Compute: $\lim\limits_{n\to+\infty}\int\limits_{0}^1 e^{\{nx\}}x^{100}dx$ Can anyone help to compute this limit, please?
$\{nx\}$ denotes the fractional part of $nx$. 
Tried to replace $nx$ by another variable and write an entire integral as a sum of integrals: $$\sum\limits_{i=0}^{n-1}\int\limits_{i}^{i+1} $$
but it seems not to work out.
 A: We have that $e^{\{nx\}}$ is a $\frac{1}{n}$-periodic function and its mean value over a period is given by $(e-1)$.
By the Riemann-Lebesgue lemma (think to the Fourier series of $e^{\{x\}}$) it follows that:
$$ \lim_{n\to +\infty}\int_{0}^{1}e^{\{nx\}}x^{100}\,dx = (e-1)\int_{0}^{1}x^{100}\,dx = \color{red}{\frac{e-1}{101}}.$$

We do not really need it, but the explicit Fourier series of $e^{\{x\}}$ is given by:
$$ e^{\{x\}}=(e-1)+2(e-1)\sum_{m\geq 1}\frac{\cos(2\pi m x)-2m\pi \sin(2\pi m x)}{1+4m^2 \pi^2}$$
and the $m$-th term of the series is bounded by $\frac{1}{\sqrt{1+4m^2\pi^2}}$ due to the Cauchy-Schwarz inequality.
Additionally, for any $h\in\mathbb{N}^+$ we have
$$ \left|\int_{0}^{1}x^h \sin(2\pi n x)\,dx\right|\sim \left|\int_{0}^{1}x^h \cos(2\pi n x)\,dx\right|\sim\frac{1}{\pi n} $$
for large $n\in\mathbb{N}$.
A: An alternative approach. We have $$I=\int_{0}^{1}e^{\left\{ xn\right\} }x^{100}dx=\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}e^{\left\{ xn\right\} }x^{100}dx
 $$ $$\stackrel{y=nx}{=}\frac{1}{n}\sum_{k=0}^{n-1}\int_{k}^{(k+1)}e^{\left\{ y\right\} }\left(\frac{y}{n}\right){}^{100}dy
 $$ then by the mean value theorem exists some $z_{k}\in\left[k,k+1\right]
 $ such that $$I=\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100}\int_{k}^{(k+1)}e^{\left\{ y\right\} }dy
 $$ and since $e^{\left\{ x\right\} }
 $ has period $1$ $$ I=\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100}\int_{0}^{1}e^{y}dy=\left(e-1\right)\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100}
 $$ now if we take the limit we have a Riemann sum, so $$\begin{align}
\lim_{n\rightarrow\infty}\int_{0}^{1}e^{\left\{ xn\right\} }x^{100}dx= & \left(e-1\right)\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100} \\
 = & \left(e-1\right)\int_{0}^{1}x^{100}dx=\color{red}{\frac{e-1}{101}}.
\end{align}$$
Addendum: I realized that this approach works for any periodic function with period $1$. So it is easily generalizable.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{\lim_{n \to \infty}\int_{0}^{1}\expo{\braces{nx}}x^{100}\,\dd x}
=
\lim_{n \to \infty}{1 \over n^{101}}\int_{0}^{n}\expo{\braces{x}}x^{100}\,\dd x
\end{align}

With Stolz-Ces$\mathrm{\grave{a}}$ro Theorem:
\begin{align}
&\color{#f00}{\lim_{n \to \infty}\int_{0}^{1}\expo{\braces{nx}}x^{100}\,\dd x}
\\[3mm] = &\
\lim_{n \to \infty}{1 \over \pars{n + 1}^{101} - n^{101}}\pars{%
\int_{0}^{n + 1}\expo{\braces{x}}x^{100}\,\dd x - \int_{0}^{n}\expo{\braces{x}}x^{100}\,\dd x}
\\[3mm] = &\
{1 \over 101}\lim_{n \to \infty}{1 \over n^{100}}
\int_{n}^{n + 1}\expo{x - n}x^{100}\,\dd x
\end{align}

Integrating by parts:
\begin{align}
&\color{#f00}{\lim_{n \to \infty}\int_{0}^{1}\expo{\braces{nx}}x^{100}\,\dd x}
\\[3mm] = &\
{1 \over 101}\lim_{n \to \infty}\braces{%
\bracks{\expo{}\pars{1 + {1 \over n}}^{100} - 1} -
{100 \over n^{100}}\int_{n}^{n + 1}\expo{x - n}x^{99}\,\dd x}\tag{1}
\end{align}

$$
\begin{array}{|c|}\hline\mbox{}\\
\quad\mbox{Note that}\ds{\quad%
0 < \verts{{1 \over n^{100}}\int_{n}^{n + 1}\expo{x -n}x^{99}\,\dd x} <
{\expo{}\pars{n + 1}^{99} \over n^{100}}
\stackrel{n\ \to\ \infty}{\longrightarrow} 0\quad}
\\[4mm]
\mbox{such that ( see (1) )}\ds{\quad
\lim_{n \to \infty}\ds{\pars{%
{100 \over n^{100}}\int_{n}^{n + 1}\expo{x - n}x^{99}\,\dd x} =
\color{#f00}{0}}}
\\ \mbox{}\\ \hline
\end{array}
$$

$$
\mbox{Then ( see (1) ),}\quad
\color{#f00}{\lim_{n \to \infty}\int_{0}^{1}\expo{\braces{nx}}x^{100}\,\dd x}
=
\color{#f00}{\expo{} - 1 \over 101}
$$
