The General Legendre differential equation as we know, $$ (1-x^2)y''-2xy'+n(n+1)y=0$$ is solved using the frobenius power series method and the solutions are obtained to be the general Legendre polynomials, where suitable normalization constants and boundary conditions are used.
With this result, we proceed further to completely exploit all the possibile solutions can be obtained from this equation.
Before that, we will have to use the orthonormalizability property of our solution which is proved as,
From the generating function, $$ \frac{1}{\sqrt{1-2xz+z^2}}=\sum_n P_n(x) z^n$$ or $$ \frac{1}{1-2xz+z^2} = \sum_n \sum_m z^{n+m} P_n(x) P_m(x) \\ \int\frac{1}{1-2xz+z^2} \,dx = \sum_{m,n} z^{n+m} \int{P_n(x)}P_m(x) \,dx$$ The limits can change maximum from -1 to +1 as in Legendre polynomials and summation is usually implied from 0 to infinity.
Left hand side term is evaluated to give,$$ \frac{-1}{2z}\int_{(1+z)^2}^{(1-z)^2}\frac{1}{u} du = \frac{-1}{2z} \ln{\frac{(1-z)^2}{(1+z)^2}}\\where\,\, u=1-2xz+z^2 and \,\,du = -2z\,dx$$
and $$ \frac{-1}{2z}\ln{\left(\frac{1-z}{1+z}\right)^2} = \frac{-2}{2z} \ln{\frac{1-z}{1+z}} = \frac{1}{z}\ln{\frac{(1+z)}{(1-z)}}$$
$$\ln{\frac{(1+z)}{(1-z)}} = \ln(1+z) - \ln(1-z)$$
using taylor expansion,
$$ f(x) = f(a) + f'(a)\, (x-a) + \frac{f''(a)}{2!}\,(x-a)^2+...$$
where the function should be infinitely differentiable at x=a.
So, $$ \ln(1+z) = \ln(1+a) + \frac{1}{1+a}(z-a) - \frac{1}{2!(1+a)^2} (z-a)^2 +...$$
where we can take a=0, $$ \ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} +... $$ Similarly, $$ \ln(1-z) = -z - \frac{z^2}{2} - \frac{z^3}{3} -..$$ This gives, $$ \ln(1+z) - \ln(1-z) = 2 \left[z+\frac{z^3}{3} + \frac{z^5}{5}+...\right] = 2\sum_{n} \frac{z^{2n+1}}{2n+1}$$
Thus we have our integral equal to, $$ \sum_{m,n} z^{n+m} \int_{-1}^{1} P_n(x)\,P_m(x) \,dx = \frac{1}{z} \sum_{n} \frac{z^{2n+1}}{2n+1} = 2 \sum_n \frac{z^{2n}}{2n+1}$$
From this, the left side terms will equal the right side terms only when n=m, so we get the result,
$$ when\,\, n=m\rightarrow\,\, \sum_n z^{2n} \int_{-1}^{+1}P_n(x)\,P_m(x) \,dx = 2\sum_n \frac{z^{2n}}{2n+1}\\ \rightarrow\,\,\,\, \int_{-1}^{1} P_n(x)\,P_m(x)\, dx = \frac{2}{2n+1}$$ $$ when \,\, n\neq{m}\rightarrow\,\, \int_{-1}^{1}P_n(x)\,P_m(x) \,dx = 0 \\ or \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \delta_{nm} \frac{2}{2n+1}$$ which is our orthogonality relation.
We can also prove the completeness relation - that any arbitrary function can be expanded in terms of the linear combination of these Legendre polynomials.
$$ f(x) = \sum_n a_n P_n(x)\\ \int_{-1}^{1} f(x)\,P_m(x)\,dx = \sum_n a_n \frac{2}{2n+1}\delta{nm} = \sum_n a_n \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \frac{2}{2m+1}a_m $$ amd $$ a_m = \frac{2m+1}{2} \int_{-1}^{1} f(x)\,P_m(x)\,dx$$
Now, we will get back to our differential equation where we substitute our solution and differentiate it "m" times w.r.t. x
(m < n), $$ (1-x^2)P_n''-2xP_n'+ n(n+1)P_n=0$$
$$ \frac{d^m}{dx^m}\left((1-x^2)\frac{d^2P_n}{dx^2}\right) = (1-x^2) \frac{d^{m+2}P_n}{dx^{m+2}} - m.2x.\frac{d^{m+1}P_n}{dx^{m+1}} - \frac{m(m-1)}{2!}2 \frac{d^mP_n}{dx^m}$$
where we made use of the Leibniz formula, $$ \frac{d^m}{dx^m} A(x) \,B(x) = \sum_{k=0}^m \frac{m!}{k!(m-k)!} \frac{d^kA}{dx^k} \frac{d^{n-k}B}{dx^{n-k}}$$
Similarly, $$ -2\frac{d^m}{dx^m}\left[x\frac{dP_n}{dx}\right] = -2x \frac{d^{m+1}P_n}{dx^{m+1}} - 2m\frac{d^mP_n}{dx^m}$$
Which finally gives, $$ \frac{d^m}{dx^m}\left[Legendre\, eqn.\right] \\= (1-x^2)\frac{d^2}{dx^2}\left(\frac{d^mP_n}{dx^m}\right) - 2(m+1)x \frac{d}{dx}\left(\frac{d^mP_n}{dx^m}\right) + \frac{d^mP_n}{dx^m}\left[n(n+1)-m(m+1)\right] = 0 $$
If we assume, $$ \frac{d^mP_n}{dx^m} = V = \frac{W}{(1-x^2)^{\frac{m}{2}}}$$ So, $$ \frac{dV}{dx} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{dW}{dx} +\frac{mxW}{1-x^2}\right]$$ and $$ \frac{d^2V}{dx^2} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{d^2W}{dx^2} + \frac{2mx}{1-x^2}\frac{dW}{dx}+W\left(\frac{m}{1-x^2}+ \frac{mx^2(m+2)}{(1-x^2)^2}\right)\right]$$ Substituting and doing the necessary manipulation (multiply the whole equation by $(1-x^2)^{\frac{m}{2}}$, We get our associated Legendre equation, $$ (1-x^2)\frac{d^2W}{dx^2} - 2x\frac{dW}{dx} + \left[n(n+1) - \frac{m^2}{1-x^2}\right]W = 0 $$
With this result, we proceed further to completely exploit all the possibile solutions can be obtained from this equation.
Before that, we will have to use the orthonormalizability property of our solution which is proved as,
From the generating function, $$ \frac{1}{\sqrt{1-2xz+z^2}}=\sum_n P_n(x) z^n$$ or $$ \frac{1}{1-2xz+z^2} = \sum_n \sum_m z^{n+m} P_n(x) P_m(x) \\ \int\frac{1}{1-2xz+z^2} \,dx = \sum_{m,n} z^{n+m} \int{P_n(x)}P_m(x) \,dx$$ The limits can change maximum from -1 to +1 as in Legendre polynomials and summation is usually implied from 0 to infinity.
Left hand side term is evaluated to give,$$ \frac{-1}{2z}\int_{(1+z)^2}^{(1-z)^2}\frac{1}{u} du = \frac{-1}{2z} \ln{\frac{(1-z)^2}{(1+z)^2}}\\where\,\, u=1-2xz+z^2 and \,\,du = -2z\,dx$$
and $$ \frac{-1}{2z}\ln{\left(\frac{1-z}{1+z}\right)^2} = \frac{-2}{2z} \ln{\frac{1-z}{1+z}} = \frac{1}{z}\ln{\frac{(1+z)}{(1-z)}}$$
$$\ln{\frac{(1+z)}{(1-z)}} = \ln(1+z) - \ln(1-z)$$
using taylor expansion,
$$ f(x) = f(a) + f'(a)\, (x-a) + \frac{f''(a)}{2!}\,(x-a)^2+...$$
where the function should be infinitely differentiable at x=a.
So, $$ \ln(1+z) = \ln(1+a) + \frac{1}{1+a}(z-a) - \frac{1}{2!(1+a)^2} (z-a)^2 +...$$
where we can take a=0, $$ \ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} +... $$ Similarly, $$ \ln(1-z) = -z - \frac{z^2}{2} - \frac{z^3}{3} -..$$ This gives, $$ \ln(1+z) - \ln(1-z) = 2 \left[z+\frac{z^3}{3} + \frac{z^5}{5}+...\right] = 2\sum_{n} \frac{z^{2n+1}}{2n+1}$$
Thus we have our integral equal to, $$ \sum_{m,n} z^{n+m} \int_{-1}^{1} P_n(x)\,P_m(x) \,dx = \frac{1}{z} \sum_{n} \frac{z^{2n+1}}{2n+1} = 2 \sum_n \frac{z^{2n}}{2n+1}$$
From this, the left side terms will equal the right side terms only when n=m, so we get the result,
$$ when\,\, n=m\rightarrow\,\, \sum_n z^{2n} \int_{-1}^{+1}P_n(x)\,P_m(x) \,dx = 2\sum_n \frac{z^{2n}}{2n+1}\\ \rightarrow\,\,\,\, \int_{-1}^{1} P_n(x)\,P_m(x)\, dx = \frac{2}{2n+1}$$ $$ when \,\, n\neq{m}\rightarrow\,\, \int_{-1}^{1}P_n(x)\,P_m(x) \,dx = 0 \\ or \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \delta_{nm} \frac{2}{2n+1}$$ which is our orthogonality relation.
We can also prove the completeness relation - that any arbitrary function can be expanded in terms of the linear combination of these Legendre polynomials.
$$ f(x) = \sum_n a_n P_n(x)\\ \int_{-1}^{1} f(x)\,P_m(x)\,dx = \sum_n a_n \frac{2}{2n+1}\delta{nm} = \sum_n a_n \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \frac{2}{2m+1}a_m $$ amd $$ a_m = \frac{2m+1}{2} \int_{-1}^{1} f(x)\,P_m(x)\,dx$$
Now, we will get back to our differential equation where we substitute our solution and differentiate it "m" times w.r.t. x
(m < n), $$ (1-x^2)P_n''-2xP_n'+ n(n+1)P_n=0$$
$$ \frac{d^m}{dx^m}\left((1-x^2)\frac{d^2P_n}{dx^2}\right) = (1-x^2) \frac{d^{m+2}P_n}{dx^{m+2}} - m.2x.\frac{d^{m+1}P_n}{dx^{m+1}} - \frac{m(m-1)}{2!}2 \frac{d^mP_n}{dx^m}$$
where we made use of the Leibniz formula, $$ \frac{d^m}{dx^m} A(x) \,B(x) = \sum_{k=0}^m \frac{m!}{k!(m-k)!} \frac{d^kA}{dx^k} \frac{d^{n-k}B}{dx^{n-k}}$$
Similarly, $$ -2\frac{d^m}{dx^m}\left[x\frac{dP_n}{dx}\right] = -2x \frac{d^{m+1}P_n}{dx^{m+1}} - 2m\frac{d^mP_n}{dx^m}$$
Which finally gives, $$ \frac{d^m}{dx^m}\left[Legendre\, eqn.\right] \\= (1-x^2)\frac{d^2}{dx^2}\left(\frac{d^mP_n}{dx^m}\right) - 2(m+1)x \frac{d}{dx}\left(\frac{d^mP_n}{dx^m}\right) + \frac{d^mP_n}{dx^m}\left[n(n+1)-m(m+1)\right] = 0 $$
If we assume, $$ \frac{d^mP_n}{dx^m} = V = \frac{W}{(1-x^2)^{\frac{m}{2}}}$$ So, $$ \frac{dV}{dx} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{dW}{dx} +\frac{mxW}{1-x^2}\right]$$ and $$ \frac{d^2V}{dx^2} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{d^2W}{dx^2} + \frac{2mx}{1-x^2}\frac{dW}{dx}+W\left(\frac{m}{1-x^2}+ \frac{mx^2(m+2)}{(1-x^2)^2}\right)\right]$$ Substituting and doing the necessary manipulation (multiply the whole equation by $(1-x^2)^{\frac{m}{2}}$, We get our associated Legendre equation, $$ (1-x^2)\frac{d^2W}{dx^2} - 2x\frac{dW}{dx} + \left[n(n+1) - \frac{m^2}{1-x^2}\right]W = 0 $$
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