Sum of all sine harmonics I was discussing this with my calculus teacher, but she didn't come up with anything.
I would like to take an infinite sum of functions (sine specifically) but don't know how to do that.  I would like to sum every $\sin(nx)$ where $n$ is a positive multiple of $1/2$, where the amplitude of each sine is infinitesimal.
Something like this, although I would guess this isn't correct form to do this sort of problem.
$$
f(x) = \sum_{n=1}^\infty \sin\left(\frac{nx}{2}\right) dx
$$
 A: Consider the following series:
$$ f (x)= \sum_{n=-\infty}^\infty \epsilon\delta (x-n/2)$$ Where $\delta (x)$ is the Dirac delta distribution with the property $$\int_{-\infty}^{\infty} g(x)\delta (x-a)dx=\int_{a-\epsilon}^{a+\epsilon} g(x)\delta (x-a)dx=g (a) $$.  Taking the Fourier transform of this sum one obtains $$F (k) = \sum_{n=-\infty}^\infty \epsilon e^{-ikn/2}$$ which is a Fourier series with coefficients of $\epsilon$. Because $f(x)$ repeats over periods of 1, we can also take the Fourier series of the initial function which results in $$ f (x)=\sum_{n=-\infty}^\infty \epsilon e^{-inx/2} $$ So $f $ is an eigenfunction of the Fourier transform/decomposition operation. Thus trying to synthesize the function from its series gives the same function times a constant (1/$\pi$ because of the orthogonality of sines and cosines over symmetric intervals), making the function independent of the amplitude  $\epsilon $ of each term given they are all equal. Then splitting the sum, and taking the fact that the amplitude of each term is arbritary we arrive at  $$f(x)=\frac {1}{2}+\sum_{n=1}^\infty  \frac {e^{inx/2}+e^{-inx/2}}{2}=\frac {1}{2}+\sum_{n=1}^\infty\cos(nx/2) $$ This also happens to be known as the Dirac-Comb distribution.
