show that
lim
I have see this similar problem
\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin^n{x}dx=0
poof:
\forall \xi>0,0<\delta<\xi/2,and there is N,such 0<\sin^n{\pi/2-\delta}<\xi/\pi(n\ge N)
then we have
\int_{0}^{\pi/2}\sin^n{x}dx=\left(\int_{0}^{\pi/2-\delta}+\int_{\pi/2-\delta}^{\pi/2}\right)\sin^n{x}dx=I_{1}+I_{2}
then
|I_{1}|\le\left(\sin{\pi/2-\delta}\right)^n(\pi/2-\delta)<\xi/\pi\cdot\pi/2=\xi/2
and
|I_{2}|\le\left(\pi/2-(\pi/2-\delta)\right)=\delta<\xi/2
and This problem have many other methods,
But for this \lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin{x^n}dx=0
I can't prove it,Thank you
No comments:
Post a Comment