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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.

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?

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    Incidentally, do you know that not only is this integral convergent, but that its exact value is $\sqrt{\pi/8}$? You can relate this to the Gaussian integral of $e^{-x^2}$ through the relation $e^{ix}=\cos(x)+i\sin(x)$. See [http://en.wikipedia.org/wiki/Fresnel_integral](http://en.wikipedia.org/wiki/Fresnel_integral).2012-12-09

2 Answers 2

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$\lim\limits_{x\to\infty}\int_0^x \sin(t^2)\,dt$ 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. $

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    If you cut up the integral over intervals $[\sqrt{n\pi},\sqrt{(n+1)\pi}]$, then at least every subintegral is over an entirely positive or entirely negative region, and these alternate. So if you can show that they decrease in absolute value and converge to zero, then firstly, you have met the conditions of the alternating series test. Secondly, you would have bounded $\int_0^x$ between $\int_0^{\sqrt{n\pi}}$ and $\int_0^{\sqrt{(n+1)\pi}}$ for all $x$ in $[\sqrt{n\pi},\sqrt{(n+1)\pi}]$, and the concern in my previous comment is alleviated.2012-12-09
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We know that $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, $\lim_{y\to\infty} F(y) = A$.