$X$ are an iid draw from $(-\infty, \infty)$ according to $F$ with mean $\mu$. Further let $A = a(x, \theta)/\cos (\alpha)$ and $B = ((1- \cos(\alpha) - \sin (\alpha))\mu + \sin (\alpha)x)/\cos(\alpha)$, where $a(x,\theta)$ is determined by \begin{equation}\tag{1} 1 - F(A-B) = f(A-B)A. \end{equation} It is assumed that that $F$ has an increasing hazard rate, so there is a unique solution. Note that both $A$ and $B$ are functions of $x$ and $\alpha \in [0,\pi/2]$. Show that the following integral is convex: \begin{equation} \int_{-\infty}^{\infty}(1-F(A-B))A \cos(\alpha)dF(x) \end{equation}
Attempt: The regular method of differenting the function twice does not work. I thought it may have something to do with Brunn-Minkowski theorem