# Prove that $\int_0^{\infty}\sin(x^2)dx$ converges. [duplicate]

Prove that $\int_0^{\infty}\sin(x^2)dx$ converges by making the change of variable $u=x^2$, and applying integration by parts to the resulting integral.

Prove the integral does not converge absolutely.(Use the $u$ form.)

$\int_0^{\infty}\sin(x^2)dx = \int_0^\infty \sin u \frac{1}{2\sqrt u}du = \sin u \sqrt u|_0^\infty - \int_0^\infty \cos u \sqrt udu$. Then I don't know how to proceed.

## marked as duplicate by Jack D'Aurizio real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 14 '18 at 16:28

\begin{align*} \int_{1}^{\infty}\dfrac{\sin u}{2\sqrt{u}}du&=\dfrac{-\cos u}{2\sqrt{u}}\bigg|_{u=1}^{u=\infty}-\dfrac{1}{4}\int_{1}^{\infty}\dfrac{\cos u}{u^{3/2}}du\\ &=\dfrac{\cos 1}{2}-\dfrac{1}{4}\int_{1}^{\infty}\dfrac{\cos u}{u^{3/2}}du, \end{align*} where \begin{align*} \int_{1}^{\infty}\left|\dfrac{\cos u}{u^{3/2}}\right|du\leq\int_{1}^{\infty}\dfrac{1}{u^{3/2}}du<\infty. \end{align*} Note that \begin{align*} \int_{0}^{1}\sin(x^{2})dx \end{align*} exists by the continuity of $x\rightarrow\sin(x^{2})$ on $[0,1]$.