This question is about whether the Fourier transform of an odd function equals zero, if and only if the odd function equals zero.
Edit: in my course we only tackle functions so that there exists a finite row of $x_1,x_2,..,x_n$ so that f(x) is continuous over $(-\infty,x_1)$, $(x_1,x_2)$, ..., $(x_n,+\infty)$. And the integrals over these intervals are all finite. My professor called these type of functions "Riemann-integrable".
After I rewrote the equation I stumbled on how to prove that if $$ i\cdot \int_{-\infty}^{+\infty} sin(2\pi k x)\cdot f(x) dx=0$$ for a function $f(x)$, $f(x)$ not depending on $k$ and for every $k \in $ $R$, (real numbers) then $f(x)$ must equal zero.
What I tried to do, was to show that if it holds for $k=k_0$ then $\delta \in R$ and I tried to show that :$$ i\cdot \int_{-\infty}^{+\infty} sin(2\pi (k_0+\delta) x)\cdot f(x) dx=0$$
$$ i\cdot \int_{-\infty}^{+\infty} sin(2\pi k_0 x)\cdot cos(2\pi \delta \cdot x)\cdot f(x) +cos(2\pi k_0 x)\cdot sin(2\pi \delta x)\cdot f(x)dx=0$$
However this only seems to complicate the problem even more.