Evaluate $\lim\limits_{n \rightarrow \infty}\frac1{n^2}\sum\limits_{k=1}^n\sin\left (\frac{\pi k}n\right)\varphi(k)$ 
Evaluate $$\lim_{n \rightarrow \infty}\frac{1}{n^2} \sum_{k=1}^{n}   \sin \left (\frac{\pi k}{n} \right)  \varphi(k)$$ where $\varphi$ denotes Euler's totient function.

I first tried simplifying the sum $$\sum_{k=1}^{n}   \sin \left (\frac{\pi k}{n} \right)  \varphi(k)$$ by converting to exponentials: $\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$, so the sum is $$\sum_{k=1}^{n}   \sin \left (\frac{\pi k}{n} \right)  \varphi(k) = \sum_{k=1}^n \dfrac{e^{i \cdot \frac{\pi k}{n}}-e^{-i \cdot \frac{\pi k}{n}}}{2i} \varphi(k),$$ but I didn't see how to use this to simplify the sum. How can we calculate the limit?
 A: We know that $$\sum_{k\leq n}\phi\left(k\right)=\frac{3}{\pi^{2}}n^{2}+O\left(n\log\left(n\right)\right)$$ hence using Abel's summation $$\sum_{k=1}^{n}\sin\left(\frac{\pi k}{n}\right)\phi\left(k\right)=-\frac{\pi}{n}\int_{1}^{n}\left(\sum_{k\leq t}\phi\left(k\right)\right)\cos\left(\frac{\pi t}{n}\right)dt$$ $$=-\frac{3}{n\pi}\int_{1}^{n}t^{2}\cos\left(\frac{\pi t}{n}\right)dt+O\left(\frac{1}{n}\int_{1}^{n}t\log\left(t\right)dt\right).$$ Now we can observe, integrating by parts, that $$-\frac{3}{n\pi}\int_{1}^{n}t^{2}\cos\left(\frac{\pi t}{n}\right)dt=\frac{3\pi^{2}\sin\left(\frac{\pi}{n}\right)-6n^{2}\sin\left(\frac{\pi}{n}\right)+6\pi n^{2}+6\pi n\cos\left(\frac{3\pi}{n}\right)}{\pi^{4}}$$ and $$O\left(\frac{1}{n}\int_{1}^{n}t\log\left(t\right)dt\right)=O\left(n\left(2\log\left(n\right)-1\right)\right)$$ hence $$\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}\sin\left(\frac{\pi k}{n}\right)\phi\left(k\right)=\color{red}{\frac{6}{\pi^{3}}}.$$
A: By this paper by Omran Kouba we get the following theorem.
Theorem. Let $\alpha$ be a positive real number, and let $(a_n)_{n \geq 1}$ be a sequence of real numbers such that $$\lim_{n \to \infty}\dfrac{1}{n^{\alpha}} \sum_{k=1}^n a_k = L.$$ Then, for every continuous function $f$ on the interval $[0,1]$, we have $$\lim_{n \to \infty}\dfrac{1}{n^{\alpha}}\sum_{k=1}^n f\left(\dfrac{k}{n}\right)a_k = L \int_{0}^1 \alpha x^{\alpha-1}f(x)dx.$$
Then $$\lim_{n \to \infty}\dfrac{1}{n^2}\sum_{k=1}^n \sin\left(\dfrac{\pi k}{n}\right)\varphi(k) = \left(\lim_{n \to \infty}\dfrac{1}{n^2}\sum_{k=1}^n \varphi(k)\right)\left(\int_{0}^1 2x\sin(x)dx\right).$$ It is well-known that $\displaystyle \lim_{n \to \infty}\dfrac{1}{n^2}\sum_{k=1}^n \varphi(k) = \dfrac{3}{\pi^2}$, so the limit in question is $$\dfrac{6}{\pi^2}\int_{0}^1 x\sin(\pi x)dx = \dfrac{6}{\pi^3}.$$
