For finding the generating function $$F(t) \;=\; \sum_{n=0}^\infty P_n(z)t^n$$ of the sequence of the Legendre polynomials
$P_0(z) \;=\; 1$
$P_1(z) \;=\; z$
$P_2(z) \;=\; \frac{1}{2}(3z^2\!-\!1)$
$P_3(x) \;=\; \frac{1}{2}(5z^3\!-\!3z)$
$P_4(z) \;=\; \frac{1}{8}(35z^4\!-\!30z^2\!+\!3)$
$P_5(z) \;=\; \frac{1}{8}(63z^5\!-\!70z^3\!+\!15z)$
$\cdots \qquad\;\; \cdots$
we have to present $P_n(z)$ as the general coefficient of Taylor series in $t$ , i.e. as the $n$ th derivative of some $F(t)$ in the origin, divided by the factorial $n!$ . The Cauchy integral formula offers the chance to implement that.

Starting from the Rodrigues formula of Legendre polynomials, we may write $$P_n(z) \;=\; \frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2\!-\!1)^n \;=\; \frac{1}{2^nn!}\frac{n!}{2i\pi}\oint_c\frac{(\zeta^2\!-\!1)^n}{(\zeta\!-\!z)^{n+1}}d\zeta\;=\; \frac{1}{2i\pi}\oint_c\left(\frac{1}{2}\frac{\zeta^2\!-\!1}{\zeta\!-\!z}\right)^n\!\frac{d\zeta}{\zeta\!-\!z},$$ where the contour $c$ runs anticlockwise once around the point $z$ . The change of variable $$\frac{\zeta^2\!-\!1}{2(\zeta\!-\!z)} \;=\; \frac{1}{t}, \quad d\zeta \;=\; \frac{zt\!-\!1\!-\!\sqrt{1\!-\!zt\!+\!t^2}}{t^2\sqrt{1\!-\!zt\!+\!t^2}}dt$$ gives $$P_n(z) \;=\; -\frac{1}{2i\pi}\oint_{c'}\frac{dt}{t^nt\sqrt{1\!-\!zt\!+\!t^2}}$$ where $t$ must go round the origin clockwise, but in $$P_n(z) \;=\; \frac{1}{n!}\cdot\frac{n!}{2i\pi}\oint_{c'}\frac{dt}{\sqrt{1\!-\!zt\!+\!t^2}\cdot(t\!-\!0)^{n+1}}$$ anticlockwise. This is, by Cauchy integral formula again, $$P_n(z) \;=\; \frac{1}{n!}\left[\frac{d^n}{dt^n}\frac{1}{\sqrt{1\!-\!zt\!+\!t^2}}\right]_{t=0}.$$ This means that $$F(t) \;:=\; \frac{1}{\sqrt{1\!-\!zt\!+\!t^2}}$$ is the searched generating function of the Legendre polynomials: $$\frac{1}{\sqrt{1\!-\!zt\!+\!t^2}} \;=\; P_0(z)+P_1(z)t+P_2(z)t^2+P_3(z)t^3+\ldots$$

A similar method can be used also for finding e.g. the generating function of the Hermite polynomials. Cf. the generating function of the Bessel functions.