express $a\sin(\omega t)$ as a fourier series. Is the solution $f(\omega t)=\frac{1}{2\pi}a$?

Question

To express $a\sin(\omega t)$ as a fourier series.

let, ${\omega t}=\theta$

$f(\theta)= a_{0}+\sum_{n=1}^{n=\infty}a_{n}{\cos( n\theta)}+\sum_{n=1}^{n=\infty}b_{n}{\sin( n\theta)}$

$$a_{0}=\frac{1}{2\pi}\int_{0}^{2\pi}f(\theta)d{\theta}$$

$$a_{0}=\frac{1}{2\pi}a$$

$$a_n=\frac{1}{\pi}\int_{0}^{2\pi}f(\theta)\cos(n\theta)d{\theta}$$ $$a_{n}= 0, \forall n >0$$

$$b_n=\frac{1}{\pi}\int_{0}^{2\pi}f(\theta)\sin(n\theta)d{\theta}$$ $$b_{n}= 0, \forall n$$

Answer 1

You have assumed that the integrals for $a_n$ and $b_n$ evaluate to zero, but this is only true if omega is an integer.

Note that if $n\in\mathbb{Z}$, then

$\large\frac{1}{\pi}\int_0^{2\pi}sin(\omega t)sin(nt)dt = \frac{n sin(2\pi\omega)}{\omega^2 - n^2}$

Answer 2

This might be absurd to the point of being fascinating... Note that $a\sin(\omega t)=a\sin(\omega t)$ hence $$ a\sin(\omega t)=a_0+\sum_na_n\cos(n\omega t)+\sum_nb_n\sin(n\omega t), $$ with $$ a_n=0,\quad b_n=0\ (n\ne1),\quad b_1=a.$$ Note that to expand the function as $$ a\sin(\omega t)=a_0+\sum_na_n\cos(nt)+\sum_nb_n\sin(nt), $$ is an entirely different cup of tea when $\omega$ is not an integer (although one would know that $a_n=0$ for every $n$ by symmetry).