An Amazing Integral!

First we use the Feynman Technique to differentiate under the integral sign: $g'(y) = \frac{d}{dy}\int_0^{\infty} e^{-xy}\frac{\sin(2018x)}{x} \,dx = \int_0^{\infty}\frac{\partial}{\partial y} e^{-xy}\frac{\sin(2018x)}{x}\,dx = -\int_0^{\infty} e^{-xy}\sin(2018x)\,dx .$ This integral is easily evaluated with integration by parts, doing this gives $g'(y) = -\frac{2018}{y^2+2018^2},$ and integrating again will give us $g$ : $g(y) = -\int \frac{2018}{y^2+2018^2}\,dy = -\arctan\Big(\frac{y}{2018}\Big) + C .$ Now, notice that $\lim_{y\to\infty} g(y) = 0 \Rightarrow 0 = -\lim_{y\to \infty}\arctan\Big(\frac{y}{2018}\Big) + C \Rightarrow C = \frac{\pi}{2},$ so we can finally compute $g(2018)$ : $g(2018) = -\arctan(1) + \frac{\pi}{2} = \boxed{\frac{\pi}{4}}.$