Starting from the Rodrigues formula, derive the orthonormality condition for Legendre polynomials:

$\underset{\mathbf{-}\mathbf{1}}{\overset{\mathbf{1}}{\mathbf{\int }}}{\mathbf{P}}_{\mathbf{l}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}\mathbf{=}\left(\frac{2}{2l+1}\right){\mathbf{\delta }}_{\mathbf{II}}\mathbf{.}$

Hint: Use integration by parts.

The Legendre polynomials:

${\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\left[\mathrm{n}/2\right]}\frac{(2\mathrm{n}-2\mathrm{k})!}{2^{\mathrm{n}}(\mathrm{n}-\mathrm{k})!\mathrm{k}!(\mathrm{n}-2\mathrm{k})!}(-1{)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{n}-2\mathrm{k}}$

As an explicit sum, the Legendre polynomials could be stated as follows:

${\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\left[\mathrm{n}/2\right]}\frac{(2\mathrm{n}-2\mathrm{k})!}{2^{\mathrm{n}}(\mathrm{n}-\mathrm{k})!\mathrm{k}!(\mathrm{n}-2\mathrm{k})!}(-1{)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{n}-2\mathrm{k}}...........(1)$

If , $\mathrm{k}\le \left[\mathrm{n}/2\right]$ we can write:

$\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}{\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}}=\frac{(2\mathrm{n}-2\mathrm{k})!}{(\mathrm{n}-2\mathrm{k})!}{\mathrm{x}}^{\mathrm{n}-2\mathrm{k}}$

Note that if $\mathrm{k}>\left[\mathrm{n}/2\right]$ , the derivative of${\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}}$ is zero, since after the$\mathrm{n}/2$ derivatives, the term${\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}}$ is reduced to a constant. Therefore:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\left[\mathrm{n}/2\right]}\frac{{(-1)}^k}{2^{\mathrm{n}}(\mathrm{n}-\mathrm{k})!\mathrm{k}!}\frac{d^n}{d{x}^n}{\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}} \\ =\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2^{\mathrm{n}}(\mathrm{n}-\mathrm{k})!\mathrm{k}!}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}{\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}} \\ =\frac{1}{2^{\mathrm{n}}\mathrm{n}!}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\left(-1\right)}^{\mathrm{k}}\mathrm{n}!}{\left(\mathrm{n}-\mathrm{k}\right)!\mathrm{k}!}{\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}} \\ =\frac{1}{2^{\mathrm{n}}\mathrm{n}!}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\left(-1\right)}^{\mathrm{k}}\mathrm{n}!}{\left(\mathrm{n}-\mathrm{k}\right)!\mathrm{k}!}{\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}} \end{gathered}$$

In the last line the sum is the binomial expansion of ${\left({\mathrm{x}}^2-1\right)}^{\mathrm{n}}$, that is:

$$\begin{gathered} ({\mathrm{x}}^2-1{)}^{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{(-1{)}^{\mathrm{k}}\mathrm{n}!}{(\mathrm{n}-\mathrm{k})!\mathrm{k}!}{\mathrm{x}}^{2\mathrm{n}-2\mathrm{k}} \\ {\mathrm{P}}_{\mathrm{n}}(\mathrm{x})=\frac{1}{2^{\mathrm{n}}\mathrm{n}!}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}({\mathrm{x}}^2-1{)}^{\mathrm{n}} \end{gathered}$$

Using this formula, we can write:

$\underset{-1}{\overset{1}{\int }}{\mathrm{P}}_{\mathrm{m}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2^{\mathrm{m}+\mathrm{n}}\mathrm{m}!\mathrm{n}!}\underset{-1}{\overset{1}{\int }}\frac{{\mathrm{d}}^{\mathrm{m}}}{{\mathrm{dx}}^{\mathrm{m}}}({\mathrm{x}}^2-1{)}^{\mathrm{m}}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx}$

Assume that $\mathrm{m}<\mathrm{n}$, then do the integral by parts, as:

$$\begin{gathered} ={\int }_{-1}^1\frac{{\mathrm{d}}^{\mathrm{m}}}{{\mathrm{dx}}^{\mathrm{m}}}({\mathrm{x}}^2-1{)}^{\mathrm{m}}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{n}}}({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx} \\ =\frac{{\mathrm{d}}^{\mathrm{n}-1}}{{\mathrm{dx}}^{\mathrm{n}-1}}{({\mathrm{x}}^2-1)}^{\mathrm{n}}\frac{\mathrm{d}\mathrm{m}}{\mathrm{dx}\mathrm{m}}({\mathrm{x}}^2-1{)}^{\mathrm{m}}\overline{)1\overline{){}_{-1}^1}}-{\int }_{-1}^1\frac{{\mathrm{d}}^{\mathrm{m}+1}}{{\mathrm{dx}}^{\mathrm{m}+1}}({\mathrm{x}}^2-1{)}^{\mathrm{m}}\frac{{\mathrm{d}}^{\mathrm{n}-1}}{{\mathrm{dx}}^{\mathrm{n}-1}}({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx} \end{gathered}$$

We see that the first term is zero at 1 and -1 , so the first term vanishes. Now we can continue by integrating the remaining integral by parts, note that the first term always vanishes, after we have done n integrations, we will get:

$\underset{-1}{\overset{1}{\int }}{\mathrm{P}}_{\mathrm{m}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2^{\mathrm{m}+\mathrm{n}}\mathrm{m}!\mathrm{n}!}{\left(-1\right)}^{\mathrm{n}}\underset{-1}{\overset{1}{\int }}({\mathrm{x}}^2-1{)}^{\mathrm{n}}\frac{{\mathrm{d}}^{\mathrm{m}+\mathrm{n}}}{{\mathrm{dx}}^{\mathrm{m}+\mathrm{n}}}({\mathrm{x}}^2-1{)}^{\mathrm{m}}\mathrm{dx}$ ……. (2)

Since $\mathrm{m}<\mathrm{n}$, then the derivative inside the integral is zero, because the largest power of x in${\left({\mathrm{x}}^2-1\right)}^{\mathrm{m}}$ is${\mathrm{x}}^{2\mathrm{m}}$ and $2\mathrm{m}<\mathrm{m}+\mathrm{n}$, thus:

${\int }_{-1}^1{\mathrm{P}}_{\mathrm{m}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=0$

This means for$\mathrm{m}\ne \mathrm{n}$ (it doesn't matter if $\mathrm{n}>\mathrm{m}\mathrm{or}\mathrm{m}>\mathrm{n}$, it is the same case since we can swap the indexes) the polynomials are orthogonal.

Now, set $\mathrm{n}=\mathrm{m}$in equation (2), so we get:

${\int }_{-1}^1{\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=\frac{(-1{)}^{\mathrm{n}}}{(\mathrm{n}!{)}^2{2}^2\mathrm{n}}{\int }_{-1}^1({\mathrm{x}}^2-1{)}^{\mathrm{n}}\frac{{\mathrm{d}}^{{}^{2\mathrm{n}}}}{{\mathrm{dx}}^{2\mathrm{n}}}({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx}$

The 2n^th derivative inside the integral will cause all the terms in${\left({\mathrm{x}}^2-1\right)}^{\mathrm{n}}$ vanishes except ${\mathrm{x}}^{2\mathrm{n}}$since 2n^th derivative cancels all the terms with a power less than the rank of the derivative. That is:

$$\begin{gathered} \frac{{\mathrm{d}}^{2\mathrm{n}}}{{\mathrm{dx}}^{2\mathrm{n}}}{\left({\mathrm{x}}^2-1\right)}^{\mathrm{n}}=\frac{{\mathrm{d}}^{2\mathrm{n}}}{{\mathrm{dx}}^{2\mathrm{n}}}{\mathrm{x}}^{2\mathrm{n}} \\ =\left(2\mathrm{n}\right)! \end{gathered}$$

Thus, we can write:

${\int }_{-1}^1{\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=\frac{(-1{)}^{\mathrm{n}}(2\mathrm{n})!}{(\mathrm{n}!{)}^2{2}^2\mathrm{n}}{\int }_{-1}^1({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx}\dots \dots .(3)$

let ,role="math" localid="1656393894301" $\mathrm{x}=\sin (\mathrm{\theta }),\mathrm{so}\mathrm{dx}=\cos (\mathrm{\theta })\mathrm{d\theta }$ so and the limits are:

$-1\to -\frac{\mathrm{\pi }}{2}\hspace{1em}1\to \frac{\mathrm{\pi }}{2}$

Substitute the limits:

role="math" localid="1656394169842" ${\int }_{-1}^1{\left({\mathrm{x}}^2-1\right)}^{\mathrm{n}}\mathrm{dx}={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}{\left({\sin }^2(\mathrm{\theta })-1\right)}^{\mathrm{n}}\cos (\mathrm{\theta })\mathrm{d\theta }$

But, ${\sin }^2(\mathrm{\theta })-1={\cos }^2(\mathrm{\theta })$, thus:

${\int }_{-1}^1{\left({\mathrm{x}}^2-1\right)}^{\mathrm{n}}\mathrm{dx}={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}{\cos }^{2\mathrm{n}+1}(\mathrm{\theta })\mathrm{d\theta }$

The value of the integral is the beta function $\mathrm{B}\left(\frac{1}{2},\mathrm{n}\right)$, so:

${\int }_{-1}^1({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx}=(-1{)}^{\mathrm{n}}\mathrm{B}(\frac{1}{2},\mathrm{n}+1)$

The beta function is given by:

$\mathrm{B}(\mathrm{a},\mathrm{b})=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }\left(\mathrm{b}\right)}{\left(\mathrm{\Gamma }\right(\mathrm{a}+\mathrm{b})}$

It can also be written as:

${\int }_{-1}^1({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx}=(-1{)}^{\mathrm{n}}\frac{\mathrm{\Gamma }\left({\displaystyle \frac{1}{2}}\right)\mathrm{\Gamma }(\mathrm{n}+1)}{\mathrm{\Gamma }\left(\mathrm{n}+\frac{3}{2}\right)}$

The gamma function is given by$\mathrm{\Gamma }\left(\mathrm{n}\right)=(\mathrm{n}-1)$ ! where $\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\mathrm{\pi }}$, so:

${\int }_{-1}^1({\mathrm{x}}^2-1{)}^{\mathrm{n}}\mathrm{dx}=(-1{)}^{\mathrm{n}}\frac{\left(\mathrm{n}\right)!\sqrt{\mathrm{\pi }}}{\mathrm{\Gamma }\left(\mathrm{n}+\frac{1}{2}\right)}$

It can also be written as:

$\mathrm{\Gamma }\left(\mathrm{n}+\frac{3}{2}\right)=\mathrm{\Gamma }\left((\mathrm{n}+1)+\frac{1}{2}\right)=\frac{\sqrt{\mathrm{\pi }}(2\mathrm{n}+1)!}{2^{1-2\mathrm{n}}\mathrm{n}!}$

As a result, the integral's value will be:

${\int }_{-1}^1{\left({\mathrm{x}}^2-1\right)}^{\mathrm{n}}\mathrm{dx}={\left(-1\right)}^{\mathrm{n}}\frac{{\left(\mathrm{n}!\right)}^2{2}^{1+2\mathrm{n}}}{\left(2\mathrm{n}+1\right)!}$

Substitute value in equation (3),

role="math" localid="1656394760208" $$\begin{gathered} {\int }_{-1}^1{\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{{\left(\mathrm{n}!\right)}^2{2}^{2\mathrm{n}}}\left(2\mathrm{n}\right)!{\left(-1\right)}^{2\mathrm{n}}\frac{{\left(\mathrm{n}!\right)}^2{2}^{1+2\mathrm{n}}}{\left(2\mathrm{n}+1\right)!} \\ =\frac{2}{2\mathrm{n}+1} \end{gathered}$$

Combine this result with the case when $\mathrm{m}\ne \mathrm{n}$, so we can write:

role="math" localid="1656394529070" ${\int }_{-1}^1{\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{m}}\left(\mathrm{x}\right)\mathrm{dx}=\frac{2}{2\mathrm{n}+1}{\mathrm{\delta }}_{\mathrm{nm}}$