$\lim_{n\to\infty} \int_0^1 n^k |\sin nx |^2 dx = 0$, $k$? How can I show that
$$
\lim_{n\to\infty} \int_0^1 n^k |\sin nx |^2 \, dx = 0
$$
if and only if
$$
k<0
$$
I proved that if $k<0$ then the above one holds. But I could not know how the other part holds.
 A: Can you show that
$$
\int \sin^2(x) dx = \frac 12 (x - \frac 12 \sin(2x)) + C?
$$
If so, then
$$
\int_0^1 \sin^2(nx) dx = \frac 1{2n} (n - \frac 12\sin(2n)) = \frac 12 - \frac{\sin(2n)}{4n}.
$$
Hence
$$
\int_0^1 n^k \sin^2(nx) dx = \frac{n^k}{2} - \frac 14 n^{k-1} \sin(2n) \sim \frac {n^k}2
$$
as $n \to +\infty$.
A: Asssuming $\text{n}$ and $x$ are positive, we can set $\left|\sin\left(\text{n}x\right)\right|^2=\sin^2\left(\text{n}x\right)$, so we get:
$$\lim_{\text{n}\to\infty}\int_0^1\text{n}^k\left|\sin\left(\text{n}x\right)\right|^2\space\text{d}x=\lim_{\text{n}\to\infty}\text{n}^k\int_0^1\sin^2\left(\text{n}x\right)\space\text{d}x=$$
$$\lim_{\text{n}\to\infty}\frac{\text{n}^k}{2}\left\{\int_0^11\space\text{d}x-\int_0^1\cos\left(2\text{n}x\right)\space\text{d}x\right\}=\lim_{\text{n}\to\infty}\frac{\text{n}^k}{2}\left\{1-\frac{\sin\left(2\text{n}\right)}{2\text{n}}\right\}$$
Using:


*

*$$\sin^2\left(\text{n}x\right)=\frac{1-\cos\left(2\text{n}x\right)}{2}$$

*$$\int_0^11\space\text{d}x=\left[x\right]_0^1=1-0=1$$

*Substitute $u=2\text{n}x$ and $\text{d}u=2\text{n}\space\text{d}x$:
$$\int_0^1\cos\left(2\text{n}x\right)\space\text{d}x=\frac{1}{2\text{n}}\int_0^{2\text{n}}\cos\left(u\right)\space\text{d}u$$

*$$\int_0^{2\text{n}}\cos\left(u\right)\space\text{d}u=\left[\sin(u)\right]_0^{2\text{n}}=\sin\left(2\text{n}\right)-\sin\left(0\right)=\sin\left(2\text{n}\right)$$


Now, for the Limit:
$$\lim_{\text{n}\to\infty}\frac{\text{n}^k}{2}\left\{1-\frac{\sin\left(2\text{n}\right)}{2\text{n}}\right\}=\lim_{\text{n}\to\infty}\frac{2\text{n}\cdot\text{n}^k-\text{n}^k\sin\left(2\text{n}\right)}{4\text{n}}=\frac{1}{4}\lim_{\text{n}\to\infty}\left(2\text{n}^k-\text{n}^{k-1}\sin\left(2\text{n}\right)\right)$$
