$\lim_{n \to \infty }n\int_{0}^{\pi}\left \{ x \right \}^{n}dx$ Calculate
$$\lim_{n \to \infty }n\int_{0}^{\pi}\left \{ x \right \}^{n}dx$$
My try: $x-1<[x]<x$
$-x<-[x]<1-x$
$0<x-[x]<1$ then I applied an integral but I get the result pi which is the wrong answer.
 A: $$\begin{align}
\int_0^\pi\{x\}^n\mathrm{d}x
&=\int_0^1x^n\mathrm{d}x+\int_1^2(x-1)^n\mathrm{d}x+\int_2^3(x-2)^n\mathrm{d}x+\int_3^\pi(x-3)^n\mathrm{d}x\\
&=\left[\frac1{n+1}x^{n+1}\right]_0^1+\left[\frac1{n+1}(x-1)^{n+1}\right]_1^2+\left[\frac1{n+1}(x-2)^{n+1}\right]_2^3+\left[\frac1{n+1}(x-3)^{n+1}\right]_3^\pi\\
&=\frac1{n+1}+\frac1{n+1}+\frac1{n+1}+\frac{(\pi-3)^{n+1}}{n+1}\\
&=\frac3{n+1}+\frac{(\pi-3)^{n+1}}{n+1}\\
\end{align}$$
So the limit in question is
$$\lim_{n\to\infty}n\left(\frac3{n+1}+\frac{(\pi-3)^{n+1}}{n+1}\right)=\lim_{n\to\infty}\frac{n}{n+1}\left(3+(\pi-3)^{n+1}\right)=\boxed3$$
A: You have to split your integral.
$\displaystyle I=\int_0^\pi \{x\}^n\mathop{dx}=\int_0^1 x^n\mathop{dx}+\int_1^2 (x-1)^n\mathop{dx}+\int_2^3 (x-2)^n\mathop{dx}+\int_3^\pi (x-3)^n\mathop{dx}$
By substitution $u=x-k,\ \mathop{du}=\mathop{dx}\quad$ you can equal the first three integrals.
$\displaystyle I=3\int_0^1 u^n\mathop{du}+\int_0^{\pi-3} u^n\mathop{du}=\dfrac 3{n+1}+\dfrac{(\pi-3)^{n+1}}{n+1}$
Since $|\pi-3|<1$ the quantity $(\pi-3)^{n+1}\to 0$ 
Thus $\lim\limits_{n\to\infty}nI=3$

Another interesting solution would be to prove that in distribution sense, $n\int_0^1 x^n\mathop{dx}\to\delta_1$ so that $nI$ would be $\delta_1([0,\pi])+\delta_2([0,\pi])+\delta_3([0,\pi])=1+1+1=3$
A: Break the integral into four parts; from $0$ to $1$, from $1$ to $2$, from $2$ to $3$ and from $3$ to $\pi$. Then $\{x\}=x,\ \{x\}=x-1,\ \{x\}=x-2,\ \{x\}=x-3$ in those intervals respectively. Then, you can easily calculate the three integrals to find that 
$$n\int_{0}^{\pi}\{x\}^ndx=\frac{3n}{n+1}+\frac{n(\pi-1)^{n+1}}{n+1}.$$ 
Now you can take it from here. Note that $0<\pi-3<1,$ so that $(\pi-1)^{n+1}\rightarrow 0.$
With time delay. I apologize for that, because all the answers here are the same.
A: $\int_0^1 x^n + \int_1^2 (x-1)^n + \int_2^3 (x-2)^n +\int_3^\pi (x-3)^n$ then 
$\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+1}+\frac{(\pi-3)^{n+1}}{n+1}$ 
$\lim \frac{3n+n(\pi-3)^{n+1}}{n+1}$
