Show that:
$$\int_{0}^{\infty}\frac{\sin x^{2}}{\pi-x^{2}}dx=\frac{\sqrt{\pi}}{2}\left[C(\sqrt{2})+S(\sqrt{2})\right]$$
where $C$ and $S$are the Fresnel integrals defined as:
$$\displaystyle C(u)=\int_{0}^{u}\cos(\frac{\pi}{2}x^{2})dx, \;\ S(u)=\int_{0}^{u}\sin(\frac{\pi}{2}x^{2})dx$$
In this case, $u=\sqrt{2}$
Here is what I have attempted so far
$$\displaystyle{\int\limits_0^\infty {\frac{{\sin {x^2}}}{{\pi – {x^2}}}dx} = \mathop = \limits^{x \to \sqrt \pi \cdot x} = \frac{1}{{\sqrt \pi }}\int\limits_0^\infty {\frac{{\sin \pi {x^2}}}{{1 – {x^2}}}dx} = \frac{1}{{\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi {x^2}}}{{1 – {x^2}}}dx} – \int\limits_1^\infty {\frac{{\sin \pi {x^2}}}{{{x^2} – 1}}dx} } \right) = \mathop = \limits^{{x^2} \to \sqrt x } = }$$
$$\displaystyle{ = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi x}}{{\sqrt x \left( {1 – x} \right)}}dx} – \int\limits_1^\infty {\frac{{\sin \pi y}}{{\sqrt y \left( {y – 1} \right)}}dy} } \right) = \mathop = \limits_{y \to 1 + y}^{x \to 1 – x} = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi x}}{{x\sqrt {1 – x} }}dx} + \int\limits_1^\infty {\frac{{\sin \pi y}}{{y\sqrt {1 + y} }}dy} } \right)}$$
Best Answer
Note that \begin{align} I=&\int_0^\infty {\frac{{\sin {x^2}}}{{\pi - {x^2}}}dx} \\ =&\ \frac{1}{{2\sqrt \pi }}\int_{-\infty}^\infty {\frac{{\sin \pi {x^2}}}{{1 - {x^2}}}dx} =\frac{1}{4{\sqrt \pi }}\int_{-\infty}^\infty \frac{\sin \pi {x^2} }{\underset{1-x\to x}{1 - x}} +\frac {\sin \pi {x^2} }{\underset{1+x\to x}{1 +x}}\ dx\\ =& \ \frac{1}{{2\sqrt \pi }}\int_{-\infty}^\infty \frac{\sin \pi {(1-x)^2} }x dx = \frac{1}{{2\sqrt \pi }}\int_{-\infty}^\infty \frac{\sin [\pi x(2-x) ]}x dx \end{align} Let $K(a) = \int_{-\infty}^\infty \frac{\sin[a^2x(2-x)]}x dx$ \begin{align} K’(a) =& \ 2a\int_{-\infty}^\infty \cos [a^2-a^2(1-x)^2 ](2-x)\ \overset{1-x\to x}{dx}\\ = & \ 2a\int_{-\infty}^\infty \cos [a^2(1-x^2)]{dx}\\ = & \ 2a \int_{-\infty}^\infty [\cos (ax)^2 \cos a^2 + \sin (ax)^2 \sin a^2 ] dx \\ = &\ \sqrt{2\pi}\left({\cos a^2} + {\sin a^2} \right) \end{align} Then \begin{align} I= &\ \frac1{2\sqrt\pi} K(\sqrt{\pi})= \frac1{2\sqrt\pi}\int_0^{\sqrt{\pi}}K’(a)da\\ =&\ \frac{1}{\sqrt2}\int_0^{\sqrt{\pi }}\left({\cos a^2} + {\sin a^2}\right)da =\frac{\sqrt{\pi}}2\left[C(\sqrt{2})+S(\sqrt{2})\right] \end{align}