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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.

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Hongyan
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1 Answers1

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\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 $[0,1]$.

It is not absolutely convergent since \begin{align*} \int_{1}^{\infty}\dfrac{|\sin u|}{\sqrt{u}}du&\geq\sum_{n=1}^{\infty}\int_{n\pi+\pi/6}^{(n+1)\pi-\pi/6}\dfrac{|\sin u|}{\sqrt{u}}du\\ &\geq\sum_{n=1}^{\infty}\int_{n\pi+\pi/6}^{(n+1)\pi-\pi/6}\dfrac{1}{2\sqrt{u}}du\\ &\geq\sum_{n=1}^{\infty}\int_{n\pi+\pi/6}^{(n+1)\pi-\pi/6}\dfrac{1}{2\sqrt{(n+1)\pi-\pi/6}}du\\ &\geq\sum_{n=1}^{\infty}\dfrac{2\pi}{3}\dfrac{1}{2\sqrt{(n+1)\pi-\pi/6}}\\ &=\infty. \end{align*}

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