I'm aware of the two questions [1] and [2] but I'm trying to proof the orthongonality of the Legendre polynomials without integration by parts. $$ \int_{-1}^1 P_\ell(x)P_{\ell'}(x) dx = \frac{2}{2\ell+1}\delta_{\ell \ell'} $$ defined by the Rodrigues Formula $$ P_\ell(x) = \frac{1}{2^\ell \ell!} \left(\frac{d}{dx} \right)^\ell (x^2-1)^\ell$$
I want to know how can I continue if I start as follows $$ P_l(x) = \frac{1}{2^\ell \ell!} \left(\frac{d}{dx} \right)^\ell \sum_{k=0}^\ell \binom{\ell}{k}(-1)^{\ell-k}x^{2k} \\ = \frac{1}{2^\ell \ell!} \sum_{k=0}^\ell \binom{\ell}{k}(-1)^{\ell-k}\binom{2k}{\ell}\ell!\ x^{2k - \ell} $$ The last integral can be summed starting from $k \geq \ell/2$ instead of zero. Now substitute the in the integral the suitable expressions $$ \frac{1}{2^{\ell+\ell'} \ell! \ell'!}\int_{-1}^1\left(\sum_{k=0}^\ell \binom{\ell}{k}(-1)^{\ell-k}\binom{2k}{\ell}\ell!\ x^{2k - \ell}\right)\left(\sum_{k'=0}^{\ell'} \binom{\ell'}{k'}(-1)^{\ell'-k'}\binom{2k'}{\ell'}\ell'!\ x^{2k' - \ell'}\right) dx \\ =\frac{1}{2^{\ell+\ell'} \ell! \ell'!}\int_{-1}^1 \left( \sum_{k,k'=0}^{\ell,\ell'} (-1)^{\ell + \ell'-(k+k')} \binom{\ell}{k}\binom{\ell'}{k'}\binom{2k}{\ell}\binom{2k'}{\ell'}\ell!\ell'!\ x^{2(k+k') - (\ell+\ell')}\right) dx \\ =\frac{1}{2^{\ell+\ell'} \ell! \ell'!}\left[ \sum_{k,k'=0}^{\ell,\ell'} (-1)^{\ell + \ell'-(k+k')} \binom{\ell}{k}\binom{\ell'}{k'}\binom{2k}{\ell}\binom{2k'}{\ell'}\ell!\ell'!\ \frac{x^{2(k+k') - (\ell+\ell')+1}}{2(k+k') - (\ell+\ell')+1} \right]_{-1}^1 \\ $$ In the sum above only non-negative powers of $x$ count, since $\ell$-th and $\ell'$-th derivatives were taken. Now if $\ell + \ell'$ is odd, then the integral vanishes, since the powers of $x$ are even. But if $\ell + \ell'$ is even then it is not clear ... I don't know yet how to conclude the rest.