# Compute $\lim\limits_{n\to \infty} \int\limits_0^1 x^{2019} \{nx\} dx$

Compute $$\lim\limits_{n\to \infty} \int\limits_0^1 x^{2019} \{nx\} dx,$$ where $$\{a\}$$ denotes the fractional part of the real number $$a$$.
I firstly tried to apply the substitution $$nx=t$$, but the computations didn't look nice, so I couldn't make any further progress. I also tried to use the mean value theorem for integrals, but it was also a dead end.

Here is another approach which is somewhat simpler than the one given in another answer here.

I establish that $$\int_{0}^{1}f(x)\{nx\}\,dx\to\frac{1}{2}\int_{0}^{1}f(x)\,dx$$ as $$n\to\infty$$. The integral on left of above equation can be split as sum of $$n$$ integrals $$\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}f(x)\{nx\}\,dx=\frac{1}{n}\sum_{k=0}^{n-1}\int_{k}^{k+1}f(t/n)\{t\}\,dt$$ Using mean value theorem for integrals the right hand side of the above equation can be written as $$\frac{1}{n}\sum_{k=0}^{n-1}f(t_k/n)\int_{k}^{k+1}\{t\}\,dt$$ where $$t_k\in[k,k+1]$$ and since $$\{t\}$$ is periodic with period $$1$$ the above reduces to $$\left(\int_{0}^{1}\{t\}\,dt\right)\cdot\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{t_k}{n}\right)$$ The integral above is $$1/2$$ as $$\{t\} =t$$ if $$t\in[0,1)$$ and the next factor is Riemann sum for $$f$$ or $$[0,1]$$. Thus the above tends to $$\frac{1}{2}\int_{0}^{1}f(x)\,dx$$ Above derivation assume that $$f$$ is continuous on $$[0,1]$$. Putting $$f(x) =x^{2019}$$ we get the desired limit as $$1/4040$$.

More generally we can use same method to prove that $$\lim_{n\to\infty} \int_{0}^{1}f(x)g(\{nx\})\,dx=\left(\int_{0}^{1}f(x)\,dx\right)\left(\int_{0}^{1}g(x)\,dx\right)$$ where $$f$$ is continuous on $$[0,1]$$ and $$g$$ is of constant sign and Riemann integrable on $$[0,1]$$.

Going further we can also note that if $$g$$ is periodic with period $$T$$ and of constant sign and Riemann integrable on $$[0,T]$$ and $$f$$ is continuous on $$[0,T]$$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$

Based on suggestion in comments, one can prove that the above result holds for Riemann integrable $$f, g$$ and $$g$$ also being periodic with period $$T$$.

The idea is to express the integral on left as a sum $$\frac{1}{n}\sum_{k=0}^{n-1}\int_{kT}^{(k+1)T}f(x/n)g(x)\,dx$$ which can be further rewritten as $$\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{T}f((x+kT)/n)g(x+kT)\,dx$$ And since $$g$$ is periodic it follows that the above can be written as $$\frac{1}{T}\int_{0}^{T}\left(\frac{T}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kT}{n}\right)g(x)\right)\,dx\tag{1}$$ Since $$f$$ is Riemann integrable on $$[0,T]$$ with integral $$I=\int_{0}^{T}f(x)\,dx$$ we can see that if $$P_n=\{0,T/n,2T/n,\dots,(n-1)T/n,T\}$$ is a partition of $$[0,T]$$ and $$U(f, P_n), L(f, P_n)$$ be corresponding upper and lower Darboux sums then we have $$L(f, P_n) \leq S(f, P_n) \leq U(f, P_n)$$ where $$S(f, P_n)$$ is any Riemann sum for $$f$$ over $$P_n$$. Since the integral $$I$$ is also sandwiched between both upper and lower sums we have $$|S(f, P_n) - I|\leq U(f, P_n) - L(f, P_n)$$ We can now observe that the integrand in equation $$(1)$$ is of the form $$S(f, P_n) g(x)$$ and hence $$\left|\int_{0}^{T}S(f,P_n)g(x)\,dx-I\int_{0}^{T}g(x)\,dx\right|\leq (U(f, P_n) - L(f, P_n)) \int_{0}^{T}|g(x)|\,dx$$ and clearly the right hand side above tends to $$0$$ so that the left hand side also does the same. It follows that the desired limit is $$\frac{1}{T}\int_{0}^{T}f(x)\,dx\int_{0}^{T}g(x)\,dx$$ Credit for the idea of above proof must go to the user WE Tutorial School.

If the integral $$\int_{0}^{T}g(x)\,dx=0$$ then the above can be used as a proof of Riemann-Lebesgue Lemma for Riemann integrable functions and therefore the above is a generalization of it.

• Nice generalization at the end. Commented Jan 30, 2020 at 16:53
• I am not sure why $g$ must be of constant sign (although I understand why you put this condition in your answer). I think being Riemann integrable is sufficient. Perhaps $f$ needs to only be Riemann integrable as well, but I understand that your proof requires continuity of $f$. Commented Jan 31, 2020 at 0:38
• @WETutorialSchool: The mean value theorem requires that the second function $g$ be of constant sign. Commented Jan 31, 2020 at 0:39
• After thinking about this carefully, I am very certain that $f$ only needs to be Riemann integrable, and $g$ only needs to be Lebesgue integrable. This can be proved by DCT. Commented Jan 31, 2020 at 1:04
• @Jules_99: its a substitution $x=u+kT$ and then further replace the symbol $u$ by $x$. This does not involve any periodic properties of $g$. Commented Nov 17, 2020 at 8:05

\begin{align} &\int_0^1x^{2019}\{nx\}\,\mathrm{d}x\\ &=\frac1{n^{2020}}\int_0^nx^{2019}\{x\}\,\mathrm{d}x\tag1\\ &=\frac1{n^{2020}}\sum_{k=0}^{n-1}\int_0^1(k+x)^{2019}((k+x)-k)\,\mathrm{d}x\tag2\\ &=\frac1{n^{2020}}\sum_{k=0}^{n-1}\left(\frac{(k+1)^{2021}-k^{2021}}{2021}-k\frac{(k+1)^{2020}-k^{2020}}{2020}\right)\tag3\\ &=\frac1{n^{2020}}\sum_{k=0}^{n-1}\left(\frac{(k+1)^{2021}-k^{2021}}{2021}-\frac{(k+1)^{2021}-(k+1)^{2020}-k^{2021}}{2020}\right)\tag4\\ &=\frac1{n^{2020}}\left(\frac{n^{2021}}{2021}-\frac{n^{2021}}{2020}+\sum_{k=0}^{n-1}\frac{(k+1)^{2020}}{2020}\right)\tag5\\ &=\frac1{n^{2020}}\left(-\frac{n^{2021}}{2021\cdot2020}+\frac{n^{2021}}{2021\cdot2020}+\frac12\frac{n^{2020}}{2020}+O\!\left(n^{2019}\right)\right)\tag6\\[6pt] &=\frac1{4040}+O\!\left(\frac1n\right)\tag7 \end{align} Explanation:
$$(1)$$: substitute $$x\mapsto x/n$$
$$(2)$$: break into integer intervals; $$x\mapsto k+x$$ and $$\{x\}\mapsto x$$
$$(3)$$: integrate
$$(4)$$: $$k(k+1)^{2020}=(k+1)^{2021}-(k+1)^{2020}$$
$$(5)$$: sum the telescoping parts
$$(6)$$: use the first two terms of Faulhaber's Formula
$$(7)$$: simplify

Thus, $$\lim_{n\to\infty}\int_0^1x^{2019}\{nx\}\,\mathrm{d}x=\frac1{4040}\tag8$$

Faulhaber's Formula \begin{align} \sum_{k=1}^nk^m &=\int_0^nx^m\,\mathrm{d}\lfloor x\rfloor\tag9\\ &=\int_0^nx^m\,\mathrm{d}\!\left(x-\{x\}\right)\tag{10}\\ &=\tfrac1{m+1}n^{m+1}-\int_0^nx^m\,\mathrm{d}\!\left(\{x\}-\tfrac12\right)\tag{11}\\ &=\tfrac1{m+1}n^{m+1}+\tfrac12n^m+m\int_0^nx^{m-1}\left(\{x\}-\tfrac12\right)\,\mathrm{d}x\tag{12}\\[6pt] &=\tfrac1{m+1}n^{m+1}+\tfrac12n^m+O\!\left(n^{m-1}\right)\tag{13} \end{align} Explanation:
$$\phantom{1}(9)$$: write the sum as a Stieltjes integral
$$(10)$$: $$\lfloor x\rfloor=x-\{x\}$$
$$(11)$$: integrate
$$(12)$$: integrate by parts
$$(13)$$: use the estimate of the error below \begin{align} \left|\,m\int_0^nx^{m-1}\left(\{x\}-\tfrac12\right)\,\mathrm{d}x\,\right| &=\left|\,m\sum_{k=0}^{n-1}\int_k^{k+1}\left(x^{m-1}-k^{m-1}\right)\left(\{x\}-\tfrac12\right)\,\mathrm{d}x\,\right|\tag{14}\\ &\le\frac{m}2\sum_{k=0}^{n-1}\int_k^{k+1}\left(x^{m-1}-k^{m-1}\right)\,\mathrm{d}x\tag{15}\\ &=\frac{m}2\sum_{k=0}^{n-1}\left(\frac{(k+1)^m-k^m}m-k^{m-1}\right)\tag{16}\\ &\le\frac{m}2\sum_{k=0}^{n-1}\left((k+1)^{m-1}-k^{m-1}\right)\tag{17}\\[6pt] &=\frac{m}2n^{m-1}\tag{18} \end{align} Explanation:
$$(14)$$: partition the domain at the integers; $$\{x\}-\frac12$$ has mean value $$0$$ over each interval
$$(15)$$: $$\left|\{x\}-\tfrac12\right|\le\frac12$$
$$(16)$$: integrate
$$(17)$$: Mean Value Theorem
$$(18)$$: sum the telescoping series

For a finite value of $$n$$ our equation The graph of our function looks like a saw tooth, that touches the curve $$x^{2019}$$ when $$x$$ is a multiple of $$\frac {1}{n}$$

The area under the curve is the red area.

As $$n$$ approaches infinity, the red area becomes $$\frac 12$$ the area under the curve.

$$\frac 12 \int_0^1 x^{2019} dx = (\frac 12) (\frac 1{2020})$$

• As $n$ approaches infinity, the red area becomes $\frac 12$ the area under the curve. I get that this is intuitively obvious, but how would you show it rigorously? Commented Jan 29, 2020 at 23:08
• I show it in my answer. Commented Jan 30, 2020 at 0:21
– user454960
Commented Jan 30, 2020 at 0:31
• @FaradayPathak: one should be cautious while using intuition in analysis. Most of the intuitive approaches are provided by those who back it up with rigorous proof. Commented Jan 30, 2020 at 3:19

Here's a proof that $$\lim_{n \to \infty} \int\limits_0^1 f(x) \{nx\} dx =\dfrac12 \int_0^1 f(x) dx$$.

If $$f(x) = x^m$$, then $$\lim_{n \to \infty} \int\limits_0^1 f(x) \{nx\} dx =\dfrac12 \int_0^1 x^m dx =\dfrac1{2(m+1)}$$.

Let

\begin{align}\\ g(n) &=\int\limits_0^1 f(x) \{nx\} dx\\ &=\sum_{k=0}^{n-1}\int\limits_{k/n}^{(k+1)/n} f(x) \{nx\} dx\\ &=\sum_{k=0}^{n-1}\dfrac1{n}\int\limits_{k}^{k+1} f(y/n) \{y\} dy \qquad y = nx, dx = dy/n\\ &=\sum_{k=0}^{n-1}\dfrac1{n}\int\limits_{0}^{1} f((z+k)/n) \{z+k\} dz \qquad z = y-k\\ &=\sum_{k=0}^{n-1}\dfrac1{n}\int\limits_{0}^{1} f((z+k)/n) \{z\} dz\\ &=\dfrac1{n}\sum_{k=0}^{n-1}\int\limits_{0}^{1} f((z+k)/n) z dz\\ \\ &\text{(uses IBP } \int zf = \frac12 z^2f-\frac12\int z^2f' \\ &=\dfrac1{n}\sum_{k=0}^{n-1}(\dfrac12 (z^2f((z+k)/n)))_0^1-\dfrac1{2n}\int\limits_{0}^{1} f'((z+k)/n) z^2 dz)\\ &=\dfrac1{n}\sum_{k=0}^{n-1}\dfrac12 (z^2f((z+k)/n)))_0^1-\dfrac1{n}\sum_{k=0}^{n-1}\dfrac1{2n}\int\limits_{0}^{1} f'((z+k)/n) z^2 dz\\ &=\dfrac1{2n}\sum_{k=0}^{n-1}(f((1+k)/n)))-\dfrac1{2n^2}\sum_{k=0}^{n-1}\int\limits_{0}^{1} f'((z+k)/n) z^2 dz\\ &=\dfrac1{2n}\sum_{k=1}^{n}(f(k/n)))-\dfrac1{2n^2}\int\limits_{0}^{1} z^2f'(z) dz\\ &\to \frac12 \int_0^1 f(z) dx\\ \end{align}

• You need to replace $f$ by $f'$ in second term after IBP. +1 Commented Jan 30, 2020 at 2:51
• You are right. Thanks. Upvoted and fixed. Commented Jan 30, 2020 at 6:51
• I have given a somewhat simplified proof in another answer here. Commented Jan 30, 2020 at 8:40
• I think integration by parts can be avoided so that the claim holds for all Riemann integrable functions $f$ defined on $[0,1]$ (which are not necessarily differentiable). Note that $$g(n)=\frac{1}{n} \sum_{k=0}^{n-1}\int_0^1 f\left(\frac{z+k}{n}\right) zdz=\int_0^1\left(\sum_{k=0}^{n-1}\frac{1}{n}f\left(\frac{z+k}{n}\right)\right)zdz.$$ If $f$ is Riemann integrable, then $$\lim_{n\to \infty}\sum_{k=0}^{n-1}\frac{1}{n}f\left(\frac{z+k}{n}\right)=\int_0^1 f(x)dx.$$ So $$\lim_{n\to \infty} g(n)=\int_0^1 \left(\int_0^1f(x)dx\right)zdz=\frac12\int_0^1f(x)dx.$$ Commented Jan 31, 2020 at 0:26
• However, I have not checked which conditions would allow me to swap the limit with the integral, so maybe I need a stronger assumption on $f$. At least for continuous $f$, there seems to be no problems. Commented Jan 31, 2020 at 0:32

We have $$f_n(x)=\int_0^x\{nu\}du=\begin{cases} {nx^2\over 2}&,\quad 0\le x< {1\over n}\\ {1\over 2n}+{n\left(x-{1\over n}\right)^2\over 2}&,\quad {1\over n}\le x< {2\over n}\\ {2\over 2n}+{n\left(x-{2\over n}\right)^2\over 2}&,\quad {2\over n}\le x< {3\over n}\\ {3\over 2n}+{n\left(x-{3\over n}\right)^2\over 2}&,\quad {3\over n}\le x< {4\over n}\\ {4\over 2n}+{n\left(x-{4\over n}\right)^2\over 2}&,\quad {4\over n}\le x< {5\over n}\\ \vdots \end{cases}$$we know that $${x\over 2}-{1\over 8n}\le {k\over 2n}+{n\left(x-{k\over n}\right)^2\over 2}\le{x\over 2}\quad,\quad {k\over n}\le x<{k+1\over n}$$therefore$${x\over 2}-{1\over 8n}\le\int_0^x\{nu\}du\le{x\over 2}\quad,\quad 0\le x<1$$By using Integration by parts we obtain$$\int_0^1 x^{2019}\{nx\}dx{= x^{2019}f_n(x)\Big|_0^1-\int_0^1 2019x^{2018}f_n(x)dx \\={1\over 2}-\int_0^1 2019x^{2018}f_n(x)dx }$$where the latter integral can be bounded as$${1\over 4040}\le {1\over 2}-\int_0^1 2019x^{2018}f_n(x)dx\le {1\over 4040}+{1\over 8n}$$therefore$$\lim\limits_{n\to \infty} \int\limits_0^1 x^{2019} \{nx\} dx={1\over 4040}$$

• Thank you! But why can you apply IBP? $f_n$ is not differentiable Commented Jan 30, 2020 at 7:54
• You are right but because of the continuity, the error becomes negligible and tending to zero in discontinuities of the 1st order differentiation (where the original function is not differentiable). Commented Jan 30, 2020 at 14:05
• I think the answer is incorrect. Commented Jan 30, 2020 at 14:34

I have a slightly different approach. Might not be the best. We have $$\int\limits_{0}^{1}x^{2019}\{nx\}dx$$.

Say $$x\in[\frac{r-1}{n},\frac{r}{n})$$, then $$nx\in[0,1)$$. Hence, $$[nx]\in [r-1,r)$$. We can, therefore, write the integral as follows: $$I=\lim\limits_{n\to\infty}\frac{1}{n^{2019}}\int\limits_{0}^{1}(nx)^{2019}\{nx\}dx=\lim\limits_{n\to\infty}\frac{1}{n^{2020}}\int\limits_{0}^{\infty}t^{2019}\{t\}dt$$ For $$t\in[r-1,r),\{t\}=t-(r-1)$$ \begin{aligned} I=&\lim_{n\to\infty}\frac{1}{n^{2020}}\left[\int_{0}^{1}t^{2020}dt+\int_{1}^{2}t^{2019}(t-1)dt\cdots\int_{n-1}^{n}t^{2019}(t-(n-1))dt\right]\\ =&\lim_{n\to\infty}\frac{1}{n^{2020}}\left[\int_{0}^{n}t^{2020}dt-\left\{\int_{1}^{2}t^{2019}dt+2\int_{2}^{3}t^{2019}dt\cdots(n-1)\int_{n-1}^{n}t^{2019}dt\right\}\right] \\ =&\lim_{n\to\infty}\frac{1}{n^{2020}}\left[\frac{n^{2021}}{2021}-\left\{\int_{1}^{n}t^{2019}dt+\int_{2}^{n}t^{2019}dt\cdots+\int_{n-1}^{n}t^{2019}dt\right\}\right]\\ =&\lim_{n\rightarrow \infty}\left( \frac{n}{2021}-\frac{1}{n^{2020}}\left\{ \left( \frac{n^{2020}-1^{2020}}{2020} \right) +\left( \frac{n^{2020}-2^{2020}}{2020} \right) \cdots +\left( \frac{n^{2020}-\left( n-1 \right) ^{2020}}{2020} \right) \right\} \right) \\ =&\lim_{n\to\infty}\left[\frac{n}{2021}-\frac{1}{2020\cdot n^{2020}}\left\{(n-1)n^{2020}-1^{2020}-2^{2020}\cdots-(n-1)^{2020}\right\}\right] \\ =&\lim_{n\to\infty}\left[\frac{n}{2021}-\frac{n-1}{2020}+\frac{1}{2020}\sum_{r=1}^{n-1}\left(\frac{r}{n}\right)^{2020}\right] \\ =&\lim_{n\to\infty}\left[\frac{1}{2020}-\frac{n}{2020\cdot2021}+\frac{n}{2020}\int_{0}^{1}x^{2020}dx\right]\rightarrow\text{(Summation as integration)}\\ =&\lim_{n\to\infty}\left[\frac{1}{2020}\right]=\frac{1}{2020}\\ \end{aligned}

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[5px,#ffd]{\lim_{n\to \infty} \int_{0}^{1}x^{2019}\braces{nx}\dd x} \,\,\,\stackrel{\large nx\ \mapsto x}{=}\,\,\, \lim_{n\to \infty} {1 \over n^{2020}}\int_{0}^{n}x^{2019}\braces{x}\dd x \\[5mm] = &\ \lim_{n\to \infty} {1 \over \pars{n + 1}^{2020} - n^{2020}}\ \times \\[2mm] &\ \phantom{\lim_{n\to \infty}\,\,\,\,}\pars{% \int_{0}^{n + 1}x^{2019}\braces{x}\dd x - \int_{0}^{n}x^{2019}\braces{x}\dd x} \\[5mm] = &\ \lim_{n\to \infty} {1 \over \pars{n + 1}^{2020} - n^{2020}} \int_{n}^{n + 1}\pars{x^{2020} - nx^{2019}}\dd x\label{1}\tag{1} \end{align} where I used the Stolz-Ces$$\mrm{\grave{a}}$$ro Theorem.

Indeed the integration is an elemental one and it's $$\ds{\sim \color{red}{n^{2019} \over 2}}$$ while the denominator is $$\ds{\sim \color{red}{2020\, n^{2019}}}$$ as $$\ds{n \to \infty}$$ such that \begin{align} &\bbox[5px,#ffd]{\lim_{n\to \infty} \int_{0}^{1}x^{2019}\braces{nx}\dd x} = {1/2 \over 2020} = \bbx{\large{1 \over 4040}} \\ & \end{align}