Evaluate : $\int _0^{\infty } \sin (ax^2)\cos \left(2bx\right)dx$ How can I solve 
$$\int_{0}^{\infty} \sin  \left (ax^2 \right) \cos \left (2bx \right)dx$$
Thanks!

Concerning the antiderivative, start writing$$\sin  \left (ax^2 \right) \cos \left (2bx \right)=\frac12\left(\sin(ax^2+2bx)+\sin(ax^2-2bx)\right)$$ Now $$ax^2+2bx=\left(\sqrt a x +\frac{b}{\sqrt a}\right)^2-\frac{b^2}a$$ $$ax^2-2bx=\left(\sqrt a x -\frac{b}{\sqrt a}\right)^2-\frac{b^2}a$$ Now, consider $$I=\int \sin(ax^2+2bx)\,dx$$ and change variable accordingly for each sine. Expand the sine and you will arrive to a linear comination of sine and cosine Fresnel integrals. Back to $x$, you should get $$I=\frac{\sqrt{\frac{\pi }{2}} \left(\cos \left(\frac{b^2}{4 a}\right) S\left(\frac{b+2
   a x}{\sqrt{a} \sqrt{2 \pi }}\right)-\sin \left(\frac{b^2}{4 a}\right)
   C\left(\frac{b+2 a x}{\sqrt{a} \sqrt{2 \pi }}\right)\right)}{\sqrt{a}}$$ Similarly  $$J=\int \sin(ax^2-2bx)\,dx$$ $$J=-\frac{\sqrt{\frac{\pi }{2}} \left(\sin \left(\frac{b^2}{4 a}\right) C\left(\frac{2 a
   x-b}{\sqrt{a} \sqrt{2 \pi }}\right)-\cos \left(\frac{b^2}{4 a}\right)
   S\left(\frac{2 a x-b}{\sqrt{a} \sqrt{2 \pi }}\right)\right)}{\sqrt{a}}$$ I suppose that you have all pieces to finish your problem.
