Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
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?


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?

share|cite|improve this question

marked as duplicate by DonAntonio, Henry T. Horton, tomasz, Brandon Carter, Micah Dec 9 '12 at 5:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The easiest way is just to use the alternating series test (some details need to be filled in). – Potato Dec 9 '12 at 2:38
@Potato I was trying to prove that "For any strictly increasing sequence $\{c_n\}$ in $\mathbb{R}$ such that $c_1=a$, $\int_{a}^{\infty} f d\alpha$ converges iff $\sum_{n=1}^{\infty} \int_{c_n}^{c_{n+1}} f d\alpha$ converges and their limits are the same", but i thought this is false. I have no idea how to change this limit to limit of series. – Katlus Dec 9 '12 at 2:58
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 – alex.jordan Dec 9 '12 at 5:22

2 Answers 2

$\lim\limits_{x\to\infty}\int_0^x \sin(t^2)\,dt$ 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: $$ \sum_{n=0}^\infty \int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2) \, dt. $$

share|cite|improve this answer
I was trying to prove the if statement before i post this question, and i thought it is false. What is that theorem called? – Katlus Dec 9 '12 at 2:54
It's not true that convergence of $\sum_{n=0}^{\infty}\int_n^{n+1}f(t)\,dt$ implies convergence of $\int_0^{\infty}f(t)\,dt$. $\int_0^{x}f(t)\,dt$ could oscillate as a function of $x$ while behaving better at the integral values. So there'd need to be more specific information about this function used. – alex.jordan Dec 9 '12 at 5:03
@alex.jordan would you please check if my argument is correct? – Katlus Dec 9 '12 at 5:09
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. – alex.jordan Dec 9 '12 at 5:16

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, $\lim_{y\to\infty} F(y) = A$.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.