Evaluating $\int_0^a \lfloor x^n \rfloor \,\mathrm{d}x$ 
Evaluate $\int_0^a \lfloor x^n \rfloor \,\mathrm{d}x$ (where $ \lfloor \,\cdot\, \rfloor $ denotes greatest integer function).

Can anyone please give a detailed explanation of how to do this? 
This is my first question on MathStack Exchange.
Thank You
 A: If you mean $$ \begin{align} 
\int_{0}^{\infty}\lfloor x^n\rfloor dx,
\end{align} $$
then this integral diverges to infinity for all $ n\in\mathbb{R} $. To see this, I suggest making a comparison between $ \int_{0}^{\infty}\lfloor x^n\rfloor dx $ and $ \int_{0}^{\infty} x^n dx $, then using the integral test for convergence.
A: For a given $a\geq0$ there should $\exists k \in \mathbb{N}: \color{red}{k}\leq a^{n}<\color{red}{k}+1$ (i.e. $k=\left \lfloor  a^n\right \rfloor$) then 
$$\int\limits_0^a \lfloor x^n \rfloor dx=
\int\limits_0^1 \lfloor x^n \rfloor dx+
\int\limits_1^{\sqrt[n]{2}} \lfloor x^n \rfloor dx+
\int\limits_{\sqrt[n]{2}}^{\sqrt[n]{3}} \lfloor x^n \rfloor dx+...+
\int\limits_{\sqrt[n]{\color{red}{k}}}^{a} \lfloor x^n \rfloor dx\tag{1}$$
and we have


*

*for $x\in[0,1) \Rightarrow x^n\in[0,1) \Rightarrow \lfloor x^n \rfloor=0 \Rightarrow \int\limits_0^1 \lfloor x^n \rfloor dx=0$

*for $x\in[1,\sqrt[n]{2}) \Rightarrow x^n\in[1,2) \Rightarrow \lfloor x^n \rfloor=1 \Rightarrow \int\limits_1^{\sqrt[n]{2}} \lfloor x^n \rfloor dx=\sqrt[n]{2}-1$

*for $x\in[\sqrt[n]{2},\sqrt[n]{3}) \Rightarrow x^n\in[2,3) \Rightarrow \lfloor x^n \rfloor=2 \Rightarrow \int\limits_{\sqrt[n]{2}}^{\sqrt[n]{3}} \lfloor x^n \rfloor dx=2\left(\sqrt[n]{3}-\sqrt[n]{2}\right)$

*...

*for $x\in[\sqrt[n]{t},\sqrt[n]{t+1}) \Rightarrow x^n\in[t,t+1) \Rightarrow \lfloor x^n \rfloor=t \Rightarrow \int\limits_{\sqrt[n]{t}}^{\sqrt[n]{t+1}} \lfloor x^n \rfloor dx=t\left(\sqrt[n]{t+1}-\sqrt[n]{t}\right)$

*...

*for $x\in[\sqrt[n]{\color{red}{k}},a) \Rightarrow x^n\in[\color{red}{k},a^n) \Rightarrow \lfloor x^n \rfloor=\color{red}{k} \Rightarrow \int\limits_{\sqrt[n]{\color{red}{k}}}^{a} \lfloor x^n \rfloor dx=\color{red}{k}\left(a-\sqrt[n]{\color{red}{k}}\right)$


then $(1)$ becomes 
$$\int\limits_0^a \lfloor x^n \rfloor dx=
\left(\sum\limits_{t=1}^{\color{red}{k}-1}t\left(\sqrt[n]{t+1}-\sqrt[n]{t}\right)\right) + \color{red}{k}\left(a-\sqrt[n]{\color{red}{k}}\right)$$
A: Try to understand what this integral means. Assume for simplicity that $a$ is a positive integer, say $a=3$, and that, e.g. $n=2$. Then draw the graph of the integrand
$$f(x):=\lfloor x^n\rfloor\qquad(0\leq x\leq a)\ ,$$
and shade the area you want to compute. You shall see that you don't need the fundamental theorem of calculus, since $f$ is a step function, and this area is a sum of rectangular pieces. The final result will not be a simple expression in the parameters $a$ and $n$, but a  sum of about $a^n$ terms.
