Find $\lim\limits_{n \to \infty} n \int\limits_0^\pi \{ -x\}^n dx$.

Question

I have to find the limit:

$$\lim_{n \to \infty} n \int_0^\pi \{ -x \}^n dx$$

Where $\{ x \}$ is the fractional part of the real number $x$.

I know that:

$$\{ x \} = x - \lfloor x \rfloor$$

where $\lfloor x \rfloor$ is the integer part of the real number $x$, so I'd have something like:

$$\lim_{n \to \infty} n \int_0^\pi (x - \lfloor x \rfloor )^n dx$$

But I don't know how to continue from this point.

Answer 1

You can do the integral explicitly by breaking the interval into subintervals: $$[0,\pi] = [0,1) \cup [1,2) \cup [2,3) \cup [3, \pi].$$ Note that for $0 < \epsilon \le 1$ and $k \in \mathbb N$, if $x \in [k,k+\epsilon)$, then $-x \in (-k-\epsilon, -k]$ and so $\{-x\} = -x + (k+1).$ Thus \begin{align*} \int^{k+\epsilon}_k \{-x\}^n dx &= \int^{k+\epsilon}_k (-x+k+1)^ndx \\ &= -\int^{1-\epsilon}_1 y^n dy = \frac{1- (1-\epsilon)^{n+1}}{n+1}.\end{align*} Then \begin{align*}\int^\pi_0 \{-x\}^ndx &= \int^1_0 (1-x)^n dx + \int^2_1 (2-x)^ndx +\int^3_2 (3-x)^ndx + \int^\pi_3 (4-x)^ndx\\&= \frac{1}{n+1} + \frac 1 {n+1} + \frac 1 {n+1} + \frac{1-(4-\pi)^{n+1}}{n+1}.\end{align*} The last term exhibits exponential decay, so in the limit we see $$n\int^\pi_0 \{-x\}^ndx = \frac{4n}{n+1} - \frac{n}{n+1}(4-\pi)^{n+1} \to 4.$$

Answer 2

\begin{align*} \lim_{n\to\infty}\int_0^\pi\{-x\}^n\,dx&=\lim_{n\to\infty}n\left(\int_0^1\{-x\}^n\,dx+\int_1^2\{-x\}^n\,dx+\int_2^3\{-x\}^n\,dx+\int_3^\pi\{-x\}^n\,dx\right)\\ &=\lim_{n\to\infty}n\left(\int_0^1 (-x-\lfloor -x\rfloor)^n\,dx+\int_1^2(-x-\lfloor -x\rfloor)^n\,dx+\int_2^3(-x-\lfloor -x\rfloor)^n\,dx+\int_3^\pi(-x-\lfloor -x\rfloor)^n\,dx\right)\\ &=\lim_{n\to\infty}n\left(\int_0^1 (-x+1)^n\,dx+\int_1^2(-x+2)^n\,dx+\int_2^3(-x+3)^n\,dx+\int_3^\pi(-x+4)^n\,dx\right)\\ \end{align*}

Can you take it from there?