# Showing that $\int_0^\infty \frac{\sin x^2}{\sqrt{x}}dx$ converges

I want to show that this integral converges:

$$I := \int_0^\infty \frac{\sin x^2}{\sqrt{x}}dx$$

My attempt is to say that $$I < \int_0^\infty {\sin x^2}dx := J$$, and this one converges ( Prove: $\int_0^\infty \sin (x^2) \, dx$ converges. the first answer is awesome).

Do you have any other idea to show that $I$ converges? I tried to use substitution and int. by part but in vain :/. I was lucky to find that $J$ converges, but I would never have been able to guess this result alone.

• No, that doesn't work at all. If $J$ converged absolutely this would be fine. But it's clear that $\int|\sin(x^2)|=\infty$. You can't say $\sin(x^2)/{\sqrt t}<\sin(x^2)$ because $\sin(x^2)$ is not positive. – David C. Ullrich Jun 17 '18 at 14:03
• ohhh yes you're totally right... – Marine Galantin Jun 17 '18 at 14:04

Seems to me integration by parts does work. You can't write down a formula for the antiderivative of $\sin(x^2)$, but that doesn't mean it doesn't have an antiderivative. Define $$F(x)=\int_0^x\sin(t^2)\,dt.$$Now $$\int\frac{\sin(x^2)}{\sqrt x}\,dx=\frac{F(x)}{\sqrt x}+\frac12\int\frac{F(x)}{x^{3/2}}.$$

And whether you realize it or not, you already know that $\lim_{x\to\infty} F(x)$ exists, hence $F$ is bounded...

I would not consider my method an elegant one.

By the substitution $\sqrt u=x$, we get $$\frac12\int^\infty_0 u^{-3/4}\sin u du=\frac12\text{Im}\int^\infty_0 u^{-3/4}e^{iu}du$$

Substitute $t=-iu$, we get $$-i^{1/4}\int^{-i\infty}_0 t^{-3/4}e^{-t}dt$$

Let $I(a,b)=\int^b_a t^{-3/4}e^{-t}dt$.

By Cauchy’s theorem, $$I(i\epsilon,-i\infty)+\text{large arc}_\infty+I(\infty,\epsilon)+\text{small arc}_{\epsilon}=0$$($\epsilon\to0^+$)

The arc integrals vanish. Thus, our integral equals $\Gamma (\frac14)$(because $I(\infty,\epsilon)=-\Gamma(\frac14)$) which converges. Therefore, the original integral converges.

Please notify me if you spot any error.

Convergence is ensured by the substitution $x\mapsto \sqrt{x}$ and integration by parts / Kronecker's lemma.

By the Laplace transform, for any $\alpha\in(0,1)$ we have $$\int_{0}^{+\infty}\frac{\sin x}{x^\alpha}\,dx = \frac{1}{\Gamma(a)}\int_{0}^{+\infty}\frac{s^{\alpha-1}}{s^2+1}\,ds=\Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)\tag{1}$$ where the last equality follows from Euler's Beta function and the reflection formula for the $\Gamma$ function.
Your integral is just the instance $\alpha=\frac{3}{4}$: $$\int_{0}^{+\infty}\frac{\sin x^2}{\sqrt{x}}\,dx = \Gamma\left(\tfrac{5}{4}\right)\sqrt{2-\sqrt{2}} .\tag{2}$$