I've been asked to calculate: $$ \int_{0}^{1} P_{\ell}(x)dx, $$ where $P_{\ell}(x)$ is a Legendre polynomial by using:
i)The generating function: $$ \sum_{\ell=0}^{\infty}P_{\ell}(x)t^{\ell}=\frac{1}{\sqrt{1+t^{2}-2xt}} $$
ii)The recurrence rule: $$ xP^{\prime}_{\ell}(x)=\ell P_{\ell}(x)+P_{\ell-1}^{\prime}(x) $$ iii)The recurrence rule: $$ P^{\prime}_{\ell+1}(x)-P^{\prime}_{\ell-1}(x)=(2 \ell+1)P_{\ell}(x) $$
My attempt:
Replacing in the legendre equation $$ (1-x^{2})\frac{d^2}{dx^2}[P_{\ell}(x)]-2x\frac{d}{dx}[P_{\ell}(x)]+\ell(\ell+1)[P_{\ell}(x)]=0 $$ I get the expression $$ \frac{d}{dx}\left[ (1-x^2) \frac{d}{dx}P_{\ell}(x)\right]+\ell(\ell+1)P_{\ell}(x)=0. $$
Integrating the previous expression $$ \left[(1-x^2)\frac{d}{dx}P_{\ell}(x)\right]_{0}^1=-\ell(\ell+1)\int_{0}^{1}P_{\ell}(x)dx $$ and finally $$ \frac{P^{\prime}_{\ell}(0)}{\ell(\ell+1)}=\int_{0}^{1}P_{\ell}(x)dx. $$ But using Wolfram Alpha I get $$ \int_{0}^{1}P_{\ell}(x)dx=\frac{P_{\ell-1}(0)-P_{\ell+1}(0)}{2\ell+1}. $$
The expression is correct but mine is not, any help is welcome
