Who's up to the challenge? 8

Let the integral be $I$ , then we have:

$\begin{aligned} I & = \int_0^\infty \frac{\color{#3D99F6}{\sin \sqrt{x}}}{e^x} dx \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \int_0^\infty e^{-x} \color{#3D99F6}{\sum_{n=0}^\infty \frac{(-1)^n x^{n+\frac{1}{2}}}{(2n+1)!}} dx \\ & = \sum_{n=0}^\infty \frac{(-1)^n \int_0^\infty e^{-x} x^{n+\frac{1}{2}} dx}{(2n+1)!} \\ & = \sum_{n=0}^\infty \frac{(-1)^n \Gamma \left(n+\frac{3}{2} \right)}{(2n+1)!} \\ & = \sum_{n=0} ^\infty \frac{(-1)^n (2n+1)!! \sqrt{\pi}}{(2n+1)! 2^{n+1}} \\ & = \sqrt{\pi} \sum_{n=0} ^\infty \frac{(-1)^n (2n+1)!}{(2n+1)! 2^{n+1}2^n n!} \\ & = \sqrt{\pi} \sum_{n=0} ^\infty \frac{(-1)^n}{2^{2n+1}n!} \\ & = \frac{\sqrt{\pi}}{2} \sum_{n=0} ^\infty \frac{(-1)^n}{n!}\left(\frac{1}{4} \right)^n = \frac{\sqrt{\pi}}{2} e^{-\frac{1}{4}} = \frac{\sqrt{\pi}}{2\sqrt[4]{e}} \end{aligned}$

$\Rightarrow A+B+C = 2+2+4 = \boxed{8}$