A squared sum and integral

Prove that:

$${\large \int_{-\pi}^{\pi}}\left ( \sum_{k=1}^{2014}\sin (kx) \right )^2\,dx=2014\pi$$

Solution

We are using the identity:

$$\int_{-\pi}^{\pi} \sin n x\sin m x \ \mathrm{d}x = \frac{1}{2} \int_{-\pi}^{\pi} [\cos((n-m) x) +\cos ((n+m)x) ]\ \mathrm{d}x = 0$$

whenever $m\neq n$ where $m, n \in \mathbb{Z}$.

Hence:

$$\begin{align*}{\large \int_{-\pi}^{\pi }}  \left(\sum_{k=1}^{2014} \sin k x\right)^2 \ \mathrm{d}x &= \int_{-\pi}^{\pi} \left[ \left(\sum_{k=1}^{2014} \sin^2 kx\right)+2 \left(\sum_{1\leq i<j\leq 2014} \sin ix \sin jx\right) \right] \ \mathrm{d}x \\ &=\sum_{k=1}^{2014} \int_{-\pi}^{\pi} \sin^2 kx \ \mathrm{d}x \\ &= \frac{1}{2} \cdot \sum_{k=1}^{2014} \int_{-\pi}^{\pi} (1-\cos kx) \ \mathrm{d}x \\ &= 2014\pi\end{align*}$$

ending the exercise.