I already asked a similar question on this topic, but after a small discussion and a new finding, I noted that I did must boil down the problem such that the solution space shrinks so to say, to maybe have a concrete answer. I am still not a complex analysis or measure theory expert, so I hope that this simplification helps to solve the problem...
Suppose I have two rational, complex functions, mapping from the real numbers to the complex plane $$f(ix) = \frac{p(ix)}{q(ix)} \qquad \text{and}\qquad g(ix) = \frac{r(ix)}{s(ix)}$$ Furthermore, the polynomials $q$ and $s$ do not have zeros on the imaginary axis (or if it simplifies the problem, the right half complex plane), and I know that $|g(ix)|\le 1 \ \forall x$. I can also assume that $\mathrm{deg}(p)\le\mathrm{deg}(q)$ and $\mathrm{deg}(r)\le\mathrm{deg}(s)$, and the polynomials do not have common factors.
Is it then true to state that $$\left|\int_{-\infty}^\infty f(ix)g(ix)\mathrm{d}x \right|\le \left| \int_{-\infty}^\infty f(ix)\mathrm{d}x\right| \quad \forall x \in \mathbb{R}\quad?$$
I have the idea that this must be true, due to the fact that the functions are holomorphic on the imaginary axis (correct me if I'm wrong), and the functions are proper. But again, I cannot prove this... I tried to find a counter example, but up to now this inequality holds in simulation :P
If this is not enough information to solve the problem, I may have another relaxation that can solve the problem... In my application it is also possible to formulate an additional function $h(ix)$ for which we know that $|h(ix)|=1 \ \forall x$ and $\angle h(ix)=\angle g(ix) \ \forall x$. Then the inequality I would like to hold becomes $$\left|\int_{-\infty}^\infty f(ix)g(ix)\mathrm{d}x \right|\le \left| \int_{-\infty}^\infty f(ix)h(ix)\mathrm{d}x\right| \quad \forall x \in \mathbb{R}$$
When I use the above, I think it is true to conclude that $$\angle\left\{ \int_{-\infty}^\infty f(ix)g(ix)\mathrm{d}x \right\} = \angle\left\{ \int_{-\infty}^\infty f(ix)h(ix)\mathrm{d}x \right\},$$ since $\angle h(ix)=\angle g(ix)$. Suppose this angle is $\theta$, then if we write $g$ and $h$ as a complex exponential, we have $$\left|\int_{-\infty}^\infty f(ix)g(ix)\mathrm{d}x \right| = e^{-i\theta}\int_{-\infty}^\infty f(ix)g(ix)\mathrm{d}x = e^{-i\theta}\int_{-\infty}^\infty f(ix)G(x)e^{i\tilde{g}(x)}\mathrm{d}x$$ $$\left|\int_{-\infty}^\infty f(ix)h(ix)\mathrm{d}x \right| = e^{-i\theta}\int_{-\infty}^\infty f(ix)h(ix)\mathrm{d}x = e^{-i\theta}\int_{-\infty}^\infty f(ix)\cdot1\cdot e^{i\tilde{h}(x)}\mathrm{d}x,$$ with $\tilde{g},\tilde{h}$ functions that describe the phase of $g$ and $h$, respectively, and $G(x)$ describing the modulus of $g$. Since I assumed that $\angle h(ix)=\angle g(ix)$, we have $$\mathbb{R}\ni\left|\int_{-\infty}^\infty f(ix)g(ix)\mathrm{d}x \right| = \int_{-\infty}^\infty \underbrace{G(x)}_{\in[0,1]}f(ix)e^{i(\tilde{h}(x)-\theta)}\mathrm{d}x$$ $$\mathbb{R}\ni\left|\int_{-\infty}^\infty f(ix)h(ix)\mathrm{d}x \right| = \int_{-\infty}^\infty f(ix) e^{i(\tilde{h}(x)-\theta)}\mathrm{d}x,$$ Then you would say that the lower term is an upperbound for the upperterm?! Or what does not hold in this analysis?