Consider a function $f(t)$ with $f(t>0)>0$ and $f(-t)=-f(t)$. Can I make any statement about the positivity of the Sine transform $$\hat{f}(\omega) = \int_{0}^{\infty} \sin(\omega t) f(t) \mathrm{d}t$$ of such a function? I guess that $\hat{f}(\omega>0)>0$, but I am not sure why.
Edit: Apparently, my guess was wrong. However, I would still be interested, if there exist additional conditions for $f(t)$, which lead to $\hat{f}(\omega>0)>0$. So far, I have only found the condition $f''(t)>0$ for all $t>0$, which has been proven in E. Tuck, Bull. Austral. Math Soc. Vol. 74 (2006), p. 133.
Background: In statistical physics, there emerge kernels $f(t)$ with the above-mentioned properties. Conservation of energy implies that $\hat{f}(\omega>0)>0$ must hold true.
No. Consider for instance $f(t)=t\,|t|\,e^{-|t|}$. Then $$ \hat f(\omega)=\int_0^\infty t^2e^{-t}\sin(\omega\,t)\,dt=\frac{2\,\omega\left(3-\omega^2\right)}{\left(\omega^2+1\right)^3} $$ is $<0$ for $\omega>\sqrt 3$.