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