- #1

Another1

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\(\displaystyle {P}_{n}(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n\)

my thoughts

\(\displaystyle \int_{-1}^{1} \,{P}_{n}(x){P}_{n}(x)dx = \frac{1}{2^{2n}(n!)^2}\int_{-1}^{1} \,\frac{d^n}{dx^n}(x^2-1)^n\frac{d^n}{dx^n}(x^2-1)^ndx\)

let

\(\displaystyle u = \frac{d^n}{dx^n}(x^2-1)^n\) and \(\displaystyle du =\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^ndx\)

\(\displaystyle dv = \frac{d^n}{dx^n}(x^2-1)^ndx\) and \(\displaystyle v = \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\)

so

\(\displaystyle uv=\frac{d^n}{dx^n}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\)

\(\displaystyle -vdu=-\int_{-1}^{1} \,\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx\)

\(\displaystyle uv-vdu = \frac{d^n}{dx^n}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n-\int_{-1}^{1} \,\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx\)

what should I do?