I've been trying to figure out a general rule for integrating functions of the form $\lfloor x^n \rfloor$. I don't have any ideas other than trying some kind of clever substitution but I have no idea what that would look like.

I know because of some theorems that $\int \lfloor x^n \rfloor dx = x * \lfloor x^n \rfloor + f(\lfloor x^n \rfloor) + c$ but I don't know what $f$ actually looks like or whether or not $f$ has a closed form. I just know that's the form the result will look like.

  • $\begingroup$ Just to clarify: $x$ is any real? or $n$ is any real? Typically, $x$ is used for reals, while $n$ is used for integers. $\endgroup$ – JavaMan Nov 29 '18 at 3:30
  • $\begingroup$ @JavaMan $x$ is a variable. $\endgroup$ – The Great Duck Nov 29 '18 at 3:31
  • $\begingroup$ That doesn't answer my question. So $n$ is a fixed real number? Or is $n$ a fixed integer? And $x$ is a variable, but presumably it is a real number? $\endgroup$ – JavaMan Nov 29 '18 at 3:33
  • $\begingroup$ @JavaMan we're integrating a function. I would believe the correct specification is that $x$ is neither? And if I said $n$ is a real number then $n$ is a real number. My choice of letter shouldn't matter to you. $\endgroup$ – The Great Duck Nov 29 '18 at 3:39
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    $\begingroup$ So, you want $\int_{}\lfloor x^n\rfloor dx$ for given integer/real $n$ ? Or $\int_{}\lfloor x^n\rfloor dn$ for given integer/real $x$? $\endgroup$ – Jimmy R. Nov 29 '18 at 3:43

You have to find the regions where $k \le x^n \lt k+1$ for each integer $k$. These are $k^{1/n} \le x \lt (k+1)^{1/n}$.

Then replace the integral with a sum over these regions.

In particular, I don't see how you can get an indefinite integral - you have to integrate over an actual region.

(added later)

If $I(a, b) =\int_a^b \lfloor x^n \rfloor dx $, then, without worrying about possible leftover intervals at the beginning at end, $I(a, b) =\sum_{k =\lfloor a^{1/n} \rfloor}^{\lfloor b^{1/n} \rfloor}\int_{k}^{k+1} k^n dx =\sum_{k =\lfloor a^{1/n} \rfloor}^{\lfloor b^{1/n} \rfloor} k^n $.

This is a first attempt. There are problems at the ends of the intervals, but this is enough for me now.

  • $\begingroup$ Can you elaborate on that last sentence? $\endgroup$ – The Great Duck Nov 29 '18 at 5:32
  • $\begingroup$ So what is the final function? Does it have no closed form? $\endgroup$ – The Great Duck Nov 30 '18 at 3:22

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