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Could you tell me how to calculate the following limit?

$$\lim_{n \rightarrow \infty} \int_{2\pi n}^{(2n+1)\pi}\!\left(x^t - \left \lfloor x^t \right \rfloor\right) \sin x \,dx$$ for a given parameter $t$?

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Try to use the technique used in this problem. –  Mhenni Benghorbal Jun 6 '13 at 19:17

1 Answer 1

Suppose first that $t>1$.

Let $f_t(x)=\sin y$, where $y$ is the largest number less than $x$ with $y^t$ an integer. Let $[a_n,b_n]\subset [2n\pi,(2n+1)\pi]$ be the largest interval such that $(a_n)^t$ and $(b_n)^t$ are integers. Since $t>1$, the length of $[2n\pi,(2n+1)\pi] \setminus [a_n,b_n]$ goes to $0$ as $n \to \infty$. Moreover, the integrand is uniformly bounded. So we can disregard the portion of the integral contained in the interval $[a_n,b_n]$. Now:

  • $\displaystyle\int_{a_n}^{b_n} f_t(x) \, dx$ is (up to translation and extension by zero) a Riemann sum approximating $\displaystyle\int_0^\pi \sin x \, dx$. Since $t>1$, the mesh size of this sum goes to zero as $n \to \infty$; thus $\lim_{n \to \infty} \displaystyle\int_{a_n}^{b_n} f_t(x) \, dx = \displaystyle\int_0^\pi \sin x \, dx = 2$.
  • $\displaystyle\int_{a_n}^{b_n} \left(x^t-\lfloor x^t \rfloor\right)f_t(x) \, dx=\frac{1}{t+1}\displaystyle\int_{a_n}^{b_n} f_t(x) \, dx$ (because it consists of a bunch of translated/scaled copies of $\int_0^p x^t \, dt$, where the step function $f$ determines the scaling factor of each copy). Thus $\lim_{n \to \infty} \displaystyle\int_{a_n}^{b_n} \left(x^t-\lfloor x^t \rfloor\right) f_t(x) \, dx = \frac{2}{t+1}$.
  • $\displaystyle{\lim_{n \to \infty} \int_{a_n}^{b_n} \left(x^t-\lfloor x^t \rfloor\right)(f_t(x)-\sin x) \, dx}=0$, since $x^t-\lfloor x^t \rfloor$ is uniformly bounded and $f_t(x)-\sin x$ uniformly approaches $0$ as $n \to \infty$. It follows that the limit we're looking for is equal to $\dfrac{2}{t+1}$.

Next, suppose $t<1$. Then we can find infinitely many intervals $[2n\pi, (2n+1)\pi]$ on which $x^t-\lfloor x^t \rfloor>\frac{2}{3}$ everywhere, and infinitely many on which $x^t-\lfloor x^t \rfloor<\frac{1}{3}$ everywhere. Thus the limit diverges.

Finally, if $t=1$, numerics suggest that the limit is $1$ (as you would expect if you "extended by continuity" from the $t>1$ case), but I don't have a proof of this.

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