Lower bound of a trigonometric integral

Question

Let $\alpha$, $\beta$, and $\gamma$ be non-zero real numbers. Further, suppose that $f$ and $g$ are probability density functions defined on $\mathbb{R}$. I'm interested in computing a lower bound of the quantity $\alpha\int\limits_{0}^{\infty}\cos(vx)f(v)dv+\int\limits_{0}^{\infty}\left(\beta\cos(vx)+\gamma\sin(vx)\right)g(v)dv$, where $x\in\mathbb{R}$. I'm aware of the fact that both $\sin$ and $\cos$ functions have a range of $[-1,1]$ and that the probability density functions have integrals equal to 1. I suspect that the inequality $x\geq -|x|$, for all $x\in\mathbb{R}$, will be helpful in handling $\alpha$, $\beta$, and $\gamma$. For example, $\alpha\geq -|\alpha|$, etc. In fact, I've established the lower bound $-(|\alpha|+|\beta|+|\gamma|)$. However, I'm not convinced that my reasoning is watertight. All suggestions are welcome.

Answer 1

Denote $$\sin\left(\theta \right):=\frac{\beta}{\sqrt{\beta^2 +\gamma^2}}$$ then $$I= \alpha\int\limits_{0}^{\infty}\cos(vx)f(v)dv+\sqrt{\beta^2 +\gamma^2}\int\limits_{0}^{\infty}\sin(vx +\theta)g(v)dv$$ $$\implies \color{red}{-|\alpha|- \sqrt{\beta^2 +\gamma^2}\le I \le |\alpha|+ \sqrt{\beta^2 +\gamma^2}} \tag{1}$$

Given the fact that only $\alpha,\beta$ and $\gamma$ are known, the bounds $(1)$ are tightest in the sense that they can be reached:

• For the lower bound, let us define $$\begin{align} &f(v)=\delta_{\{v=\pi \}} \iff\cases{ \mathbb{P}(V=\pi)=1\\ \mathbb{P}(V\ne \pi)=0} \\ &g(u)=\delta_{\{u=\pi/2 - \theta \}} \iff\cases{ \mathbb{P}(U=\pi/2 - \theta)=1\\ \mathbb{P}(U\ne \pi/2 - \theta)=0} \\ &x = 0 \end{align}$$ then the lower bound is reached.

• For the upper bound, similarly let us define $$\begin{align} &f(v)=\delta_{\{v=0 \}} \iff\cases{ \mathbb{P}(V=0)=1\\ \mathbb{P}(V\ne 0)=0} \\ &g(u)=\delta_{\{u=-\pi/2 - \theta \}} \iff\cases{ \mathbb{P}(U=-\pi/2 - \theta)=1\\ \mathbb{P}(U\ne -\pi/2 - \theta)=0} \\ &x = 0 \end{align}$$ then the upper bound is reached.