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?
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ See for example this answer which uses real methods and avoid the gamma function. This question is very very close to being a duplicate of math.stackexchange.com/questions/187729/… $\endgroup$– WintherMay 18, 2016 at 22:33
-
$\begingroup$ @Winther, could you tell me where $e^{-\lambda x^2}$ in $\left(\int_0^\infty{\sin}(x^2) e^{-\lambda x^2}\,\mathrm{d}x\right)^2$ in this solution comes from? $\endgroup$– CachetejackMay 18, 2016 at 22:59
-
3$\begingroup$ This is a common trick (sometimes called Feynman integration). If we want to evaluate an integral $\int f(x){\rm d}x$ it's sometimes useful to generalize the problem by adding a function with a free parameter to the integrand like e.g. $\int f(x)e^{-\lambda x^2}{\rm d}x $. This allows us to perform manipulations like differentiate with respect to this free parameter $\lambda$. Taking this parameter to $0$ gives us the integral we really want. This is not derived from something; it's just "seen" (from exerience) that this might work. $\endgroup$– WintherMay 18, 2016 at 23:05
-
3$\begingroup$ See for example en.wikipedia.org/wiki/… for some examples of how this trick can be applied. Notice the different functions needed for the different integrals; these are basically found by trial-and-error. $\endgroup$– WintherMay 18, 2016 at 23:07
-
1$\begingroup$ Is the Laplace (inverse) transform considered a real-analytique technique? In such a case, through $\mathcal{L}(\sin x)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)=\frac{1}{\sqrt{\pi s}}$, the problem boils down to computing $$\int_{0}^{+\infty}\frac{ds}{\sqrt{s}(1+s^2)}=\int_{0}^{+\infty}\frac{2\,dt}{1+t^4}.$$ $\endgroup$– Jack D'AurizioMay 19, 2016 at 0:48
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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 $$ \begin{aligned} I&=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{\frac{1}{v^2} }{v^2+\frac{1}{v^2}} d v \\ &=\frac{1}{2 \sqrt{\pi}} \int_0^{\infty} \frac{\left(1+\frac{1}{v^2}\right)-\left(1-\frac{1}{v^2}\right)}{v^2+\frac{1}{v^2}} d v \\ &=\frac{1}{2 \sqrt{\pi}}\left[\int_0^{\infty} \frac{d\left(v-\frac{1}{v}\right)}{\left(v-\frac{1}{v}\right)^2+2}- \underbrace{\int_0^{\infty} \frac{d\left(v+\frac{1}{v}\right)}{\left(v+\frac{1}{v}\right)^2-2}}_{=0}\right] \\ &=\frac{1}{2 \sqrt{2} \cdot \sqrt{\pi}}\left[\tan ^{-1}\left(\frac{v-\frac{1}{v}}{\sqrt{2}}\right)\right]_0^{\infty} \\ &=\frac{\pi}{2 \sqrt{2} \sqrt{\pi}} \\ &=\sqrt{\frac{\pi}{8}} \\ & \end{aligned} $$
