This is somewhat standard being a Gaussian integral. You will have $$ \int_{-\infty}^\infty e^{-i\Omega t}\sum_{k=1}^N\sin(\omega t+k\beta t^2)dt. $$ This reduces to the evaluation of the two integrals $$ \int_{-\infty}^\infty e^{-i\Omega t\pm i(\omega t+k\beta t^2)}dt. $$ Now, complete the square in the exponent as $\left[\pm\sqrt{k}\beta t+\frac{(\Omega\pm\omega)}{2\sqrt{k}\beta}\right]^2-\frac{(\Omega\pm\omega)^2}{4k\beta^2}$ and you are left with $$ e^{i\frac{(\Omega\pm\omega)^2}{4k\beta^2}}\int_{-\infty}^\infty e^{-i\left[\pm\sqrt{k}\beta t+\frac{(\Omega\pm\omega)}{2\sqrt{k}\beta}\right]^2}dt. $$ Finally, $$ e^{i\frac{(\Omega\pm\omega)^2}{4k\beta^2}}\frac{1}{\sqrt{k}\beta}\int_{-\infty}^\infty e^{-i\tau^2}d\tau=e^{i\frac{(\Omega\pm\omega)^2}{4k\beta^2}}\frac{1}{\sqrt{k}\beta}(1-i)\sqrt{\frac{\pi}{2}}. $$ Turning back to your sum, you are done.