Question

clarification

Answer 1

\[\int\limits_{-1}^{1}dx[\frac{ d^m }{ dx^m }(x^2-1)^m][\frac{ d^n }{ dx^n }(x^2-1)^n]\]

Answer 2

how is that equal to \[\int\limits_{-1}^{1}dx(x^2-1)^m \frac{ d^{n+m} }{ dx^{n+m} }(x^2-1)^n\]

Answer 3

Integration by parts

Answer 4

@phi

Answer 5

interesting, I've not seen that before. Can you post how to show that ?

Answer 6

integrate by parts: $$\int_a^b f(x) g'(x)\, dx=f(x)g(x)\bigg|_a^b-\int_a^b f'(x) g(x)\, dx$$
in our case, \(f(x) g(x)\biggr|_a^b=0\) each time we use it due to symmetry.

repeated \(n\) times gives: $$\int_a^b f(x)g^{(n)}(x)\, dx=(-1)^n\int_a^b f^{(n)}(x) g(x)\, dx$$

Answer 7

so take \(f(x)=\frac{d^n}{dx^n}(x^2-1)^n\) and \(g(x)=(x^2-1)^m\) so we have $$\int_{-1}^1 f(x) g^{(m)}(x)\, dx=(-1)^m\int_{-1}^1 f^{(m)}(x) g(x)\, dx$$
where clearly $$f^{(m)}(x)=\frac{d^{m+n}}{dx^{m+n}} (x^2-1)^n$$

so technically what you have only holds in the case \(m\) is even otherwise there's a \(-1\) out in front

Answer 8

To me, it is definition of integration by parts. Assume \((x^2-1)^m\in C^m\) and \((x^2-1)^n \in C^n\), then integration by parts says
\[\int_\Omega f(x)\partial g dx= -\int_\Omega \partial f*g dx+\int_{\partial \Omega} f.g*\nu d\nu\]
but your boundary is just a number (-1,1), hence it is =0

Answer 9

So, if you do it m times, then you get what you want. As @oldrin.bataku stated above, I think you missed some constraint to force the expression positive.