Letting $I$ denote the integral
$$I = \int_{-\infty}^{\infty} \frac{\sin(a\cdot x) \cdot \cos(b \cdot x)}{x} \cdot dx$$ for some real $a,b$, I want to find a lower bound on it depending on $a,b$. We may assume that these are strictly positive.
Edit after J.G. comment:
$$ I = \frac{1}{2}\cdot \int_{-\infty}^{\infty} \frac{\sin((a+b)\cdot x)}{x} \cdot dx + \frac{1}{2}\cdot \int_{-\infty}^{\infty} \frac{\sin((a-b)\cdot x)}{x} \cdot dx = I_1 + I_2 $$ I think we can now see $I_1$ (for instance) as the inverse Fourier transform of $L(x) = \frac{\sin((a+b)\cdot x)}{(a+b)\cdot x}$ in the point $t = 0$ up to a multiplicative constant. Note that $L(x)$ is itself the fourier transform of the indicator function $1_{[-(a+b), a+b]}$
Edit 2
It seems that $$ \int_{-\infty}^{\infty} \frac{\sin(a\cdot x)}{x} \cdot dx = \pi \cdot sgn(a) $$ i.e. a knonw Dirichlet integral. Therefore $I = \frac{\pi}{2} \cdot \left( sgn(a+b) + sgn(a-b) \right)$
