Norm of Legendre Polynomials $P_m(x)$

Question

While studying to prove the norm of Legendre polynomials $P_m(x)$ is $\sqrt{\frac{2}{2m+1}}$, I faced $\int_{-1}^{1} [D^m (x^2-1)^m]^2 dx = (2m)! \int_{-1}^{1} (1-x^2)^m dx.$ $D^m$ stands for derivative of $m$-th degree. It seems to me I shoud use integration by parts of LHS $m$ times to arrive RHS. Please show me what is the detailed steps to do integration by parts of this case. I am confused by the existence of $D^m$ in this case. I am puzzled how to handle $D^m.$ Thanks in advance.

Answer 1

Inductively integrate by parts \begin{align*}\require{cancel} &\int_{-1}^1 [D^{m-k}(x^2-1)^m][D^{m+k}(x^2-1)^m]\,\mathrm{d}x\\ &=\cancel{\bigg[[D^{m-k-1}(x^2-1)^m][D^{m+k}(x^2-1)^m]\bigg]_{-1}^1}\\ &\quad-\int_{-1}^1 [D^{m-k-1}(x^2-1)^m][D^{m+k+1}(x^2-1)^m]\,\mathrm{d}x \end{align*} where $u=[D^{m+k}(x^2-1)^m]$, $\mathrm{d}v=[D^{m-k}(x^2-1)^m]\,\mathrm{d}x$ so $\mathrm{d}u=[D^{m+k+1}(x^2-1)^m]\,\mathrm{d}x$, $v=D^{m-k-1}(x^2-1)^m$ and $v=0$ at $x=\pm 1$.