$\mathcal{I}=\int\limits_0^0 \{x\}^{\lfloor x\rfloor}\,\mathrm dx=0\textrm{ or undefined ?}$

Question

Consider the following integral:

$$\mathcal{I}=\int\limits_0^0 \{x\}^{\lfloor x\rfloor}\,\mathrm dx$$

Now, my concern is that at $x=0$, the value of the integrand is $0^0$ which is undefined.

It's obvious that the Cauchy principal value of $\mathcal I$ is $0$, but what would the value of the integral be in general sense? $0$ or undefined?

Clarification: $\{\cdot\}$ is the fractional part function and $\lfloor \cdot\rfloor$ is the floor function.

Thanks in advance! :)

Answer 1

$\displaystyle\int_L^Lf(x)~dx=0~$ if $~f(L)~$ is bounded, since it represents the area of a finite straight line

segment; i.e., $~\displaystyle\int_0^0\sin\bigg(\frac1x\bigg)~dx=0.~$ So, unless you can come up with a situation of

$\displaystyle\lim_{a\to0^+}\lim_{b\to0^+}a^b=\infty$, it would appear that the non-definability of the integrand does not

influence the definability of the integral.

Answer 2

In my opinion, you need a some regularization of it. Let's denote $$J(x) = \int\limits_0^x \{x\}^{\lfloor x\rfloor}\,dx,$$ and $$J(x) = \lim_{\epsilon\to0}\int\limits_{\epsilon}^x \{x\}^{\lfloor x\rfloor}\,dx.$$ But if $0 < \epsilon \le x < 1$, $\{x\}=x$, $\lfloor x\rfloor = 0$, and $\{x\}^{\lfloor x\rfloor}=1$. So we have $$J(x)=\lim_{\epsilon\to 0} (x-\epsilon)=x$$ for $0<x<1$. Your integral is $\lim\limits_{x\to 0} J(x) = 0$. It's long and sound; I prefer argument with Lebesgue measure.