# $\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{cos(x/n)-1}{x^4}dx$

Show that the following limit exists and compute it: $$\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx$$

Attempt: By using the integration by parts, I get the following result:

\begin{align*}\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx&=\lim_{k\to\infty}\int_{1}^{k}\frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\\\\ &=\lim_{k\to\infty}\left\{\left(\frac{k}{n}\right)^4\sin\left(\frac{k}{n}\right)+4\left(\frac{k}{n}\right)^3\cos\left(\frac{k}{n}\right)\right.\\ &\quad\quad\quad\quad-12\left(\frac{k}{n}\right)^2\sin\left(\frac{k}{n}\right)-24\left(\frac{k}{n}\right)\cos\left(\frac{k}{n}\right)\\ &\quad\quad\quad\quad+24\sin\left(\frac{k}{n}\right)+\frac{n^3}{3k^3}\\ &\quad\quad\quad\quad\left.-\left[\frac{1}{n^4}\sin\left(\frac{1}{n}\right)+\frac{4}{n^3}\cos\left(\frac{1}{n}\right)\right.\right.\\ &\quad\quad\quad\quad\left.\left.-12\left(\frac{1}{n}\right)^2\sin\left(\frac{1}{n}\right)-\frac{24}{n}\cos\left(\frac{1}{n}\right)\right.\right.\\ &\quad\quad\quad\quad\left.\left.+24\sin\left(\frac{1}{n}\right)+\frac{n^3}{3}\right]\right\}\end{align*}

But I could not compute this limit. And how can we show the existence of the limit $\displaystyle\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx$? Thanks!

• Why exactly can't you compute the limit as $k\to\infty$? Have you tried distributing the limit over the sum and evaluating them individually? Commented Jan 17, 2015 at 11:52
• Have you studied about theorems which allow you to change limits with integrations? Commented Jan 17, 2015 at 11:59
• @DavidH $lim_{k\to\infty}(k/n)^4sin(k/n)-lim_{k\to\infty}-12(k/n)^2sin(k/n)=\infty -\infty$. Here I have confusion. Commented Jan 17, 2015 at 12:08
• @MhenniBenghorbal I understand what you mean. Thank you Commented Jan 17, 2015 at 12:09
• @ErginSuer: You are welcome. Commented Jan 17, 2015 at 12:16

since $n^2(1-\cos(x/n)) = 2n^2\sin^2 \frac{x}{2n} \to \dfrac{x^2}{2}$ as $n \to \infty.$

so can i conclude that $$\lim_{n \to \infty}n^2\int_1^\infty \dfrac{\cos( x/n) - 1}{x^4} \ dx= -\dfrac{1}{2} \int_1^\infty \dfrac{1}{x^2} \ dx = -\dfrac{1}{2}$$

• how to justify that you can switch the order of limit and integration? Commented Jan 17, 2015 at 11:56
• @user139981, i bet there will be some dominated convergence theorem that applies in this case, but i am rusty on that.
– abel
Commented Jan 17, 2015 at 11:57
• besides, I think you missed a minus sign Commented Jan 17, 2015 at 12:01
• This "proof" is terribly incomplete. Anyone who knows what a Taylor series looks like can guess the result. But the proof is far from trivial. For example, how does one justify the the statement in the first line as $x \to \infty$, e.g., how does one find some region of interest in which the limit is true, and outside of which the integral may be neglected? Commented Jan 17, 2015 at 12:59
• @RonGordon: actually, I think a mention of Dominated Convergence would finish the proof.
– robjohn
Commented Jan 17, 2015 at 23:42

In order to avoid the use of dominated convergence theorem, we can notice that:

$$I=\int_{1}^{+\infty}\frac{1-\cos(x/n)}{x^4}\,dx = 2\int_{1}^{+\infty}\left(\frac{\sin\frac{x}{2n}}{x^2}\right)^2\,dx=2\int_{0}^{1}y^2\sin^2\frac{1}{2ny}\,dy$$ satisfies $I\leq\frac{1}{2n^2}$, since $|\sin x\,|\leq x$, while: $$I = \frac{2}{n^3}\int_{0}^{n}\left(y\sin\frac{1}{2y}\right)^2\,dy\geq\frac{2}{n^3}\int_{1}^{n}\left(y\sin\frac{1}{2y}\right)^2\,dy\geq\frac{2}{n^3(n-1)}\left(\int_{1}^{n}y\sin\frac{1}{2y}\,dy\right)^2,$$ due to Cauchy-Schwarz inequality, and: $$\int_{1}^{n}y\sin\frac{1}{2y}\,dy \geq \int_{1}^{n}\left(\frac{1}{2}-\frac{1}{48x^2}\right)\,dx \geq \frac{n}{2}-\frac{25}{48}$$ due to the fact that $\sin z \geq z-\frac{z^3}{6}$ for $z\in (0,1]$. The two inequalities together give: $$\frac{(n-25/24)^2}{n(n-1)}\leq 2n^2 I \leq 1,$$ so the wanted limit is $-\frac{1}{2}$ by squeezing.

• Last equality in first line: substitute $y=\frac1x$, $\mathrm{d}x=-\frac{\mathrm{d}y}{y^2}$, minus goes away by reversing limits, $\frac{1}{y^2}$ simplifies with $y^4$ from the $\frac{1}{x^4}$ that was there originally, giving us $y^2$, and the sine is easy to get. Cauchy-Schwartz inequality, because $L^2$, the space of square-integrable functions, is a Hilbert space, with scalar product $\langle f,g\rangle=\int fg\mathrm{d}x$, so for $f=1$ and $g=(y\sin\frac1y)^2$ we get $(n-1)\int g^2=\|f\|^2\|g\|^2\geq\langle fg\rangle^2=(\int y\sin y)^2$, divide by $n-1$ on both sides. Commented Jan 17, 2015 at 14:17
• ALso, last inequality is removing a $\frac{1}{6n^2}$, hence its validity. Just making these passages more explicit. But the equality chain has both the leftmost and rightmost terms tending to 1, which implies $\frac{n^2I}{2}\to1$, so how can $n^2I\to-\frac12$? Maybe I'm missing something, but the limit seems to be 2… Commented Jan 17, 2015 at 14:18
• @MickG: fixed the last part. $\frac{n^2 I}{2}\to 1$, right, hence the given limit equals $-2$. Commented Jan 17, 2015 at 14:21
• Also notice that the inequality on the third line holds, since the integral equals $n-\frac{7}{6}+\frac{1}{6n}\geq n-\frac{7}{6}$. Commented Jan 17, 2015 at 14:23
• But $n^2I$ is the given limit… ah no whoops, change of signs right at the start, so the given limit is $-n^2I$ :). That figures :). Yep that's what I meant by "removing a $\frac{1}{6n^2}$, only I had an exponent too much :). Mental calculations often lead to sign errors :). Commented Jan 17, 2015 at 14:24

$\newcommand{\abs}[1]{\left|#1\right|} \newcommand{\pa}[1]{\left(#1\right)} \newcommand{\sq}[1]{\left[#1\right]} \newcommand{\br}[1]{\left\{#1\right\}}$ I immediately thought about the dominated convergence theorem. If we move $n^2$ inside the integral, which we can, we get: $$\int_1^\infty\dfrac{\cos(\frac{x}{n})-1}{x^4}n^2\mathrm{d}x.$$ For $n\to\infty$, the numerator is asymptotic to $-\frac{x^2}{2n^2}$, which means the whole is asymptotic to $-\frac{n^2x^2}{2n^2x^4}=-\frac{1}{2x^2}$, whose integral is $-\frac12$. Let's try to find a domination then. The first thing that came to my mind was: $$\left|\frac{\cos(\frac{x}{n})-1}{x^4}\right|\leq\frac{|\cos(\frac{x}{n})|+1}{|x|^4}\leq\frac{2}{x^4}.$$ This is valid for any $n$, and this function is integrable over $[1,\infty)$ with respect to the Lebesgue measure. Of course, the question seems to be about a Riemann integral, and I'm talking about the Lebesgue integral. But then, since both exist (the domination grants the existence of the Lebesgue one, and if the Riemann one doesn't exist, well, the question is useless :)), they are equal, so if I prove things for the LI they are true for the RI as well. Well right, we have the domination, so we swap limit and integral, and we are left with the limit being: $$-\int_1^\infty\frac{1}{2x^2}\mathrm{d}x=-\frac12.$$ Just out of curiosity, I will try evaluating the limit you couldn't compute: \begin{align*} \int_{1}^{\infty} \frac{\cos\pa{\frac{x}{n}}-1}{x^4}dx={}&\lim_{k\to\infty}\int_{1}^{k}\frac{\cos\pa{\frac{x}{n}}-1}{x^4}={} \\ {}={}&\lim_{k\to\infty}\left\{\pa{\frac{k}{n}}^4\sin\pa{\frac{k}{n}}+4\pa{\frac{k}{n}}^3\cos\pa{\frac{k}{n}}-{}\right. \\ &{}\left.{}+12\pa{\frac{k}{n}}^2\sin\pa{\frac{k}{n}}-24\pa{\frac{k}{n}}\cos\pa{\frac{k}{n}}+{}\right. \\ &{}\left.{}+24\sin\pa{\frac{k}{n}}+\pa{\frac{n^3}{3k^3}}-\left[\pa{\frac{1}{n^4}}\sin\pa{\frac1n}+\pa{\frac{4}{n^3}}\cos\pa{\frac1n}-{}\right.\right. \\ &{}\left.\left.{}+12\pa{\frac1n}^2\sin\pa{\frac1n}-\pa{\frac{24}{n}}\cos\pa{\frac1n}+24\sin\pa{\frac1n}+\frac{n^3}{3}\right]\right\}. \end{align*} First of all, there are a bunch of terms independent of $k$. I suggest we forget about those. Also, there is $\frac{n^3}{k^3}$ (or a constant times that), which obviously tends to 0 for $k\to\infty$. We take that away. Finally, the other terms all constantly have $\frac{k}{n}$, so I suggest we substitute it for $z=\frac{k}{n}$ to make the expression more "pretty". For $k\to\infty,z\to\infty$. Side note: do use operators (\sin, \cos) for sine and cosine; don't use the plain word in math mode; do use fractions; don't use the slash; do therefore lengthen the delimiters via \left…\right. So the limit becomes: $$\lim_{z\to\infty}\br{z^4\sin z+4z^3\cos z-12z^2\sin z-24z\cos z+24\sin z}.$$ Please check your calculations, and mine, because one of us must have made a mistake, since the dominant term here is clearly the one for $z^4$ which has no limit. Yet the limit must exist, otherwise we would be trying to compute the limit of something that doesn't exist.

Update: Following the reading of Jack's answer, I noticed I found a domiation for the integrand without $n^2$, and then applied the theorem to the integrals with $n^2$. Let me try another way. I reuse the first line of his answer by observing that: $$n^2\frac{\cos(\frac{x}{n})-1}{x^4}=-2n^2\frac{\sin^2(\frac{x}{2n})}{x^4}=-2\pa{\frac{n\sin(\frac{x}{2n})}{x^2}}^2.$$ FOr a moment, we forget about the 2 and the square, and take the modulus instead. Let's try dominating that: $$\left|\frac{n\sin\frac{x}{2n}}{x^2}\right|\leq\frac{n\frac{x}{2n}}{x^2}=\frac{1}{2x}.$$ That is not an integrable domination. However, now we remember we are integrating the square, so we apply the domination there, where it becomes $\frac{1}{4x^2}$ which is indeed integrable. So by dominated convergence theorem, the limit passes under integral, yielding: \begin{align*} \lim_{n\to\infty}\int_1^\infty n^2\frac{\cos(\frac{x}{n})-1}{x^4}\mathrm{d}x={}&-\lim_{n\to\infty}\int_1^\infty2\frac{n^2\sin^2(\frac{x}{2n})}{x^4}\mathrm{d}x=-\int_1^\infty\lim_{n\to\infty}2\frac{n^2\sin^2(\frac{x}{2n})}{x^4}\mathrm{d}x={} \\ {}={}&-2\cdot\hspace{-4pt}\int_1^\infty\frac{1}{4x^2}\mathrm{d}x=-\frac12, \end{align*} getting to the same result as before. I'm afraid Jack forgot to halve the argument of the sine when doing his equalities. I will check and notify him if so.

\begin{align} n^2\int_1^\infty\frac{\cos(x/n)-1}{x^4}\,\mathrm{d}x &\stackrel{\hphantom{n\to\infty}}=\frac1n\int_{1/n}^\infty\frac{\cos(x)-1}{x^4}\,\mathrm{d}x\tag{1}\\ &\stackrel{\hphantom{n\to\infty}}=\frac1n\left[\int_{1/n}^1\frac{\cos(x)-1}{x^4}\,\mathrm{d}x +\int_1^\infty\frac{\cos(x)-1}{x^4}\,\mathrm{d}x\right]\tag{2}\\ &\stackrel{\hphantom{n\to\infty}}=\frac1n\left[\int_{1/n}^1\left(\frac{-1}{2x^2}+O(1)\right)\,\mathrm{d}x+O(1)\right]\tag{3}\\ &\stackrel{\hphantom{n\to\infty}}=\frac1n\left[\int_{1/n}^\infty\frac{-1}{2x^2}\,\mathrm{d}x+O(1)\right]\tag{4}\\ &\stackrel{\hphantom{n\to\infty}}=\frac1n\left[-\frac n2+O(1)\right]\tag{5}\\[3pt] &\stackrel{n\to\infty}\to-\frac12\tag{6} \end{align} Explanation:
$(1)$: substitute $x\mapsto nx$
$(2)$: break up the interval of integration
$(3)$: $\cos(x)=1-\frac{x^2}2+O\left(x^4\right)$ and $\int_1^\infty\frac{\cos(x)-1}{x^4}\,\mathrm{d}x$ is a constant
$(4)$: $\int_1^\infty\frac{-1}{2x^2}\,\mathrm{d}x$ is a constant and $\int_{1/n}^1O(1)\,\mathrm{d}x$ is $O(1)$
$(5)$: evaluate the integral
$(6)$: evaluate the limit

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \lim_{n\ \to\ \infty}\, \bracks{n^{2}\int_{1}^{\infty}{\cos\pars{x/n} - 1 \over x^{4}}\,\dd x}} =\lim_{n\ \to\ \infty}\, \bracks{{1 \over n}\int_{1/n}^{\infty}{\cos\pars{x} - 1 \over x^{4}}\,\dd x} \end{align} With Stolz-Cesaro Theorem: \begin{align}&\color{#66f}{\large% \lim_{n\ \to\ \infty}\, \bracks{n^{2}\int_{1}^{\infty}{\cos\pars{x/n} - 1 \over x^{4}}\,\dd x}} \\[5mm]&=\lim_{n\ \to\ \infty}\, \braces{{1 \over \pars{n + 1} - n}\bracks{% \int_{1/\pars{n + 1}}^{\infty}{\cos\pars{x} - 1 \over x^{4}}\,\dd x -\int_{1/n}^{\infty}{\cos\pars{x} - 1 \over x^{4}}\,\dd x}} \\[5mm]&=-\lim_{n\ \to\ \infty}\, \int_{1/\pars{n + 1}}^{1/n}{2\sin^{2}\pars{x/2} \over x^{4}}\,\dd x =-\lim_{n\ \to\ \infty}\, \int_{1/\pars{n + 1}}^{1/n} {\sin^{2}\pars{x/2} \over \pars{x/2}^{2}}{1 \over 2x^{2}}\,\dd x \\[5mm]&=\lim_{n\ \to\ \infty}\,\braces{% -\int_{1/\pars{n + 1}}^{1/n}\ {1 \over 2x^{2}}\,\dd x + \int_{1/\pars{n + 1}}^{1/n}\ \bracks{1 - {\sin^{2}\pars{x/2} \over \pars{x/2}^{2}}}{1 \over 2x^{2}}\,\dd x} \\[5mm]&=-\,\half+\ \underbrace{\lim_{n\ \to\ \infty}\, \int_{1/\pars{n + 1}}^{1/n}\ \bracks{1 - {\sin^{2}\pars{x/2} \over \pars{x/2}^{2}}}{1 \over 2x^{2}}\,\dd x} _{\ds{=\dsc{0}}\,,\ \pars{~\mbox{see below}~}} \end{align} By the Mean Value Theorem, $\ds{\exists\ \xi\ \mid\ 0 < \verts{\xi} < \verts{x \over 2} \mid\ 1 - {\sin^{2}\pars{x/2} \over \pars{x/2}^{2}} =1 - \cos^{2}\pars{\xi}=\sin^{2}\pars{\xi}<\xi^{2}<{1 \over 4}\,x^{2}}$ which leads to \begin{align} 0 &< \verts{\int_{1/\pars{n + 1}}^{1/n}\ \bracks{1 - {\sin^{2}\pars{x/2} \over \pars{x/2}^{2}}}{1 \over 2x^{2}}\,\dd x} <{1 \over 8}\int_{1/\pars{n + 1}}^{1/n}\,\dd x \\[5mm]&={1 \over 8n\pars{n + 1}} \to 0\quad\mbox{when}\quad n \to \infty \end{align} Then, \begin{align}&\color{#66f}{\large% \lim_{n\ \to\ \infty}\, \bracks{n^{2}\int_{1}^{\infty}{\cos\pars{x/n} - 1 \over x^{4}}\,\dd x}} =\color{#66f}{\large -\,\half} \end{align}