 #1
 662
 16
 Homework Statement:

Show that
[tex]\int^{1}_{1}x^2\frac{d^m}{dx^m}(1x^2)^mdx=0[/tex] for ##m \geq 2##.
 Relevant Equations:

Partial integration
[tex]\int udv=uv\int vdu [/tex]
It is clear that ##1x^2## is equal to zero in both boundaries ##1## and ##1##. So for me is interesting to think like this
[tex]\frac{d^m}{dx^m}(1x^2)^m=\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)...[/tex]
and
[tex]\frac{d^{m1}}{dx^{m1}}(1x^2)^m=(1x^2)\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)...[/tex]
so there exists one ##(1x^2)## that is not differentiate. Am I right? So is it for ##dv=\frac{d^m}{dx^m}(1x^2)^mdx##, ##v=\frac{d^{m1}}{dx^{m1}}(1x^2)^m##?
[tex]\frac{d^m}{dx^m}(1x^2)^m=\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)...[/tex]
and
[tex]\frac{d^{m1}}{dx^{m1}}(1x^2)^m=(1x^2)\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)\frac{d}{dx}(1x^2)...[/tex]
so there exists one ##(1x^2)## that is not differentiate. Am I right? So is it for ##dv=\frac{d^m}{dx^m}(1x^2)^mdx##, ##v=\frac{d^{m1}}{dx^{m1}}(1x^2)^m##?