Fresnel Integral multiplied with cosine term.

Question

$$I=\int_a^b \sin(\alpha-\beta x^2)\cos(x)\, dx.$$

Can anybody tell me, how to solve this integral ? I know that this is related to Fresnel Integral if the $\cos(x)$ term is absent.

Answer 1

If $\beta>0$

$c+d=2\alpha-2\beta x^{2}$

$c-d=2x$

so $c=\alpha+x-\beta x^{2}=(\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}-\sqrt{\beta}x)^{2}$ and $d=\alpha-x-\beta x^{2}=(\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}+\sqrt{\beta}x)^{2}$

then using that $\sin(c)+\sin(d)=2\sin(\frac{1}{2}(c+d))\cos(\frac{1}{2}(c-d))$ we have:

$\int_{a}^{b}\sin(\alpha-\beta x^{2})\cos(x)dx$

$=\frac{1}{2}\int_{a}^{b}\sin((\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}-\sqrt{\beta}x)^{2})dx+\frac{1}{2}\int_{a}^{b}\sin((\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}-\sqrt{\beta} x)^{2})dx$

Notice that $\int_{s}^{t}\sin(c-u^{2})du=\int_{s}^{t}\sin(c)\cos(u^{2})du-\int_{s}^{t}\cos(c)\sin(u^{2})du$

$=\sin(c)(C(t)-C(s))-\cos(x)(S(t)-S(s))$

where $S(x)=\int_{0}^{x}\sin(p^{2})dp$ and $C(x)=\int_{0}^{x}\cos(p^{2})dp$

A similar procedure can be done for $\beta<0$ since $\sin(-x)=-\sin(x)$. If $\beta=0$ then this is just $\sin(\alpha)\int_{a}^{b}\cos(x)dx$.

Answer 2

In almost the same spirit as user71352's answer, the antiderivative of $$I=\int \sin(\alpha-\beta x^2)\cos(x)\, dx$$ can be written as $$2 \sqrt{\frac{2\beta}{\pi}} I=$$ $$\sin \left(\alpha +\frac{1}{4 \beta }\right) \left(C\left(\frac{2 x \beta -1}{\sqrt{2 \pi\beta} }\right)+C\left(\frac{2 x \beta +1}{\sqrt{2 \pi \beta} }\right)\right)-\cos \left(\alpha +\frac{1}{4 \beta }\right) \left(S\left(\frac{2 x \beta -1}{\sqrt{2 \pi \beta} }\right)+S\left(\frac{2 x \beta +1}{\sqrt{2 \pi \beta} }\right)\right)$$ where $C$ and $S$ are the Fresnel integrals already mentioned by user71352.