Upper bound for the sine integral
$$\frac{\pi}{2}-\text{Si}(x) = \int_{x}^{+\infty}\frac{\sin(t)}{t}\,dt = \int_{0}^{+\infty}\frac{\sin(x)\cos(t)+\cos(x)\sin(t)}{x+t}\,dt $$ due to the Laplace transform, can be written as $$ \int_{0}^{+\infty}\frac{\cos(x)+s\sin(x)}{1+s^2}\,e^{-sx}\,ds $$ which by the Cauchy-Schwarz inequality is bounded by $$ \int_{0}^{+\infty}\frac{e^{-sx}}{\sqrt{1+s^2}}\,ds < \int_{0}^{+\infty}e^{-sx}\,ds=\frac{1}{x},$$ hence your conjecture is correct.