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In the book (Beyond Integrals and Series) I ran into this:

Prove that: $$\int_{\left [ -\pi, \pi \right ]}\left ( \sum_{k=1}^{2014}\sin (kx) \right )^2\,dx=2014\pi$$



I guess we can generalize that. I have not a found something like that, anyway.

A possible first step is to use Lagrange's trigonometric identity: \begin{align*}
\mathop{\sum}\limits_{k=1}^n {\sin(kx)}&=\frac{1}{2}\frac{\cos\frac{x}{2}-\cos\bigl({\bigl({n +\frac{1}{2}}\bigr) x}\bigr)}{\sin\frac{x}{2}}\\
&=........................\\
&=\frac{\sin\bigl({({n+1})\frac{x}{2}}\bigr)\,\sin\bigl({n\frac{x}{2}}\bigr)}{\sin\frac{x}{2}}\,.
\end{align*}

Replied by ex-member aziiri:

Grigorios Kostakos wrote:A possible first step is to use Lagrange's trigonometric identity: \begin{align*}
\mathop{\sum}\limits_{k=1}^n {\sin(kx)}&=\frac{1}{2}\frac{\cos\frac{x}{2}-\cos\bigl({\bigl({n +\frac{1}{2}}\bigr) x}\bigr)}{\sin\frac{x}{2}}\\
&=........................\\
&=\frac{\sin\bigl({({n+1})\frac{x}{2}}\bigr)\,\sin\bigl({n\frac{x}{2}}\bigr)}{\sin\frac{x}{2}}\,.
\end{align*}

It is easier that that, note that for two integers \(n\neq m \) : \[\int_{-\pi}^{\pi} \sin n x\sin m x \ \mathrm{d}x = \frac{1}{2} \int_{-\pi}^{\pi} \cos((n-m) x) +\cos ((n+m)x) \ \mathrm{d}x = 0\]
Then : \[\begin{align*}\int_{-\pi}^{\pi } \left(\sum_{k=1}^{2014} \sin k x\right)^2 \ \mathrm{d}x &=\int_{-\pi}^{\pi} \left(\sum_{k=1}^{2014} \sin^2 kx\right)+2 \left(\sum_{1\leq i<j\leq 2014} \sin ix \sin jx\right) \ \mathrm{d}x \\ &=\sum_{k=1}^{2014} \int_{-\pi}^{\pi} \sin^2 kx \ \mathrm{d}x \\ &= \frac{1}{2} =\sum_{k=1}^{2014} \int_{-\pi}^{\pi} 1-\cos kx \ \mathrm{d}x \\ &= 2014\pi.\end{align*}\]

Azirii, can we generalize that? That is, does the identity $$\int_{-\pi}^{\pi}\left ( \sum_{k=1}^{n}\sin \left ( kx \right ) \right )^2\,dx=n\pi$$ hold?

Replied by ex-member aziiri:

Tolaso J Kos wrote:Azirii, can we generalize that? That is, does the identity $$\int_{-\pi}^{\pi}\left ( \sum_{k=1}^{n}\sin \left ( kx \right ) \right )^2\,dx=n\pi$$ hold?

Yes, of course.

I don't know if induction works, but I believe that your solution can easily be modified for that case also.

Replied by ex-member aziiri:

Tolaso J Kos wrote:I don't know if induction works, but I believe that your solution can easily be modified for that case also.

you just need to substitute \(2014\) with \(n\) in my solution.