I just skimmed through this paper on positivity of fourier transforms where it is shown that the sine transform $$\int_0^\infty f(x) \sin(x t) \mathrm{d}x$$ is positive if $f$ is decreasing. The proof seems to be pretty straightforward, but at some point, (I assume) the following step is made: $$\begin{align} \int_0^\infty f(x) \sin(x t) \mathrm{d}x & = \frac{1}{t} \sum_{j=0}^\infty \int_0^{2\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta \\ & = \frac{1}{t} \sum_{j=0}^\infty \left[ \int_0^{\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta + \int_\pi^{2\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta \right] \\ & = \frac{1}{t} \sum_{j=0}^\infty \left[ \int_0^{\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta + \int_0^{\pi} -f\left(\frac{\theta + 2 \pi j + \pi}{t}\right) \sin(\theta) \mathrm{d}\theta \right] \\ & = \frac{1}{t} \sum_{j=0}^\infty \int_0^{\pi} \left[ f\left(\frac{\theta + 2 \pi j}{t}\right) - f\left(\frac{\theta + 2 \pi j}{t} + \frac{\pi}{t}\right) \right] \sin(\theta) \mathrm{d}\theta \end{align}$$ from which the conclusion is drawn that if $f$ is decreasing for $x \geq 0$, the sine transform will be positive.
However, by performing this step slightly differently, I found $$\begin{align} \int_0^\infty f(x) \sin(x t) \mathrm{d}x & = \frac{1}{t} \sum_{j=0}^\infty \int_0^{2\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta \\ & = \frac{1}{t} \sum_{j=0}^\infty \left[ \int_0^{\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta + \int_\pi^{2\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta \right] \\ & = \frac{1}{t} \sum_{j=0}^\infty \left[ \int_\pi^{2\pi} -f\left(\frac{\theta + 2 \pi j - \pi}{t}\right) \sin(\theta) \mathrm{d}\theta + \int_\pi^{2\pi} f\left(\frac{\theta + 2 \pi j}{t}\right) \sin(\theta) \mathrm{d}\theta \right] \\ & = \frac{1}{t} \sum_{j=0}^\infty \int_\pi^{2\pi} \left[ f\left(\frac{\theta + 2 \pi j}{t}\right) - f\left(\frac{\theta + 2 \pi j}{t} - \frac{\pi}{t}\right) \right] \sin(\theta) \mathrm{d}\theta \end{align}$$ from which it would follow that if $f$ is decreasing for $x \geq 0$, the sine transform would be negative.
My question is thus: where did I make the mistake to get this contradicting result?
It is still positive. Note that if $f$ is decreasing then
$$f\left(\frac{\theta + 2 \pi j}{t}\right)\le f\left(\frac{\theta + 2 \pi j}{t} - \frac{\pi}{t}\right)\implies f\left(\frac{\theta + 2 \pi j}{t}\right) - f\left(\frac{\theta + 2 \pi j}{t} - \frac{\pi}{t}\right)\le 0$$
And $\sin(x)\le 0$ for $x\in[\pi,2\pi]$. Hence the integrand is positive.