(l+1)\right)(1-x^2)^kz$$

$$= (k-l)(k+l+1)(1-x^2)^kz=0$$

Here we need to compensate for the following: The z used in this previous equation was built with:

$$z'=\frac{d^k}{dx^k}P_n(x) \rightarrow z=\frac{d^{k-1}}{dx^{k-1}}P_n(x)$$ (3.16)

Thus we must shift $k \rightarrow k-1$ . Thus,

$$A^k_{n,s}=\int^1_{-1}P^k_n(x)P^k_s(x)dx = (n-k+1)(n+k)\int^1_-1P^{k-1}_sP^{k-1}_ndx$$

$$A^k_{n,s} = (n-k+1)(n+k)A^{k-1}_{n,s}$$

$$A^k_{n,s} = (n-k+1)(n+k)(n-k+2)(n+k-1)A^{k-2}_{n,s}$$

$$\cdots = \cdots$$

$$A^k_{n,s}=\frac{(n+k)!}{(n-k)!}A^0_{n,s} = \frac{(n+k)!}{(n-k)!}\int^1_{-1}P_nP_sdx=\delta_{ns}\vert\vert P_n\vert\vert^2$$

We can find $\vert\vert P_n\vert\vert^2$ as another separation of variables chain:

$$\int^1_{-1}P_n(x)dx = \left(\frac{1}{2^nn!}\right)\int^1_{-1}\left(\frac{d}{dx}\right)^n(1-x^2)^n\left(\frac{d}{dx}\right)^m(1-x^2)^mdx$$

$$= \left(\frac{1}{2^nn!}\right)\left(\frac{d}{dx}\right)(1-x^2)^n\left(\frac{d}{dx}^{m+1}(1-x^2)^m \vert^1_{-1}\right)$$

$$-\left(\frac{1}{2^nn!}\right)\int^1_{-1}\left(\frac{d}{dx}\right)^{m-1}(1-x^2)^m\left(\frac{d}{dx}\right)^{n+1}(1-x^2)dx$$ (3.17)

The non-integral part equals zero, the integral component cycles up on the n derivative and down on the m derivative. Of course, $m=n$ , but we maintain the separation for the sake of derivation. The final result of the cycle gives:

$$\int^1_{-1}\vert\vert P_n(x)\vert\vert^2 dx=\frac{(2n)!}{(2^nn!)^2}\int^1_{-1}\left(\frac{d}{dx}^0(x^2-1)^m\right)dx$$ (3.18)

To solve the latter integral, we introduce two identities:

$$(1-x^2)^m = \left(\int^x_0\right)^m P_m(x')\left(dx'\right)^m$$

$$\frac{1}{2n+1}\left(P'_{n+1}(x)-P'_{n-1}\right)$$

$$= \frac{1}{2n+1}\frac{1}{2^nn!}\frac{d^{n}}{dx^{n}}\left(\frac{1}{2(n+1)}\frac{d^2}{dx^2}(x^2-1)^{n+1} -2n(x^2-1)^{n-1}\right)$$

$$\blacktriangledown \frac{1}{2(n+1)}\frac{d^2}{dx^2}(x^2-1)^{n+1}=(x^2-1)^n + 2x^2n(x^2-1)^{n-1}$$

$$\blacktriangleright \frac{1}{2n+1}\left(P'_{n+1}(x)-P'_{n-1}(x)\right)$$

$$= \frac{1}{2n+1}\frac{1}{2^nn!}\frac{d^{n}}{dx^{n}}\left((x^2-1)^n+2n(x^2-1)(x^2-1)^{n-1}\right)$$

$$= \frac{1}{2^nn!}\frac{d^{n}}{dx^{n}}(x^2-1)=P_n(x)$$ (3.19)

So we follow through,

$$P_n(x) = \frac{1}{2n+1}\left(P'_{n+1}(x)-P'_{n-1}(x)\right)$$

$$\int^x_0P_n(x)dx = \frac{1}{2n+1}\left(P_{n+1}(x)-P_{n-1}(x)\right)$$

$$\int^x_0 \int^x_0P_n(x)dx dx = \frac{1}{2n+1}\left(\frac{1}{2n+3}\left(P_{n+2}(x)-P_{n}\right)-\frac{1}{2n-1}\left(P_{n}(x)-P_{n-2}(x)\right)\right)$$

We follow through this derivation cycle n times, splitting each P value as per the above identity. The $P_n(x)$ integral becomes the $(1-x^2)^n$ as identified above and he array $P_n$ will be spread from values of $P_{2n} \rightarrow P_{0}$ . If we integrate both sides of the equation over the Legendre function range, [-1,1], all said values of $P_n$ will vanish except $P_0$ . The prefactors from the expansion give:

$$2^nn!\int^1_{-1}(1-x^2)^n dx$$

$$=\int^1_{-1}2^nn!\left(\left(\int^x_0\right)^nP_n(x)\left(dx\right)^n\right)dx=\frac{(2n)!}{(2n+1)!}\int^1_{-1}P_0(x)dx$$ (3.20)

But properly,

$$\int^1_{-1}\vert P_n\vert^2dx = \frac{(2n)!}{(2^nn!)^2}\int^1_{-1}(1-x^2)^ndx$$

$$= \frac{(2n)!}{(2^nn!)^2}\int^1_{-1}2^nn!\left(\int^x_0P_n(x)\left( dx \right)^n\right) dx$$

$$= \frac{(2n)!}{(2^nn!)^2}\int^1_{-1}2^nn!\frac{2^nn!}{(2n+1)!} dx$$

$$= \frac{2}{2n+1}$$ (3.21)

From this we conclude,

$$A^k_{n,s}=\left(\frac{(n+k)!}{(n-k)!}\frac{2}{2n+1}\right)$$ (3.22)

Of course, we cannot neglect the symmetric components of theta, which give us an additional normalization of $2\pi$ , thus the final normalization condition,

$$\int^{2\pi}_{0}\int^{\pi}_{0}\vert\vert Y^k_n\vert\sin(\theta)d\theta d\phi$$ (3.23)

Re-tagging our variables appropriately, the normalized Associated Legendre Polynomial is,

$$Y^k_n(\theta,\phi)=\sqrt{\frac{2n+1}{4\pi}\frac{(n-k)!}{(n+k)!}}e... ...^{\vert m\vert}\left(\frac{1}{2^ll!}\right)\left(\frac{d}{dx}\right)^l(x^2-1)^l$$

Or in simplified notation,

$$Y^m_n(\theta,\phi)=\sqrt{\frac{2n+1}{4\pi}\frac{(n-m)!}{(n+m)!}}e^{im\phi}P^m_l(cos(\theta))$$ (3.24)

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