I need to integrate: $$\int_{-\infty}^{+\infty}\sin\left(\alpha x^2+2\beta x+\gamma\right)dx$$ So, my steps:
- Change $\alpha=a^2$, $\beta=ab$, $\gamma=b^2+c$. $$\int_{-\infty}^{+\infty}\sin\left(a^2x^2+2abx+b^2+c\right)dx = \int_{-\infty}^{+\infty}\sin\left((ax+b)^2+c\right)dx$$
- Change $ax+b=t$ $$\frac{1}{a}\int_{-\infty}^{+\infty}\sin\left(t^2+c\right)dt = \frac{1}{a}\int_{-\infty}^{+\infty}\sin t^2\cos c+\cos t^2\sin c\; dt \\ = \frac{\cos c}{a}\int_{-\infty}^{+\infty}\sin t^2\; dt+ \frac{\sin c}{a}\int_{-\infty}^{+\infty}\cos t^2\; dt $$
- Evaluate Fresnel integrals and get $$\sqrt{\frac{\pi}{2}}\frac{\sin c +\cos c}{a}$$
- Substitute back $a=\sqrt\alpha$, $c=\gamma-\frac{\beta^2}{\alpha}$ $$\sqrt{\frac{\pi}{2}}\frac{\sin\left(\gamma-\frac{\beta^2}{\alpha}\right) +\cos \left(\gamma-\frac{\beta^2}{\alpha}\right)}{\sqrt\alpha}$$
It looks good, but first change can be done only if we have assumption that $\alpha$ is positive. And what to do if it is negative?
The integral only converges if $\alpha$ is a nonzero real number (else both of the imaginary exponentials blow up, one on each side of zero, or the integrand is periodic). The oddness of sine also means that we can assume $\alpha>0$ (else pull the minus sign out of the sine, and replace $\alpha $ with $-\alpha$ below). We also need $\beta$ to be real, for similar reasons of non-decay and exponential growth. Completing the square gives $$ \alpha x^2+2\beta x+\gamma = \alpha \left( x+\frac{\beta}{\alpha} \right)^2 + \gamma-\frac{\beta^2}{\alpha^2}. $$ Then since $\beta/\alpha$ is real, we can shift the variable to get $$ \int_{-\infty}^{\infty} \sin{(\alpha y^2 + c)} \, dy, $$ where $c$ is as you define it, and then substitute $u=\sqrt{\alpha}y$, which gives $$ \frac{1}{\sqrt{\alpha}} \int_{-\infty}^{\infty} \sin{(u^2+c)} \, du, $$ which you have evaluated. Let me know if you'd like me to explain any of that in more detail.