TITLE: on integral equations
QUESTION [5 upvotes]: Consider the equation 
$$
\int_{0}^{\infty}\sin(\omega\tau)f(\tau)d\tau =\omega\int_{0}^{\infty}\cos(\omega\tau)f(\tau)d\tau\;,
$$
where $\omega >0$ and $f$ is a probability density function. I would like to know whether this equation always has solutions $\omega$.

REPLY [5 votes]: Following @GiuseppeNegro's hint, if $f$ is the PDF of the exponential distribution with parameter $a$, one asks for solutions $\omega$ of the equation
$$
\int_0^\infty\sin(\omega\tau)(a-1)\mathrm e^{-a\tau}\mathrm d\tau=0.
$$
Writing this as the imaginary part of the integral of $\tau\mapsto(a-1)\exp((\mathrm i \omega-a)\tau)$, one gets the condition
$$
\Im\left(\frac{a-1}{a-\mathrm i \omega}\right)=0=\frac{(a-1)\omega}{a^2+\omega^2},
$$
which, if $a\ne1$, has no solution $\omega\ne0$.