Notes on Whittaker & Watson, Chapter XV, part 1

Whittaker & Watson, A Course of Modern Analysis. Chapter XV, Legendre Functions.

15.1 Definition of Legendre polynomials.

Expanding the generating function $(1-2zh+h^2)^{-\frac12} = P_0(z)+hP_1(z)+h^2P_2(z)+\cdots$, we have the Legendre polynomials $$ P_n(z)=\sum_{r=0}^{\lfloor\frac n2\rfloor}\frac{(-1)^r(2n-2r)!}{2^nr!(n-r)!(n-2r)!}z^{n-2r}. $$ 15.11 Rodrigues' formula for the Legendre polynomials.

By differentiating $(z^2-1)^n$, we have $$ P_n(z)=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n. $$ 15.12 Schläfli’s integral for $P_n(z)$.

Applying Cauchy’s integral formula to the result above, we have $$ P_n(z)=\frac{1}{2\pi i}\int_C\frac{(t^2-1)^n}{2^n(t-z)^{n+1}}dt $$ on a counterclockwise contour $C$ around $z$ in the $t$-plane.

15.13 Legendre’s differentiation formula.

Using Schläfli’s integral formula, we have $$ (1-z^2)\frac{d^2}{dz^2}P_n(z)-2z\frac{d}{dz}P_n(z)+n(n+1)P_n(z) =\frac{(n+1)}{2\pi i}\int_C\frac{d}{dt}\frac{(t^2-1)^{n+1}}{(t-z)^{n+2}}dt = 0. $$ Thus $$ \frac{d}{dz}\left[(1-z^2)\frac{d}{dz}P_n(z)\right] + n(n+1)P_n(z)=0. $$ 15.14 The integral properties (orthogonality) of the Legendre polynomials.

By repeated integration by parts, we have $$ \int_{-1}^1\frac{d^m}{dz^m}(z^2-1)^m\frac{d^n}{dz^n}(z^2-1)^ndz = (-1)^m\int_{-1}^1(z^2-1)^m\frac{d^{m+n}}{dz^{m+n}}(z^2-1)^ndz. $$ Thus, for $m>n$, $\int_{-1}^1P_m(z)P_n(z)dz=0$.

For $m=n$, $$ \begin{align*} \int_{-1}^1\frac{d^n}{dz^n}(z^2-1)^n\frac{d^n}{dz^n}(z^2-1)^ndz &= (-1)^n\int_{-1}^1(z^2-1)^n\frac{d^{2n}}{dz^{2n}}(z^2-1)^ndz\\ & = (2n)!\int_{-1}^1(1-z^2)^ndz\\ & = 2(2n)!\int_0^{\frac12\pi}\sin^{2n+1}\theta d\theta\\ & = 2(2n)!\frac{(2n)!!}{(2n+1)!!}. \end{align*} $$ Thus $$ \int_{-1}^1P_m(z)P_n(z)dz=\frac{2\delta_{mn}}{2n+1}. $$ 15.2 Legendre functions.

Legendre’s differential equation of degree $n$ with $n$ not limited to $\mathbb{N}$,
$$ \frac{d}{dz}\left[(1-z^2)\frac{d}{dz}u(z)\right] + n(n+1)u(z)= 0, $$ is satisfied by $$ u=\frac1{2\pi i}\int_C\frac{(t^2-1)^{n}}{2^n(t-z)^{n+1}}dt $$ when the integral $\int_C\frac{d}{dt}\frac{(t^2-1)^{n+1}}{(t-z)^{n+2}}dt$ on contour $C$ vanishes. Consider the branch cut connecting the singularities $t=1,z$. The integral defined on the contour from $A>1$ enclosing $t=1$ and $z$, satisfiying Legendre’s differential equation defines the Legendre function of degree $n$ of the first kind.

15.21 The Recurrance Formulae.

1º From $\int_C\frac{d}{dt}\frac{(t^2-1)^{n+1}}{(t-z)^{n+1}}dt=0$ on the contour $C$ described above, we have $$ 2\int_C\frac{t(t^2-1)^n}{(t-z)^{n+1}}dt-\int_C\frac{(t^2-1)^{n+1}}{(t-z)^{n+2}}dt=0. $$ Thus $$ \begin{align*} \frac{1}{2^{n+1}\pi i}\int_C\frac{(t^2-1)^n}{(t-z)^n}dt&=\frac{1}{2^{n+2}\pi i}\int_C\frac{(t^2-1)^{n+1}}{(t-z)^{n+2}}dt-\frac{1}{2^{n+1}\pi i}\int_C\frac{(t^2-1)^n}{(t-z)^{n+1}}dt\\ & = P_{n+1}(z)-zP_n(z). \end{align*} $$ By differentiating on both sides, we have $$ P_{n+1}'(z)-zP_n'(z)=(n+1)P_n(z). $$ 2º From $\int_C\frac{d}{dt}\frac{t(t^2-1)^{n}}{(t-z)^{n}}dt=0$, we have $$ (n+1)\int_C\frac{(t^2-1)^n}{(t-z)^n}dt+2n\int_C\frac{(t^2-1)^{n-1}}{(t-z)^n}dt-nz\int_C\frac{(t^2-1)^n}{(t-z)^{n+1}}dt=0 $$ and consequently $$ (n+1)P_{n+1}(z)-(2n+1)P_n(z)+nP_{n-1}(z)=0. $$ 3º Differentiating the result from 2º and eliminating $P_{n+1}'$ by the result from 1º, we have $$ zP_n'(z)-P_{n-1}'(z)=nP_n(z). $$ 4º Adding the results from 1º and 3º, we have $$ P_{n+1}'(z)-P_{n-1}'(z)=(2n+1)P_n(z). $$ 5º Eliminating $P_{n-1}'(z)$ in the result from 3º using 1º, we have $$ (z^2-1)P_n'(z)=nzP_n(z)-nP_{n-1}(z). $$ 15.211 The expression of any polynomial as a series of Legendre polynomials.

Let $f_n(z)$ be a polynomial of order $n$, then $f_n(z)$ can be expanded in Legendre polynomials $$ f_n(z)=a_0P_0(z)+a_1P_1(z)+\cdots+a_nP_n(z), $$ with the coefficients determined by the orthogonality of Legendre polynomials.

15.22 Murphy’s expression of $P_n(z)$​ as a hypergeometric function.

Substitute the power series expansion in $\frac{z-1}{t-1}$ for $(t-z)^{-n-1}$ in Schläfli’s integral (§15.12), we have $$ \begin{align*} P_n(z) &= \sum_{r=0}^{\infty} \frac{(z-1)^r}{2^{n+1}\pi i}\frac{(n+1)(n+2)\cdots(n+r)}{r!}\int_C\frac{(t^2-1)^n}{(t-1)^{n+1+r}}dt\\ &=\sum_{r=0}^{\infty} \frac{(z-1)^r(n+1)(n+2)\cdots(n+r)}{2^n(r!)^2}\left.\frac{d^r(t+1)^n}{dt^r}\right|_{t=1} \\ &=\sum_{r=0}^{\infty} \frac{(z-1)^r(n+1)(n+2)\cdots(n+r)(-n)(1-n)\cdots(r-1-n)}{(r!)^2}\left(\frac12-\frac12z\right)^r \\ & = F\left(n+1,-n;1;\frac12-\frac12z\right). \end{align*} $$ 15.23 Laplace’s integrals for $P_n(z)$​​​.

In Schläfli’s integral with the contour chosen as the circle of radius $|z^2-1|^\frac12$ centered at $t=z$, we have $$ P_n(z)=\frac1{2\pi}\int_{-\pi}^\pi\left(z+(z^2-1)^{\frac12}\cos\phi\right)^nd\phi. $$ When $n$​​​​ is not limited to integers, it is necessary to ensure that the branch cut connecting $t=1,z$​​​ lies inside the contour.

15.231 The Mehler-Dirichlet integral for $P_n(z)$​.

Make the change of variables $h=z+(z^2-1)^\frac12\cos\phi$​​ and $z=\cos\theta$​​​ in Laplace’s result above, we have $$ P_n(\cos\theta)=\frac i\pi\int_{e^{-i\theta}}^{e^{i\theta}}h^n(1-2hz+h^2)^{-\frac12}dh, $$ with $h=e^{i\phi}$, we get $$ P_n(\cos\theta)=\frac2\pi\int_0^\theta\frac{\cos(n+\frac12)\phi}{\sqrt2(\cos\phi-\cos\theta)^\frac12}d\phi. $$