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The question

Okay. So I'm trying to solve the problem below for a previous exam in real analysis. Thus, only such methods may be used.

The integral $\int_0^\infty\sin x^2dx$ is called a Fresnel integral and it arises in wave optics. Show that this integral converges, by proving that the sequence $a_n:=\int_0^n\sin x^2dx$ converges in $\mathbb{R}$.

The question also comes with the below hint.

Hint: Use the fact that $$ \sin x^2=-\frac{1}{2x}\frac{d}{dx}\cos x^2. $$

It makes me think of using integration by parts, and that has been the hint in similar questions. However, when I do that things don't get easier. Consequently, I'm kind of stuck.

Here are my computations

$$ \int_0^n\sin x^2dx=-\int_0^n\frac{1}{2x}\frac{d}{dx}\cos x^2dx=[-\frac{1}{2x}\cos x^2]_0^n-\int_0^n\frac{1}{2x^2}\cos x^2dx. $$

Related questions

There is a thread about evaluating the Fresnel integral called "Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?" and another one called "Trig Fresnel Integral", but none of the answers to these questions involve showing convergence as instructed in this question, and both involve the Gamma function, which wasn't included in my course on real analysis.

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    $\begingroup$ Integration by parts is the way to go. $\endgroup$ – Lord Shark the Unknown Aug 11 '18 at 17:32
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    $\begingroup$ The integrand on the RHS is bounded in magnitude by $1/2x^2$. $\endgroup$ – Bungo Aug 11 '18 at 17:40
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    $\begingroup$ The last integral fails to converge near $x=0$ @Bungo $\endgroup$ – Lord Shark the Unknown Aug 11 '18 at 17:42
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    $\begingroup$ @LordSharktheUnknown Ah right, the integration starts at $0$, not $1$. $\endgroup$ – Bungo Aug 11 '18 at 17:42
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    $\begingroup$ @OskarTegby Of course, you could use your approach for the interval $[1, \infty)$, and then obviously $\int_0^1 \sin x^2$ converges since the integrand is bounded and continuous on the compact interval $[0,1]$. $\endgroup$ – Bungo Aug 11 '18 at 17:48
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Without loss of generality, let $n>m$. Then, $$ \left| a_n-a_m \right|=\left|\int_{m}^{n}\sin{x^2}dx \right|= \left| \int^{n}_{m} {-\frac{1}{2x}\frac{d}{dx}\cos{x^2}}dx\right|=\left|\left[-\frac{1}{2x}\cos{x^2} \right]_{m}^{n}+\int_{m}^{n}-\frac{1}{2x^2}\cos{x^2}dx\right| $$

$$\leq \left|\frac{1}{2m}\cos{m^2}-\frac{1}{2n}\cos{n^2} \right|+ \int_{m}^{n} \left| \frac{1}{2x^2} \right| dx\leq \frac{1}{2m}+\frac{1}{2n}-\frac{1}{2n}+\frac{1}{2m}=\frac{1}{m}\rightarrow 0 $$ as $n,m\rightarrow \infty$.

Thus, $a_n$ is Cauchy in $\mathbb{R}$ so there is $a\in \mathbb{R}$ such that $a_n\rightarrow a$.

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$$a_n=\frac12\int_0^{n^2}\sin y\frac{dy}{\sqrt y} =\frac12\left[\frac{1-\cos y}{\sqrt y}\right]_0^{n^2} +\frac14\int_0^n\frac{1-\cos y}{y^{3/2}}dy \to\frac14\int_0^\infty\frac{1-\cos y}{y^{3/2}}dy.$$ This last integral is absolutely convergent.

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    $\begingroup$ In particular, for large $y$ the integrand $\sim y^{-3/2}$ has a $\sim y^{-1/2}$ antiderivative that vanishes at $\infty$, while for small $y$ the integrand $\sim y^{1/2}$ has a $\sim y^{3/2}$ antiderivative that vanishes at $0$. $\endgroup$ – J.G. Aug 11 '18 at 17:51

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