convergence of $ \lim_{k \to \infty} \int_0^1 |\sin^k(k\pi x) |^p dx = 0 $ Show that $$ \lim_{k \to \infty} \int_0^1 |\sin^k(k\pi x) |^p dx = 0  $$ for $p \in \mathbb{N}$ arbitary.
My try, if I subsitute $y=k\pi x, \frac{dy}{dx}=k\pi$
which leads to $$\lim_{k \to \infty} \int_0^1 |\sin^k(y)(k\pi)^{-1} |^p dy = 0 $$ and since $sin(y)$ is bounded, I get the zero function, is this correct?
Greetings.  
 A: Changing variables with $y = k \pi x$ we get
$$\begin{align}\int_0^1 |\sin^k(k\pi x) |^p \,dx &= \frac{1}{k \pi}\int_0^{k \pi} |\sin y |^{pk} \, dy \\  &= \frac{1}{k \pi} \sum_{j=1}^k \int_{j\pi - \pi}^{j \pi}|\sin y |^{pk} \, dy \\ &= \frac{1}{k \pi} \sum_{j=1}^k \int_{0}^{\pi}|\sin y |^{pk} \, dy  \\ &= \frac{1}{\pi} \int_{0}^{\pi}|\sin y |^{pk} \, dy \\ &= \frac{\sqrt{\pi}}{\pi} \frac{\Gamma\left(\frac{pk+1}{2}\right)}{\Gamma\left(\frac{pk}{2} + 1\right)} \end{align}$$
Using, for example, Stirling's approximation you can show that the limit as $k \to \infty$ is zero.
Alternatively,
$$\int_{0}^{\pi}|\sin y |^{pk} \, dy  \\ = \int_{0}^{\pi/2 - \epsilon}|\sin y |^{pk} \, dy  + \int_{ \pi/2 - \epsilon}^{\pi/2 + \epsilon}|\sin y |^{pk} \, dy + \int_{ \pi/2 + \epsilon}^{\pi}|\sin y |^{pk} \, dy \\ \leqslant |\sin(\pi/2-\epsilon) |^{pk}(\pi/2 - \epsilon) + 2\epsilon + |\sin(\pi/2+\epsilon) |^{pk}(\pi/2 - \epsilon)$$
Hence, for any $\epsilon > 0$ we get
$$0 \leqslant \liminf_{k \to \infty}\int_{0}^{\pi}|\sin y |^{pk} \, dy \leqslant \limsup_{k \to \infty}\int_{0}^{\pi}|\sin y |^{pk} \, dy  < 2 \epsilon.$$
Since $\epsilon > 0$ is arbitrary, the limit is zero.
A: The change of variables $x = y/k$ gives
$$\int_0^1 |\sin (k\pi x)|^{kp}\, dx = \frac{1}{k}\int_0^k |\sin(\pi y)|^{kp}\, dy = \int_0^1 |\sin(\pi y)|^{kp}\, dy,$$
where we've used the $1$-periodicity of $|\sin (\pi y)|.$ Now for $y\in [0,1], y\ne 1/2,$ $|\sin (\pi y)|<1.$ Hence for such $y,$ $|\sin (\pi y)|^{kp} \to 0$ as $k \to \infty.$ Since the integrands are all bounded by $1,$ the dominated convergence theorem shows the limit is $0.$  
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\lim_{k \to \infty}\int_{0}^{1}\verts{\sin^{k}\pars{k\pi x}}^{\,p}\,\dd x = 0}$

\begin{align}
\mbox{Note that} &
\lim_{k \to \infty}\int_{0}^{1}\verts{\sin^{k}\pars{k\pi x}}^{\,p}\,\dd x =
{1 \over \pi}\lim_{k \to \infty}\bracks{{1 \over k}
\int_{0}^{k\pi}\verts{\sin^{k}\pars{ x}}^{\,p}\,\dd x}
\\[5mm] = &\
{2 \over \pi}\lim_{k \to \infty}
\int_{0}^{\pi/2}\sin^{kp}\pars{x}\,\dd x =
{2 \over \pi}\lim_{k \to \infty}
\int_{0}^{\pi/2}\cos^{kp}\pars{x}\,\dd x
\\[5mm] = &\
{2 \over \pi}\lim_{k \to \infty}
\int_{0}^{\pi/2}\exp\pars{kp\ln\pars{\cos\pars{x}}}\,\dd x =
{2 \over \pi}\lim_{k \to \infty}
\int_{0}^{\infty}\expo{-kpx^{2}/2}\,\dd x
\\[5mm] = &\
{2 \over \pi}\lim_{k \to \infty}{\root{\pi/2} \over \root{kp}} = \bbx{\large 0}
\,,\qquad p > 0.
\end{align}
