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Define $$f(x)=\int_{x}^{x+1}\sin(t^2)dt$$
Find the upper and lower limits $xf(x)$, as $x\rightarrow \infty$.

I find the answer as $+1, -1$ since $|\sin(x)| \le 1$. (Of course I calculated that function)
Is that right or did I miss something?


I solved this way.
$2xf(x)=\cos(x^2)-\cos[(x+1)^2]+r(x)$ where $r(x)=\frac{\cos(x+1)^2}{x+1}-2x\int_{x^2}^{(x+1)^2}\frac{cos(u)}{4u^{3/2}}du$
Therefore $xf(x)=\frac{1}{2}{\cos(x^2)-\cos(x+1)^2}+\frac{r(x)}{2}$
Using trigonomeric formula: $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$
Rewrite $xf(x)=±\sin(x^2+x+\frac{1}{2})\sin(x+\frac{1}{2})+\frac{r(x)}{2}$.

As $x\rightarrow \infty, r(x) \rightarrow 0$.

Suppose $x^2=2k\pi$ for integer $k$.
To achieve $±1$, we have to show that $x+\frac{1}{2} \rightarrow 2n\pi+\frac{\pi}{2}$ for some $n$ as $x\rightarrow \infty$.
For each $n$, there exists $k$ such that $\sqrt{2\pi k}+\frac{1}{2} < 2n\pi+\frac{\pi}{2} <\sqrt{2\pi (k+1)} +\frac{1}{2} $
Distance between $\sqrt{2\pi k}+\frac{1}{2}$ and $2n\pi+\frac{\pi}{2}$ is at most $\sqrt{2\pi (k+1)} +\frac{1}{2} -\sqrt{2\pi k}+\frac{1}{2}$.
As $k \rightarrow \infty$ the distance becomes arbitrary small.
Therefore $ x \rightarrow \infty$, $xf(x)=±\sin(2n\pi+\frac{\pi}{2})=±1$.

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The remark that $|\sin|\leqslant1$ everywhere is certainly not enough to conclude, much less to conclude that the limsup and the liminf are $\pm1$. – Did Dec 15 '12 at 15:14
What do you mean by "Of course I calculated that function" – leo Dec 15 '12 at 15:24
By the mean value theorem (which you can use since $x$ and $x+1$ are asymptotic), you get that $f(x) = \sin(m^2)$ for some $m \in \[ x, x+1 \]$, what can you say about $f(x)$ as $x \to \pm \infty$? Then remember that you need to consider $x f(x)$ for your problem$. – Andy Dec 15 '12 at 15:30
Your solution is correct, except for some ambiguous expressions. I was also writing an answer, but iit exactly coincided with your solution, so I discarded it. – Sangchul Lee Dec 15 '12 at 15:43
In yor solution, how do you get the first inequality, notice that in $f$ the integral is respect to $t$, no $x$ – leo Dec 15 '12 at 15:47
up vote 4 down vote accepted

Here is a clarified version of your proof. (In fact, it was a part of my original solution, but I modified it by following your idea.)

Step 1. Estimation of $xf(x)$

Let $t = \sqrt{u}$. Then

\begin{align*} xf(x) &= x\int_{x}^{x+1} \sin (t^2) \, dt = x \int_{x^2}^{(x+1)^2} \frac{\sin u}{2\sqrt{u}} \, du \\ &= x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du \end{align*}


\begin{align*} x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} &= \frac{x}{2} \left(\frac{1-\cos\big((x+1)^2\big)}{x+1} - \frac{1 - \cos\big(x^2\big)}{x} \right) \\ &= \frac{1}{2} \left(\cos \left( x^2 \right) - \cos\left((x+1)^2\right) \right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} \\ &= \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)}, \end{align*}

we have

$$xf(x) = \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du. $$

It is easy to observe that

$$ - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du = O\left(\frac{1}{x}\right). $$

Indeed, it follows from the estimation

\begin{align*} \left| x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du \right| &\leq x \int_{x^2}^{(x+1)^2} \frac{\left| 1-\cos u \right| }{4u^{3/2}} \, du \\ &\leq x \int_{x^2}^{(x+1)^2} \frac{1}{2x^3} \, du \leq \frac{2x+1}{2x^2} = O\left(\frac{1}{x}\right). \end{align*}

Step 2. Evaluation of limit superior and limit inferior

The estimation above in particular implies that

$$ \limsup_{x\to\infty} xf(x) = \limsup_{x\to\infty} \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right)$$

and likewise for the liminf. To find the limsup, note from the identity above that

$$\limsup_{x\to\infty} xf(x) \leq 1. $$

We claim that it is indeed 1. To this end, it suffices to find a subsequence $(x_n)$ such that $x_n \to \infty$ and

$$ \sin \left(x_n +\frac{1}{2}\right)\sin\left(x_n^2+x_n+\frac{1}{2}\right) \to 1. $$

Let $x = \sqrt{2\pi k}$. Then we have

$$ \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) = \sin^2 \left(\sqrt{2\pi k}+\frac{1}{2}\right) $$

As $\sqrt{2\pi k} \to \infty$ and $\sqrt{2\pi(k+1)} - \sqrt{2\pi k} \to 0$ as $k \to \infty$, we can find a subsequence $(k_n)$ such that

$$ \min \left\{ \left| \sqrt{2\pi k_n} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} \to 0 \quad \text{as} \quad n \to \infty. \tag{1}$$

Thus for $x_n = \sqrt{2\pi k_n}$ we obtain $ x_n f(x_n) \to 1$ as $n \to \infty$ and therefore $\limsup_{x\to\infty} xf(x) = 1$. A slight modification of this argument also proves that $\liminf_{x\to\infty} xf(x) = -1$.

Proof of the claim $(1)$

Let $\epsilon > 0$. Then there exists $N$ such that whenever $k \geq N$ we have $0 < r_k < \epsilon$, where $r_k$ denotes $r_k = \sqrt{2\pi(k+1)} - \sqrt{2\pi k}$ for simplicity. Then for the sequence of open balls

$$ B_k = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi k} + r_k\right) = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi (k+1)}\right), $$

we have $B_k \cap B_{k+1} \neq \varnothing $ and hence

$$ \bigcup_{k=N}^{\infty} B_k = \left(\sqrt{2\pi N} - r_{N}, \infty \right). $$

Here, we used the fact that $\sqrt{2\pi k} \to \infty$ as $k \to \infty$. Then for any sufficiently large integer $m$ we have

$$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in \left(\sqrt{2\pi N} - r_{N}, \infty \right). $$

Thus we can pick some $k \geq N$ satisfing

$$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in B_k \subset \left(\sqrt{2\pi N} - \epsilon, \sqrt{2\pi N} + \epsilon \right). $$

This proves the following proposition: For any $\epsilon > 0$ there exists a positive integer $k$ such that

$$ \min \left\{ \left| \sqrt{2\pi k} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} < \epsilon.$$

Then the claim $(1)$ immediately follows.

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hi. Are you around? – I'm an artist Dec 18 '12 at 8:23
@Chris'ssister, Sorry, I saw your comment just now. It is a sad thing that the time zone differs by almost half a day. – Sangchul Lee Dec 18 '12 at 15:04
don't worry for that. I was learning things from your Best of Today's Calculation paper and wanted to ask you 2 things about the first problem. – I'm an artist Dec 18 '12 at 16:44
@Chris'ssister, Are you here now? – Sangchul Lee Dec 19 '12 at 2:02

$$ \begin{align} f(x) &=\int_x^{x+1}\sin(t^2)\,\mathrm{d}t\\ &=-\int_x^{x+1}\frac1{2t}\,\mathrm{d}\cos(t^2)\\ &=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)} -\int_x^{x+1}\cos(t^2)\frac1{2t^2}\,\mathrm{d}t\\ &=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}+O\left(\frac1{x^2}\right)\tag{1} \end{align} $$ So $$ \begin{align} xf(x) &=\frac{\cos(x^2)}{2}-\frac{\cos((x+1)^2)}{2(1+1/x)}+O\left(\frac1{x}\right)\\ &=\frac12\left(\cos(x^2)-\cos((x+1)^2)\right)+O\left(\frac1{x}\right)\tag{2} \end{align} $$ Therefore, mapping $x\mapsto\sqrt x$ on the right, we get $$ \limsup_{x\to\infty}\,xf(x) =\limsup_{x\to\infty}\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\tag{3 sup} $$ and $$ \liminf_{x\to\infty}\,xf(x) =\liminf_{x\to\infty}\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\tag{3 inf} $$ When $x$ is big, $\sqrt x$ varies much slower than $x$: $$ \begin{align} \sqrt{x+2\pi}-\sqrt x &=\frac{2\pi}{\sqrt{x+2\pi}+\sqrt x}\\ &\le\frac\pi{\sqrt x}\tag{4} \end{align} $$ Thus, for any $\epsilon>0$, choose an $x_0>\dfrac{\pi^2}{\epsilon^2}$ so that $2\sqrt{x_0}+1$ is an odd multiple of $\pi$. Then $(4)$ guarantees that for $x\in[x_0,x_0+2\pi]$, $2\sqrt x+1$ is within $2\epsilon$ of an odd multiple of $\pi$.

Choose the $x\in[x_0,x_0+2\pi]$ which is $0\bmod{2\pi}$ and we get that $$ \tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\ge1-\epsilon\tag{5 sup} $$ Choose the $x\in[x_0,x_0+2\pi]$ which is $\pi\bmod{2\pi}$ and we get that $$ \tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\le-1+\epsilon\tag{5 inf} $$ These last two inequalities show that $$ \limsup_{x\to\infty}\,xf(x)=1\tag{6 sup} $$ and $$ \liminf_{x\to\infty}\,xf(x)=-1\tag{6 inf} $$

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This does not prove that the limsup and liminf are equal to 1 and -1, respectively. – Sangchul Lee Dec 15 '12 at 16:03
@sos440: Ah, I missed that we were looking at $xf(x)$. I will add more terms, then... – robjohn Dec 15 '12 at 16:43
@sos440: That should be better. Thanks for pointing out my misunderstanding. – robjohn Dec 15 '12 at 18:02
would the downvoter care to explain what remains to be a problem? – robjohn Dec 19 '12 at 16:32

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