generating function of Legendre polynomials

For finding the generating function

$$F(t)=\sum _{n=0}^{\mathrm{\infty }}{P}_n(z)t^n$$

of the sequence of the Legendre polynomials
$P_0(z)=\mathrm{\hspace{0.33em}1}$
$P_1(z)=z$
$P_2(z)=\frac{1}{2}(3z^2-1)$
$P_3(x)=\frac{1}{2}(5z^3-3z)$
$P_4(z)=\frac{1}{8}(35z^4-30z^2+3)$
$P_5(z)=\frac{1}{8}(63z^5-70z^3+15z)$
$\mathrm{\cdots }\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}}\mathrm{\cdots }$
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 http://planetmath.org/node/11983Rodrigues formula of Legendre polynomials, we may write

$$P_n(z)=\frac{1}{2^{n}n!}\frac{d^n}{d{z}^n}{(z^2-1)}^n=\frac{1}{2^{n}n!}\frac{n!}{2i\pi }{\oint }_c\frac{{({\zeta }^2-1)}^n}{{(\zeta -z)}^{n+1}}𝑑\zeta =\frac{1}{2i\pi }{\oint }_c{\left(\frac{1}{2}\frac{{\zeta }^2-1}{\zeta -z}\right)}^n\frac{d\zeta }{\zeta -z},$$

where the contour $c$ runs anticlockwise once around the point $z$.  The change of variable

$$\frac{{\zeta }^2-1}{2(\zeta -z)}=\frac{1}{t},d\zeta =\frac{zt-1-\sqrt{1-zt+t^2}}{t^2\sqrt{1-zt+t^2}}dt$$

gives

$$P_n(z)=-\frac{1}{2i\pi }{\oint }_{c^{\prime }}\frac{dt}{t^{n}t\sqrt{1-zt+t^2}}$$

where $t$ must go round the origin clockwise, but in

$$P_n(z)=\frac{1}{n!}\cdot \frac{n!}{2i\pi }{\oint }_{c^{\prime }}\frac{dt}{\sqrt{1-zt+t^2}\cdot {(t-0)}^{n+1}}$$

anticlockwise.  This is, by Cauchy integral formula again,

$$P_n(z)=\frac{1}{n!}{\left[\frac{d^n}{d{t}^n}\frac{1}{\sqrt{1-zt+t^2}}\right]}_{t=0}.$$

This means that

$$F(t):=\frac{1}{\sqrt{1-zt+t^2}}$$

is the searched generating function of the Legendre polynomials:

$$\frac{1}{\sqrt{1-zt+t^2}}=P_0(z)+P_1(z)t+P_2(z)t^2+P_3(z)t^3+\mathrm{\dots }$$

Cf. the generating function of the Bessel functions.