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I want to show that $$\int_{-1}^{1}L_n(x)L_m(x)dx$$ is zero for $m<n$ and $\frac{2}{2n+1}$ for $m=n$. For $m<n$, I want to apply $(x^2-1)^n=(x-1)^n(x+1)^n$ and integration by parts. For $m=n$ it should be possible to use the substitution $y=\sqrt{x}$. Some relevant formulae: $$\text{Rodrigues' formula:}\hspace{.4cm} L_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}[(x^2-1)^n]$$ $$\text{Beta Integral:}\hspace{.4cm}\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}dx=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$ $$\text{Gamma identities:}\hspace{.4cm}\Gamma(x+1)=x\Gamma(x),\hspace{.2cm}\Gamma(n+1)=n!$$ For $m=n$, $$I_n=\int_{-1}^{1}L_n^2dx=(\frac{1}{2^{n}n!})^2\int_{-1}^{1}\Big(\frac{d^n}{dx^n}\big[(x^2-1)^n\big]\Big)^2dx= \big(\frac{1}{2^{n}n!}\big)^2\Big[\frac{d^{n-1}}{dx^{n-1}}[(x-1)^n(x+1)^n]\frac{d^n}{dx^n}[(x-1)^n(x+1)^n]\Big\rvert_{-1}^{\hspace{.2cm}1}-\int_{-1}^{1}\frac{d^{n-1}}{dx^{n-1}}[(x-1)^n(x+1)^n]\frac{d^{n+1}}{dx^{n+1}}[(x-1)^n(x+1)^n]\Big]$$ I'm not sure where to where to go from here. Is there something that should cancel? Is there a way to do iterative integration by parts here? How do I get to $\frac{2}{2n+1}$?

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I finally got it. Here is the proof.

We want to prove the orthogonality relation $I$ defined by

\begin{equation} I = \int_{-1}^{1}L_n(x)L_m(x)dx \end{equation}

is zero for $m\neq n$ and $\dfrac{2}{2n+1}$ for $m=n$. In $I$, we can use Rodrigues' formula to express $L_n$ as

\begin{equation} L_n(x)= \frac{1}{2^nn!}\frac{d^n}{dx^n}\big[(x^2-1)^n\big] \end{equation} and apply integration by parts, $\int u'v=uv-\int uv'$.

\begin{equation} \begin{split} I= \frac{1}{2^nn!}\int_{-1}^{1}\Big(\frac{d^n}{dx^n}\big[(x^2-1)^n\big]L_m(x)\Big)dx \\ =\frac{1}{2^nn!} \Bigg(\frac{d^{n-1}}{dx^{n-1}}\big[(x^2-1)^n\big]L_m(x) \biggr\rvert_{-1}^{1} - \int_{-1}^{1}\frac{d^{n-1}}{dx^{n-1}}\big[(x^2-1)^n\big]L_m^{(1)}(x)dx \Bigg) \\ = -\frac{1}{2^nn!} \int_{-1}^{1}\frac{d^{n-1}}{dx^{n-1}}\big[(x^2-1)^n\big]L_m^{(1)}(x)dx \end{split} \end{equation}

Repeated integration by parts $n$ times leaves only

\begin{equation} \begin{split} I= (-1)^n\frac{1}{2^nn!}\int_{-1}^{1}(x^2-1)^nL_m^{(n)}(x)dx \end{split} \end{equation}

As $L_m$ has polynomial degree $m$, it will become a constant upon differentiation $m$ times. Since $m<n$, $L_m^{(n)}=0$, so $I=0$ for $m<n$. Since the same argument can be made for $L_n$ when $n<m$, $I$ is always zero for $m\neq n$.

For the case $m=n$,

\begin{equation} \begin{split} I= (-1)^n\frac{1}{2^nn!}\int_{-1}^{1}(x^2-1)^nL_n^{(n)}(x)dx \\ = (-1)^n\frac{1}{2^n n!}\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}\big[(x^2-1)^n\big] dx \\ =(-1)^n\frac{1}{2^{2n} (n!)^2}\int_{-1}^{1}(x^2-1)^n (2n)!dx \\ =\frac{(2n)!}{2^{2n} (n!)^2}\int_{-1}^{1}(1-x^2)^ndx \end{split} \end{equation}

Considering only the integral $J=\int_{-1}^{1}(x^2-1)^ndx$, use the substitution $x=\sqrt{y}$.

\begin{equation} \begin{split} J=2\int_{0}^{1}(1-y)^n \cdot \frac{1}{2\sqrt{y}}dx \end{split} \end{equation}

Letting $\beta=n+1, \alpha=\frac{1}{2}$ and applying the Beta integral,

\begin{equation} \begin{split} J=\frac{\Gamma(\frac{1}{2})\Gamma(n+1)}{\Gamma(n+\frac{3}{2})} \end{split} \end{equation}

Using the given Gamma identity for integers,

\begin{equation} \begin{split} \Gamma(n+\frac{3}{2})=(n+\frac{1}{2})\Gamma(n+\frac{1}{2}) \\ = (n+\frac{1}{2})(n-\frac{1}{2})(n-\frac{3}{2})\cdot\cdot\cdot\frac{1}{2}\Gamma(\frac{1}{2}) \end{split} \end{equation}

and since $\Gamma(n+1)=n!$ we get

\begin{equation} \begin{split} J=\frac{n!}{(n+\frac{1}{2})(n-\frac{1}{2})(n-\frac{3}{2})\cdot\cdot\cdot\frac{1}{2}} \end{split} \end{equation}

Then

\begin{equation} \begin{split} I=\frac{(2n)!}{2^{2n} (n!)^2}\cdot\frac{n!}{(n+\frac{1}{2})(n-\frac{1}{2})(n-\frac{3}{2})\cdot\cdot\cdot\frac{1}{2}} \\ =\frac{1}{(n+\frac{1}{2})}\frac{(2n)!}{2^{2n}n!(n-\frac{1}{2})(n-\frac{3}{2})\cdot\cdot\cdot\frac{1}{2}} = \frac{1}{\frac{2n+1}{2}}= \frac{2}{2n+1} \end{split} \end{equation}

So

\begin{equation} I = \int_{-1}^{1}L_n(x)L_n(x)dx = \frac{2}{2n+1} \end{equation}

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Here is one approach to prove the orthogonality relation for the Legendre polynomials $P_n (x)$.

We wish to prove \begin{align*} \int^1_{-1} P_m (x) P_n (x) \, dx = \begin{cases} 0, & m \neq n\\[1ex] \displaystyle{\frac{2}{2n + 1}}, & m = n \end{cases} \end{align*}

The case $m \neq n$

As $P_m (x)$ and $P_n (x)$ satisfy the Legendre differential equation, namely $$(1 - x^2) \frac{d^2 P_m}{dx^2} - 2x \frac{dP_m}{dx} + m(m + 1) P_m = 0,$$ and $$(1 - x^2) \frac{d^2 P_n}{dx^2} - 2x \frac{dP_n}{dx} + n(n + 1) P_n = 0,$$ each of these equations may be rewritten as \begin{align} \frac{d}{dx} \left [(1 - x^2) \frac{dP_m}{dx} \right ] + m(m + 1) P_m &= 0, \qquad (1) \end{align} and \begin{equation} \frac{d}{dx} \left [(1 - x^2) \frac{dP_n}{dx} \right ] + n(n + 1) P_n = 0. \qquad (2) \end{equation} Multiplying Eq. (1) by $P_n (x)$ and Eq. (2) by $P_m(x)$ and subtracting we have $$P_n \frac{d}{dx} \left [(1 - x^2) \frac{dP_m}{dx} \right ] - P_m \frac{d}{dx} \left [(1 - x^2) \frac{dP_n}{dx} \right ] + [m(m + 1) - n(n + 1)]P_m P_n = 0. \,\, (*)$$

Now recognising $$\frac{d}{dx} \left [P_n (1 - x^2) \frac{dP_m}{dx} \right ] = \frac{dP_n}{dx} (1 - x^2) \frac{dP_m}{dx} + P_n \frac{d}{dx} \left [(1 - x^2) \frac{dP_m}{dx} \right ],$$ Eq. ($*$) may be rewritten as $$\frac{d}{dx} \left [P_n(1 - x^2) \frac{dP_m}{dx} \right ] - \frac{dP_n}{dx} (1 - x^2) \frac{dP_m}{dx} - \frac{d}{dx} \left [P_m(1 - x^2) \frac{dP_n}{dx} \right ] + \frac{dP_m}{dx} (1 - x^2) \frac{dP_n}{dx} + (m^2 + m - n^2 - n) P_m P_n = 0,$$ which reduces to $$\frac{d}{dx} \left [P_n(1 - x^2) \frac{dP_m}{dx} \right ] - \frac{d}{dx} \left [P_m(1 - x^2) \frac{dP_n}{dx} \right ] + (m^2 + m - n^2 - n) P_m P_n = 0.$$

On integrating up the above equation with respect to $x$ from $-1$ to 1 we have \begin{align*} \int^1_{-1} \left \{\frac{d}{dx} \left [P_n(1 - x^2) \frac{dP_m}{dx} \right ] - \frac{d}{dx} \left [P_m(1 - x^2) \frac{dP_n}{dx} \right ] \right \} dx\\ \hspace{-8.0cm}+ (m^2 + m - n^2 - n) \int^1_{-1} P_m P_n \, dx &= 0\\ \left [P_n(1 - x^2) \frac{dP_m}{dx} - P_m(1 - x^2) \frac{dP_n}{dx} \right ]^1_{-1} + (m - n)(m + n + 1) \int^1_{-1} P_m P_n \, dx &= 0\\ (m - n)(m + n + 1) \int^1_{-1} P_m P_n \, dx &= 0. \end{align*} Note the term appearing in the square brackets vanishes at both the upper and lower limits due to the presence of the factor $(1 - x^2)$ .

So when $m \neq n$, the factor out the front of the integral cancels and we obtain $$\int^1_{-1} P_m (x) P_n (x) \, dx = 0,$$ as required to prove.

The case $m = n$

To prove the $m = n$ case let us define $$A_n = \int^1_{-1} [P_n (x)]^2 \, dx.$$ From Bonnet's recurrence relation for the Legendre polynomials, namely $$(n + 1) P_{n + 1} (x) - (2n + 1)x P_n (x) + n P_{n - 1} (x) = 0,$$ if we replace $n$ with $n - 1$, after rearranging we have $$P_n (x) = \frac{2n - 1}{n} x P_{n - 1} (x) - \frac{n - 1}{n} P_{n - 2} (x).$$

Using this result, substituting for one of the $P_n(x)$ terms in the integral for $A_n$ we have \begin{align*} A_n &= \int^1_{-1} P_n (x) P_n (x) \, dx\\ &= \int^1_{-1} P_n (x) \left [\frac{2n - 1}{n} x P_{n - 1} (x) - \frac{n - 1}{n} P_{n - 2} (x) \right ] \, dx\\ &= \frac{2n - 1}{n} \int^1_{-1} x P_n (x) P_{n - 1} (x) \, dx - \frac{n - 1}{n} \int^1_{-1} P_n (x) P_{n - 2} (x) \, dx \end{align*} But by the orthogonality property, as $$\int^1_{-1} P_n (x) P_{n - 2} (x) \, dx = 0,$$ this yields $$A_n (x) = \frac{2n - 1}{n} \int^1_{-1} x P_n (x) P_{n - 1} (x) \, dx.$$

Now from Bonnet's recurrence relation, by making the term $x P_n (x)$ the subject we have $$x P_n (x) = \frac{1}{2n + 1} \left [(n + 1) P_{n + 1} (x) + n P_{n - 1} (x) \right ],$$ which on substituting into the integral for $A_n$ one obtains \begin{align*} A_n &= \frac{2n - 1}{n} \cdot \frac{n + 1}{2n + 1} \int^1_{-1} \!\! P_{n + 1} (x) P_{n -1} (x) dx + \frac{2n - 1}{2n + 1} \int^1_{-1} [P_{n - 1} (x)]^2 dx\\ &= \frac{2n - 1}{2n + 1} A_{n - 1}, \quad n = 1,2,3,\ldots \end{align*} where we have again made use of the orthogonality property.

As $P_0 (x) = 1$, a value for $A_0$ can be found. It is $$A_0 = \int^1_{-1} [P_0 (x)]^2 \, dx = \int^1_{-1} dx = 2.$$ So we have \begin{align*} A_n &= \frac{2n - 1}{2n + 1} A_{n - 1}\\ &= \frac{2n - 1}{2n + 1} \cdot \frac{2n - 3}{2n - 1} A_{n - 2}\\ & \hspace{1.0cm} \vdots\\ &= \frac{2n - 1}{2n + 1} \cdot \frac{2n - 3}{2n - 1} \cdots \frac{3}{5} \cdot \frac{1}{3} A_0\\ &= \frac{2n - 1}{2n + 1} \cdot \frac{2n - 3}{2n - 1} \cdots \frac{3}{5} \cdot \frac{1}{3} \cdot 2\\ &= \frac{2}{2n + 1} \end{align*} Hence $$\int^1_{-1} [P_n (x)]^2 \, dx = \frac{2}{2n + 1},$$ and completes the proof.

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