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.
Best Answer
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 $a_n\rightarrow a$.