# Does the improper integral $\int_{0}^{\infty}\sin(x^2)\;\mathrm dx$ converge? [duplicate]

Does the improper integral $\int_{0}^{\infty}\sin(x^2)\;\mathrm dx$ converge?

So if it converges then $\lim_{b \to\infty}\int_{0}^{b}\sin(x^2)\;\mathrm dx$ exists and our integral converges to this limit. I have tried the substitution $y=x^2$ but this does not make the calculation easier.

Another way I know is to use the integral test but it also seems to be useless now.

## marked as duplicate by GEdgar, Michael Hoppe, marwalix, Daniel W. Farlow, Michael GrantApr 23 '15 at 22:13

It is a Fresnel integral. The limit exists since $$I(b)=\int_{0}^{b}\sin t^2\,dt = \frac{1}{2}\int_{0}^{b^2}\frac{\sin x}{\sqrt{x}}\,dx$$ converges by Dirichlet's test (integral version), because $\sin x$ is a function with a bounded primitive and $\frac{1}{\sqrt{x}}$ is a monotonic function converging to zero as $x\to +\infty$. To compute it, we may use the Laplace transform, giving: $$\mathcal{L}(\sin x)=\frac{1}{1+t^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)=\frac{1}{\sqrt{\pi t}}$$ from which: $$\int_{0}^{+\infty}\sin t^2\,dt = \frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{(1+t^2)\sqrt{t}}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+u^4}=\frac{1}{2}\sqrt{\frac{\pi}{2}}.$$
• @Jack D'Aurizio I'm trying to understand the step where you take $\frac{1}{2}\int_{0}^{\infty}\mathcal{L}(\sin x)\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)dt$. This is a new technique to me. Does it have a name so I can read about it somewhere? I understand $1=e^{-st}e^{st}$ but I don't understand why can you take two infinite integrals in the middle of the integrand. – user5389726598465 Jun 30 '17 at 2:34