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

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


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

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

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


1 Answer 1


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.

  • $\begingroup$ Perfect, thank you so much. I was subbing $\zeta$ with $\sqrt{|z^2 - 1|}e^{i\theta}$ instead. $\endgroup$
    – E Werner
    Mar 19, 2020 at 23:50

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