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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! :)

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  • $\begingroup$ @b00nheT, yes, I believe so. Because this is not a homework question. It's a genuine doubt. It's trivial that $\int_a^a f(x)\,\mathrm dx=0$ when $f(a)$ is defined (finite). But does the same hold when $f(a)$ is undefined? That's my question. $\endgroup$
    – cockyguy
    Jun 4, 2015 at 11:24
  • $\begingroup$ If the integral you are considering is a Lebesgue integral, then the value will be $0$ regardless of what the integrand is. This follows from the definition of the Lebesgue integral as the limit of integrals of simple functions and from the fact that the set $[0,0]$ has measure $0$. $\endgroup$
    – Andrea
    Jun 4, 2015 at 11:27
  • $\begingroup$ @AndreaDiBiagio, I'm considering $\mathcal I$ as a simple Riemann integral (quite elementary, I guess). $\endgroup$
    – cockyguy
    Jun 4, 2015 at 11:30
  • $\begingroup$ $0^0$ is not exactly undefined - it's usually described as an indeterminate form - there's a subtle but important difference. $\endgroup$
    – PM 2Ring
    Jun 4, 2015 at 13:23

2 Answers 2

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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.

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$\displaystyle\int_L^Lf(x)~dx=0~$ 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.

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  • $\begingroup$ So, is $0^0$ bounded? $\endgroup$
    – cockyguy
    Jun 4, 2015 at 11:55
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    $\begingroup$ @integratur: Personally, I know of no situation in which it isn't, but, then again, my knowledge is not infinite. $\endgroup$
    – Lucian
    Jun 4, 2015 at 11:56

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