I have no practical reason for wanting to do this, but I was wondering why the Fourier series for $\sin x$ is the identical zero function.
I'm probably doing something wrong or missing some important condition.
Could someone help me see?
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$\begingroup$Wolfram gives the following:
$$ \frac2{\pi}\int_{0}^{\pi} \sin x \sin (nx)\ dx = -\frac{2\sin(n\pi)}{\pi(n^2-1)} $$
You are almost correct in that this is zero for all $n$ because $\sin(n\pi) = 0$ for every integer. But when $n=1$, the formula doesn't work, because the $n^2-1$ in the denominator becomes zero too. You need to consider that as a special case:
$$ \frac2{\pi}\int_{0}^{\pi} \sin x \sin (1x)\ dx = \frac2{\pi}\int_0^{\pi} \sin^2 x \ dx = \frac2{\pi} \frac{\pi}{2} = 1. $$
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