# Find the upper and lower limits of $xf(x)$, as $x\rightarrow \infty$

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?

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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$$, what can you say about $f(x)$ as $x \to \pm \infty$? Then remember that you need to consider $x f(x)$ for your problem$. – Andy Dec 15 '12 at 15:30 Your solution is correct, except for some ambiguous expressions. I was also writing an answer, but iit exactly coincided with your solution, so I discarded it. – sos440 Dec 15 '12 at 15:43 In yor solution, how do you get the first inequality, notice that in$f$the integral is respect to$t$, no$x$– leo Dec 15 '12 at 15:47 show 1 more comment ## 2 Answers Here is a clarified version of your proof. (In fact, it was a part of my original solution, but I modified it by following your idea.) Step 1. Estimation of$xf(x)$Let$t = \sqrt{u}. Then \begin{align*} xf(x) &= x\int_{x}^{x+1} \sin (t^2) \, dt = x \int_{x^2}^{(x+1)^2} \frac{\sin u}{2\sqrt{u}} \, du \\ &= x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du \end{align*} Since \begin{align*} x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} &= \frac{x}{2} \left(\frac{1-\cos\big((x+1)^2\big)}{x+1} - \frac{1 - \cos\big(x^2\big)}{x} \right) \\ &= \frac{1}{2} \left(\cos \left( x^2 \right) - \cos\left((x+1)^2\right) \right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} \\ &= \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)}, \end{align*} we have $$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.$$ Then the claim$(1)immediately follows. - hi. Are you around? – Chris's sis Dec 18 '12 at 8:23 @Chris'ssister, Sorry, I saw your comment just now. It is a sad thing that the time zone differs by almost half a day. – sos440 Dec 18 '12 at 15:04 don't worry for that. I was learning things from your Best of Today's Calculation paper and wanted to ask you 2 things about the first problem. – Chris's sis Dec 18 '12 at 16:44 @Chris'ssister, Are you here now? – sos440 Dec 19 '12 at 2:02 add comment \begin{align} f(x) &=\int_x^{x+1}\sin(t^2)\,\mathrm{d}t\\ &=-\int_x^{x+1}\frac1{2t}\,\mathrm{d}\cos(t^2)\\ &=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)} -\int_x^{x+1}\cos(t^2)\frac1{2t^2}\,\mathrm{d}t\\ &=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}+O\left(\frac1{x^2}\right)\tag{1} \end{align} So \begin{align} xf(x) &=\frac{\cos(x^2)}{2}-\frac{\cos((x+1)^2)}{2(1+1/x)}+O\left(\frac1{x}\right)\\ &=\frac12\left(\cos(x^2)-\cos((x+1)^2)\right)+O\left(\frac1{x}\right)\tag{2} \end{align} Therefore, mappingx\mapsto\sqrt x$on the right, we get $$\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}$$ - This does not prove that the limsup and liminf are equal to 1 and -1, respectively. – sos440 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