Deriving the integral form of the Legendre polynomials from the derivative form

Question

I have the Legendre polynomials $P_n(x)= \frac{1}{n!2^n}\frac{d^n}{dx^n} (x^2−1)^n$ and have to show/derive the integral form $P_n(x)=\frac{1}{\pi} \int_0^{\pi}(x+\sqrt{1−x^2}\cos t)^n dt$ for $|x|<1$ by expressing the derivative form as a complex integral

So as I understand it, I get $P_n (x) = \frac{1}{n!2^n} \int \frac{d^n}{dz^n} (z^2 - 1)^n dz$. I'm not sure if this is correct and how to proceed further? Any tip would be much appreciated

Answer 1

We start with Rodrigue's Formula \begin{eqnarray*} P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n. \end{eqnarray*} Using Cauchy's $n^{th}$ derivative formula we have Schlafli's formula \begin{eqnarray*} P_n(x) = \frac{1}{2 \pi i} \int_C \frac{(t^2-1)^n}{2^n(t-x)^{n+1}} dt \end{eqnarray*} where $C$ is any contour going around $t=x$.

Now choose this contour to be \begin{eqnarray*} t=x + (x^2-1)^{1/2} e^{i \phi} \end{eqnarray*} where $\phi$ varies from $- \pi$ to $ \pi $ ... giving \begin{eqnarray*} P_n(x) = \frac{ 1}{ 2 \pi} \int_{- \pi }^{\pi} ( x+ (x^2-1)^{1/2} \cos( \phi ) )^{n} d \phi. \end{eqnarray*} Finally note that the integrand is an even function of $\phi$, so the interval can be split in half and you have the desired formula.

Edit: Note that \begin{eqnarray*} t^2-1 &=& x^2-1 + 2x(x^2-1)^{1/2}e^{i \phi} + (x^2-1)e^{2 i \phi} \\ &=& (x^2-1)^{1/2}e^{i \phi}(2x + (x^2-1)^{1/2}(e^{ i \phi}+ e^{ -i \phi}))\\ &=& 2 (x^2-1)^{1/2}e^{i \phi}(x + (x^2-1)^{1/2}\cos \phi ).\\ \end{eqnarray*}