How to convert $\int_{0}^{\infty} \sin (t^2) dt$ to a limit of a series? 
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Prove: $\int_{0}^{\infty} \sin (x^2) dx$ converges. 

I have shown that;
$\forall \epsilon>0, \exists r\in \mathbb{R}$  such that $ \forall x,y>0, r<x,y \Rightarrow |\int_{x}^{y} \sin (t^2) \, dt| < \epsilon$.
Also, i have shown that $\forall x,y>0, |\int_{x}^{y} \sin (t^2) \, dt| < 1/x$.
How do i prove that $\lim_{x\to\infty} \int_0^x \sin (t^2) \, dt$ converges?
EDIT:
Please do not close this post. I saw Davide's answer in the link, but don't understand his argument. Why does convergence of $\int_{0}^{\infty} t^{-3/2} dt$ imply that $\int_{a}^{\infty} t^{-3/2} \cos t dt$ converges?
I think that only implies that limsup and liminf of $\int_{a}^{\infty} t^{-3/2} \cos t dt$ is finite. And that's exactly what i said at the first of the sentence in my post.
Of course, i tried to convert this integral to a limit of a series, to apply 'alternating series test'. So i was trying to prove a lemma, but i failed to prove and it is indeed false. (Check this in the comment below) (To be specific, i was trying to show that $\int_{0}^{\infty} \sin t^2 dt = \sum_{n=0}^{\infty} \int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin t^2 dt$)
To summarize, what is a theorem that is a generalization of that in Michael's post for Riemann Stieltjes Integral?
 A: $\lim\limits_{x\to\infty}\int_0^x \sin(t^2)\,dt$ converges if $\sum\limits_{n=0}^\infty \int_n^{n+1} \sin(t^2)\,dt$ converges.  What you've already proved makes it possible to apply Cauchy's convergence test to that series.
Postscript per alex.jordan's comment below:
$\sin=0$ at integer multiples of of $\pi$, so $\sin\left((\sqrt{n\pi})^2\right)$ $=\sin\left(\left(\sqrt{(n+1)\pi}\right)^2\right)$, and $\displaystyle\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2)\,dt$ s an integral over a short interval of a function whose absolute value is bounded by $1$, so it's unproblematic.  So think about convergence of the following sum and about Cauchy's criterion:
$$
\sum_{n=0}^\infty \int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2) \, dt.
$$
A: We know that $x<y \Rightarrow |\int_{x}^{y} \sin t^2 dt|<1/x$.
Define $F(x) = \int_{0}^{x} \sin t^2 dt$.
Let $\{s_n\}_{n\in \mathbb{Z}^+}$ be a sequence such that $s_n=\sqrt{n\pi}$.
Then,
$F(s_n) \\ =\sum_{i=0}^{n-1} \int_{\sqrt{i\pi}}^{\sqrt{(i+1)\pi}} \sin t^2 dt \\ ≦\pi \sum_{i=0}_{n-1} (-1)^i (\sqrt{i+1} - \sqrt{i})$.
Thus, $A\triangleq \lim_{n\to\infty} F(s_n)$ is convergent.
Now, fix $\epsilon>0$.
Then, there exists $N\in \mathbb{N}$ such that $n≧N \Rightarrow |F(s_n) - A| <\epsilon$.
Let $\frac{1}{\epsilon} < s_n$ for some $n≧N$ and $y>s_n$
Then,$|F(y)-F(s_n)|< \frac{1}{s_n} < \epsilon$.
Hence, $|F(y) - A| < \epsilon$.
Thus, $\lim_{y\to\infty} F(y) = A$.
