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Using Rodrigues' formula and integrating by parts $n$ times, prove that $$\int_{-1}^1P_n^2(x)dx=\frac{2}{2n+1}$$ where $P_n(x)$ is a Legendre polynomial.

I tried this way

Let $$f(x)=(x^2-1)^s$$ $$\int_{-1}^1f^{(s)}(x)f^{(s)}(x)dx=-\int_{-1}^1f^{(s+1)}(x)f^{(s-1)}(x)dx$$ (the identity that I have not been able to demonstrate). Then apply this formula repeatedly and obtain $$2(2s)!\int_{0}^1(1-x^2)^sdx$$ I know what to use a replacement $x=\cos\theta$ and transform this integral en $\int_{0}^{\pi/2} \sin^{2s+1} t \;dt$ I do not know how to continue.

Cristian Rodríguez
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2 Answers2

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Recall the Rodrigues' formula $$ P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left(x^2-1\right)^n, $$ then $$ \mathcal{I}(n)=\int_{-1}^1 P_n^2(x)\ dx=\frac1{4^n(n!)^2}\int_{-1}^1\frac{d^n}{dx^n}\left(x^2-1\right)^n\frac{d^n}{dx^n}\left(x^2-1\right)^n\ dx. $$ Using IBP yields \begin{align} \mathcal{I}(n)&=\frac1{4^n(n!)^2}\left[\left.\frac{d^{n-1}}{dx^{n-1}}\left(x^2-1\right)^n\right|_{-1}^1-\int_{-1}^1\frac{d^{n-1}}{dx^{n-1}}\left(x^2-1\right)^n\frac{d^{n+1}}{dx^{n+1}}\left(x^2-1\right)^n\ dx\right]. \end{align} Hence IBP $n$ times yields \begin{align} \mathcal{I}(n)&=\frac{(-1)^n}{4^n(n!)^2}\int_{-1}^1\left(x^2-1\right)^n\frac{d^{2n}}{dx^{2n}}\left(x^2-1\right)^n\ dx\\ &=\frac{(-1)^n(2n)!}{4^n(n!)^2}\int_{-1}^1\left(x^2-1\right)^n\ dx\\ \end{align} Setting $x=2t-1$ yields \begin{align} \mathcal{I}(n) &=\frac{(2n)!}{(n!)^2}\int_{0}^1t^n\left(1-t\right)^n\ dt\\ &=\frac{(2n)!}{(n!)^2}\cdot\text{B}(n+1,n+1)\\ &=\frac{(2n)!}{(n!)^2}\cdot\frac{(n!)^2}{(2n+1)!}\\ &=(2n)!\cdot\frac{1}{(2n+1)(2n)!}\\ &=\large\color{blue}{\frac1{2n+1}},\tag{Q.E.D.} \end{align} where $\text{B}(x,y)$ is a Beta function.

Cristian Rodríguez
  • 379
Tunk-Fey
  • 24,849
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$$first\ we\ know\ :\\ \\ \\ \because \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^{\infty }p_{n}(x)\ t^n\ \ \ \ \ \ \ \ \ \ \ (1)\\ \\ \\ \because \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{m=0}^{\infty }p_{m}(x)\ t^m\ \ \ \ \ \ \ \ \ \ \ \ (2)\\ \\ by\ multiplication \ (1)\ , (2)\ \ \ \ , then\ we\ have\ \\ \\ \\ \therefore \frac{1}{1-2xt+t^2}=\sum_{n=0}^{\infty }\sum_{m=0}^{\infty }\ p_{n}(x)\ p_{m}(x)\ \ \ \ \ \ \ , at\ n=m\\ \\ \\$$

$$\therefore \frac{1}{1-2xt+t^2}=\sum_{n=0}^{\infty }\ (p_{n}(x))^2\ t^{2n}\\ \\ by\ integrating\ from\ -1\ to 1\ \ then\ we\ have\ \\ \\ \\ \therefore \int_{-1}^{1}\sum_{n=0}^{\infty }(p_{n}(x))^2dx=\frac{-1}{2t}\int_{-1}^{1}\frac{-2tdt}{1-2xt+t^2}dx=\frac{-1}{2t}ln(1-2xt+t^2)_{-1}^{1}\\ \\ \\ =\frac{-1}{2t}\left [ ln(1-2t+t^2)-ln(1+2t+t^2) \right ]=\frac{-1}{2t}\left [ ln(1-t)^2-ln(1+t)^2 \right ]\\ \\ \\ =\frac{-1}{t}\left [ ln(1-t) -ln(1+t)\right ]\\ \\ \\$$ $$=\frac{-1}{t}\left [ \frac{-t}{1} -\frac{t^2}{2}-\frac{t^3}{3}-\frac{t^4}{4}-......+(\frac{t}{1}-\frac{t^2}{2}+\frac{t^3}{3}-......\right ]\\ \\ \\ =\frac{-1}{t}\left [ \frac{-2t}{1} -\frac{2t^3}{3}-.....\right ]=2\left [ 1+\frac{t^2}{3} +\frac{t^4}{5}+......\right ]\\ \\ \\ =2\sum_{n=0}^{\infty }\frac{t^{2n}}{2n+1}=\sum_{n=0}^{\infty }\int_{-1}^{1}(p_{n}(x))^2\ t^{2n}\\ \\ \\ by\ coffecients\ we\ have\ \ \\ \\ \\ \therefore \frac{2}{2n+1}=\int_{-1}^{1}\ p_{n}(x)\ p_{m}(x)\ dx \ \ \ \ \ \ \ , n=m\\ \\$$

Ahmed Hejazi
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