Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods and without gamma functions?

Question

I know that

$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$

but all of the methods I've found seem to be too complicated for an early calculus student. Is there any method of calculating this with real methods and without gamma functions?

Answer 1

We first transform the integral by putting $u=x^2$, $$ \int_0^{\infty} \sin \left(x^2\right) d x=\frac{1}{2} \int_0^{\infty} \frac{\sin u}{\sqrt{u}} d u $$

Using the Gaussian integral: $\int_0^{\infty} e^{-u v^2} d v=\frac{\sqrt{\pi}}{2 \sqrt{u}} $, where $u$ is a constant, we have $$ \begin{aligned} I &=\frac{1}{2} \int_0^{\infty} \frac{2 \sin u}{\sqrt{\pi}} \int_0^{\infty} e^{-u v^2} d v d u\\&=\frac{1}{\sqrt{\pi}} \int_0^{\infty}\left(\int_0^{\infty} \sin u \cdot e^{-u v^2} d u\right) d v \\ &=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{v^4+1} d v \end{aligned} $$

Playing a little on the integrand, we obtain $$ \begin{aligned} I&=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{\frac{1}{v^2} }{v^2+\frac{1}{v^2}} d v \\ &=\frac{1}{2 \sqrt{\pi}} \int_0^{\infty} \frac{\left(1+\frac{1}{v^2}\right)-\left(1-\frac{1}{v^2}\right)}{v^2+\frac{1}{v^2}} d v \\ &=\frac{1}{2 \sqrt{\pi}}\left[\int_0^{\infty} \frac{d\left(v-\frac{1}{v}\right)}{\left(v-\frac{1}{v}\right)^2+2}- \underbrace{\int_0^{\infty} \frac{d\left(v+\frac{1}{v}\right)}{\left(v+\frac{1}{v}\right)^2-2}}_{=0}\right] \\ &=\frac{1}{2 \sqrt{2} \cdot \sqrt{\pi}}\left[\tan ^{-1}\left(\frac{v-\frac{1}{v}}{\sqrt{2}}\right)\right]_0^{\infty} \\ &=\frac{\pi}{2 \sqrt{2} \sqrt{\pi}} \\ &=\sqrt{\frac{\pi}{8}} \\ & \end{aligned} $$