# Laplace Integral for Legendre Polynomial

Using the identity for the Legendre polynomial (where $$C$$ surrounds z):

$$P_n(z) = \frac{1}{2\pi i} \int_C \frac{(\zeta ^2 - 1)^n}{2^n (\zeta - z)^{n + 1}}~d\zeta$$

Prove

$$P_n(z) = \frac{1}{\pi} \int_0 ^\pi (z + \sqrt{z^2 - 1}\cos \theta)^n ~d\theta$$

I believe we need to take C to be a circle of radius $$\sqrt{|z^2 - 1|}$$ centered at $$z$$, but everything I get just leads to a dead end. The farthest I’ve gotten is

$$P_n (z) = \frac{1}{2^{n + 1}\pi} \int_{-\pi}^{\pi} \left( \frac{(z^2 - 1)e^{2i\theta} - 1}{\sqrt{|z^2 - 1|}e^{i\theta} - z} \right)^n \frac{\sqrt{|z^2 - 1|}e^{i\theta}~d\theta}{\sqrt{|z^2 - 1|}e^{i\theta} - z}$$

The substitution is $$\zeta=\color{blue}{z+{}}\sqrt{z^2-1}\,e^{i\theta}$$ (with any fixed value of the square root). Now $$\zeta^2\color{blue}{-1}=(z^2\color{blue}{-1})+2z\sqrt{z^2-1}\,e^{i\theta}+(z^2-1)e^{2i\theta}=2\sqrt{z^2-1}\,e^{i\theta}(z+\sqrt{z^2-1}\cos\theta),$$ hence $$(\zeta^2-1)/(\zeta-z)=2(z+\sqrt{z^2-1}\cos\theta)$$, etc.
• Perfect, thank you so much. I was subbing $\zeta$ with $\sqrt{|z^2 - 1|}e^{i\theta}$ instead. Mar 19, 2020 at 23:50