Laplace's integral for Legendre polynomials

Remarks

The choice of square root does not matter: the substitution $\varphi \mapsto \pi - \varphi$ changes the sign of $\cos \varphi$. Proof idea: multiply by $t^n$ and sum over $n$ for small $t$. The integral $\int_0^\pi \frac{d\varphi}{a - b \cos \varphi} = \frac{\pi}{\sqrt{a^2 - b^2}}$ with $a = 1 - tx$ and $b = t \sqrt{x^2 - 1}$ gives $(1 - 2xt + t^2)^{-1/2}$. This is the generating function of the Legendre polynomials, so the coefficients agree. Check for $n = 2$: the integral equals $x^2 + \frac{x^2 - 1}{2} = \frac{3x^2 - 1}{2} = P_2(x)$, since the mean of $\cos \varphi$ is $0$ and the mean of $\cos^2 \varphi$ is $\frac12$. For $-1 \le x \le 1$ the integrand is $\left( x + i \sqrt{1 - x^2} \cos \varphi \right)^n$, whose modulus is at most $\left( x^2 + (1 - x^2) \right)^{n/2} = 1$. Hence $\abs{P_n(x)} \le 1$ on $[-1, 1]$.

Going deeper

For a student

Laplace's integral is one of the standard integral representations of $P_n$, next to Rodrigues' formula and the generating function. Its main use is in estimates: it gives the bound $\abs{P_n(x)} \le 1$ on $[-1, 1]$ in one line. For $x > 1$ and the positive root, the integrand is largest at $\varphi = 0$, where it equals $\left( x + \sqrt{x^2 - 1} \right)^n$. So $P_n(x) \le \left( x + \sqrt{x^2 - 1} \right)^n$, and Laplace's method gives the precise growth as $n \to \infty$. The same idea, with Gegenbauer or Jacobi weights, gives integral representations for those families.