I have* a periodic positive function $f(\tau) = f(\tau + T) > 0$, you can assume it to be as smooth as you want. I need to prove that the integral
$$ \int_0^T \sin(\frac{2\pi a}{T} \tau)\, {\mathrm e}^{\gamma \tau / 2}\, f(\tau) \; {\mathrm d}\tau $$
is $\leq 0$ for all $a \in \mathbb N$ ($\gamma > 0$ is a constant).
Unfortunately, it is not necessarily true that $\partial_\tau \left( {\mathrm e}^{\gamma \tau / 2}\, f(\tau) \right)$ is positive for all $\tau$, but note that it has to be positive on average because $f$ is periodic and ${\mathrm e}^{\gamma T/2} > 1$.
Except for the exponential factor, this looks like Fourier sine coefficients which have to be negative. I'd be happy about any suggestions for theorems / reformulations that might help me in the proof. I am hoping for an answer like "the integral is $\leq 0$ for all $a \in \mathbb N$ if $f$ has the property X.
* $f$ is given by a quite complicated expression, I am hoping to prove the negativity of the integral using only more abstract properties of $f$ / I have not been able to see anything from the explicit expression for $f$. From a lot of numerical experiments I am quite confident that the integral is indeed always negative, but proving this has turned out to be more diffecult than expected...
The integral is $\leqslant 0$ for all $a \in \mathbb N$ if ${\mathrm e}^{\gamma \tau / 2}\, f(\tau) $ is not-decreasing.
To "prove" that consider the integral as Darboux Sums.