Sum up the powered sines

$\begin{aligned} \lim_{n\to \infty} \displaystyle\sum_{r=1}^{n^2} \left[ \dfrac{\sin(\frac{2017 r^2}{n^2})}{r} \right] & = \lim_{n\to \infty} \displaystyle\sum_{r=1}^{n^2} \left[ \dfrac{1}{n} × \dfrac{\sin(\frac{2017 r^2}{n^2})}{\frac{r}{n}} \right]\\ \\ & = \int_{0}^{\infty} \dfrac {\sin (2017 x^2)}{x} dx \\ \\ & = \int_{0}^{\infty} \dfrac {\sin (2017 x^2)}{x^2}× x dx\\ \\ & = \int_{0}^{\infty} \dfrac {\sin 2017 t}{2 \ t} dt & \color{#20A900}{\text {Let} \ x^2 \ \text{ = t, 2x dx = dt}}\\ \\ & = \dfrac {1}{2} × \dfrac{\pi}{2} & \color{#20A900}{\int_{0}^{\infty} \dfrac{\sin mx}{x} dx \ = \ \dfrac{\pi}{2} \ \forall \ m \in \mathbb {R} \setminus \{ 0\}}\\ \\ \therefore a \ + \ b & = \color{#3D99F6}{\boxed {5}} \end{aligned}$