Special Values

From the generating function
$$g(x,t)=\frac{1}{(1-2xt+t^2)^{1/2}},$$
when $x=1$ we obtain
\begin{align*}
g(1,t)&=\frac{1}{(1-2t+t^2)^{1/2}}\\
&=\frac{1}{1-t}\\
&=\sum_{n=0}^\infty t^n,
\end{align*}
since $|t|<1$. On the other hand,
$$g(1,t)=\sum_{n=0}^\infty P_n(1)t^n.$$
So by comparison we get
$$P_n(1)=1.$$ Similarly, if we let $x=-1$,
$$P_n(-1)=(-1)^n.$$
For $x=0$, the generating function results
$$(1+t^2)^{-1/2}=1-\frac{1}{2}t^2+\frac{3}{8}t^4+\cdots+(-1)^n\frac{1\cdot 3\cdots (2n-1)}{2^nn!}t^{2n}+\cdots.$$
Thus we obtain
\begin{align*}
P_{2n}(0)&=(-1)^n\frac{1\cdot 3\cdots (2n-1)}{2^nn!}=(-1)^n\frac{(2n-1)!!}{(2n)!!},\\
P_{2n+1}(0)&=0,\ n=0,1,2,\cdots.
\end{align*}
Recall that the double factorial !! is defined by
\begin{align*}
(2n)!!&=2\cdot 4\cdot 6\cdots (2n),\\
(2n-1)!!&=1\cdot 3\cdot 5\cdots (2n-1).
\end{align*}

Parity

$g(t,x)=g(-t,-x)$, that is
$$\sum_{n=0}^\infty P_n(x)t^n=\sum_{n=0}^\infty P_n(-x)(-t)^n$$
which results the parity
$$P_n(-x)=(-1)^nP_n(x).\ \ \ \ \ (1)$$
(1) tells that if $n$ is even, $P_n(x)$ is an even function and if $n$ is odd, $P_n(x)$ is an odd function.

Orthogonality

Multiply the Legendre’s diferential equation
$$\frac{d}{dx}[(1-x^2)P_n'(x)]+n(n+1)P_n(x)=0\ \ \ \ \ (2)$$ by $P_m(x)$.
$$P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)]+n(n+1)P_m(x)P_n(x)=0.\ \ \ \ \ (3)$$
Replace $n$ by $m$ in (2) and then multiply the resulting equation by $P_n(x)$.
$$P_n(x)\frac{d}{dx}[(1-x^2)P_m'(x)]+m(m+1)P_m(x)P_n(x)=0.\ \ \ \ \ (4)$$
Subtract (4) from (3) and integrate the resulting equation with respect to $x$ from $-1$ to 1.
\begin{align*}
\int_{-1}^1&\left\{P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)]-P_n(x)\frac{d}{dx}[(1-x^2)P_m'(x)]\right\}dx\\
&=[m(m+1)P_m(x)P_n(x)-n(n+1)P_m(x)P_n(x)].\end{align*}
Using integration by parts,
\begin{align*}
\int_{-1}^1P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)]dx&=\\&(1-x^2)P_m(x)P_n'(x)|_{-1}^1-\int_{-1}^1P_m(x)P_n(x)dx\\
&=-\int_{-1}^1P_m(x)P_n(x)dx.
\end{align*}
Since the integration of the second term inside $\{\ \ \}$ would have the same value, the LHS vanishes.
Hence for $m\ne n$,
$$\int_{-1}^1P_m(x)P_n(x)dx=0.\ \ \ \ \ (5)$$
That is, $P_m(x)$ and $P_n(x)$ are orthogonal for the interval $[-1,1]$.
For $x=\cos\theta$, the orthogonality (5) is given by
$$\int_0^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta d\theta=0.$$

Integrate
$$(1-2xt+t^2)^{-1}=\left[\sum_{n=0}^\infty P_n(x)t^n\right]^2$$
with respect to $x$ from $-1$ to $1$. Due to the orthogonality (5), the integration of all the crossing terms in the RHS will vanish, and so we obtain
$$\int_{-1}^1\frac{dx}{1-2xt+t^2}=\sum_{n=0}^\infty \left\{\int_{-1}^1[P_n(x)]^2dx\right\}t^{2n}.$$
\begin{align*}
\int_{-1}^1\frac{dx}{1-2xt+t^2}&=\frac{1}{2t}\int_{(1-t)^2}^{(1+t)^2}\frac{dy}{y}\\
&=\frac{1}{t}\ln\left(\frac{1+t}{1-t}\right)\\
&=\sum_{n=0}^\infty\frac{2}{2n+1}t^{2n}\ (\mbox{since $|t|<1$}).
\end{align*}
Therefore we have the normalizer of Legendre polynomial $P_n(x)$
$$\int_{-1}^1[P_n(x)]^2dx=\frac{2}{2n+1}.$$

Expansion of Functions

Suppose that
$$\sum_{n=0}^\infty a_nP_n(x)=f(x).\ \ \ \ \ (6)$$
Multiply (6) by $P_m(x)$ and integrate with respect to $x$ from $-1$ to 1:
$$\sum_{n=0}^\infty a_n\int_{-1}^1 P_n(x)P_m(x)dx=\int_{-1}^1f(x)P_m(x)dx.$$
By the orthogonality (5), we obtain
$$\frac{2}{2m+1}a_m=\int_{-1}^1f(x)P_m(x)dx\ \ \ \ \ (7)$$
and hence $f(x)$ can be written as
$$f(x)=\sum_{n=0}^\infty\frac{2n+1}{2}\left(\int_{-1}^1 f(t)P_m(t)dt\right)P_n(x).\ \ \ \ \ (8)$$
This expansion in a series of Legendre polynomials is called a Legendre series. Clearly if $f(x)$ is continuous (or integrable) on the interval $[-1,1]$, it can be expanded as a Legendre series.

(7) can be considered as an integral transform, a finite Legendre transform and (8) can be considered as the inverse transform.

Let us consider the integral operator
$$\mathcal{P}_m:=P_m(x)\frac{2m+1}{2}\int_{-1}^1P_m(t)[\ \cdot\ ]dt.\ \ \ \ \ (9)$$
Then
$$\mathcal{P}_mf(t)=a_mP_m(x).$$
The operator (9) projects out the $m$th component of the function $f(x)$.