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. 
 A: 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.
$\frac 12 \int_0^1 x^{2019} dx = (\frac 12) (\frac 1{2020})$
A: 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}
$
A: 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$$\lim\limits_{n\to \infty} \int\limits_0^1 x^{2019} \{nx\} dx={1\over 4040}$$
A: 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. 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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
$$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n\to \infty}
\int_{0}^{1}x^{2019}\braces{nx}\dd x} =
{1/2 \over 2020} = \bbx{\large{1 \over 4040}} \\ &
\end{align}
$$
