$\lim _{k\to +\infty}$ $\int_{0}^{k[x]} (kt-[kt])^k dt$ $;$ $k\in N$ is $[\frac{\lambda x} 2]$
Where $[.]$ denotes greatest Integer function then the value of $\lambda$ would be?
My approach to this question was using Leibniz integral rule which didn't work out instead of making it lengthy , even used every single property of greatest Integer function but still I'm missing something. What would be a better approach instead of Leibniz integral rule? Thanks:)
Best Answer
Using the substitution $u = kt$, we have, \begin{align*} I & = \int_{0}^{k[x]} (kt - [kt])^k \ \mathrm{d}t \\ & = \frac{1}{k} \int_{0}^{k^2[x]} (u - [u])^k \ \mathrm{d}u \\ & = \frac{1}{k} \sum_{r = 0}^{k^2[x]-1}\int_{r}^{r+1} (u - [u])^k \ \mathrm{d}u \\ & = \frac{1}{k} \sum_{r = 0}^{k^2[x]-1}\int_{0}^{1} ((u-r) - [u-r])^k \ \mathrm{d}u \\ & = \frac{1}{k} \sum_{r = 0}^{k^2[x]-1}\int_{0}^{1} ((u-r) - ([u]-r))^k \ \mathrm{d}u \\ & = \frac{1}{k} \sum_{r = 0}^{k^2[x]-1}\int_{0}^{1} (u - [u])^k \ \mathrm{d}u \\ & = \frac{k^2[x]}{k} \int_{0}^{1} (u - [u])^k \ \mathrm{d}u \\ & = {k[x]} \int_{0}^{1} u^k \ \mathrm{d}u \\ & = \frac{k[x]}{k+1}. \end{align*}
Consequently, $\displaystyle \lim_{k \to \infty} I = \lim_{k \to \infty} \frac{k[x]}{k+1} = [x]$. Hence, $\lambda = 2$.