2
$\begingroup$

I have read some posts on x mod y integrals but am unsure of how to find the integral of a constant mod x.

For example
6 mod x between 0 and 6

or

10 mod x between 0 and 10.

When I look at the graph of these functions I can see it generates a lot of lines starting at a point on the line y = x and descending to zero at some angle. So I can tell it generates an infinite number of right triangles between 0 and 1 and then n right triangles between 1 and n.

But I am struggling to even know where to begin to find their area.

$\endgroup$
1
  • $\begingroup$ Where did you read about "definite integrals in regards to $\;x\pmod n\;$" ? Perhaps it was "INTEGERS $\;x\pmod n\;$" or something of the like? $\endgroup$
    – DonAntonio
    Mar 7, 2016 at 9:28

2 Answers 2

3
$\begingroup$

You can rewrite the integral as

$$\int_0^x \left(n-t\lfloor\frac nt\rfloor\right)dt.$$

The floor factor remains constant in the ranges $t\in\left(\dfrac n{k+1},\dfrac nk\right]$ for $k$ decreasing from $\infty$ to $m=\lfloor\dfrac nx\rfloor$. The last interval is incomplete, $t\in\left(\dfrac n{m+1},x\right]$.

We have

$$I=\sum_{k=\infty}^{m}\int_{n/(k+1)}^{n/k}\left(n-tk\right)dt+\int_{n/(m+1)}^{x}\left(n-tm\right)dt,$$ giving

$$2I=\sum_{k=\infty}^{m}\left.\left(n-tk\right)^2\right|_{n/(k+1)}^{n/k}+\left.\left(n-tm\right)^2\right|_{n/(m+1)}^{x}\\ =\sum_{k=\infty}^{m}\frac{n^2}{(k+1)^2}+(n-mx)^2-\frac{n^2}{(m+1)^2}\\ =\frac{\pi^2}6-H_m^{(2)}+(n-mx)^2-\frac{n^2}{(m+1)^2},$$

where $H_m^{(2)}$ denotes the generalized harmonic numbers of degree $2$.

$\endgroup$
5
  • $\begingroup$ What do the variables t and k signify? $\endgroup$
    – ajt
    Mar 8, 2016 at 1:41
  • $\begingroup$ @ajt t is the variable as in $y\equiv n$ (mod t), where n is a constant. And it turns out y can be written as a function of t by $n-t\floor{\frac{n}{t}}$. k is the integer that represent the roots in descending order, as observed in m.wolframalpha.com/input/?i=x%3D10+%28mod+y%29&x=0&y=0 when k is 1, n/k=n is the largest root of the equation $\endgroup$
    – lEm
    Mar 8, 2016 at 4:45
  • $\begingroup$ So can t=x ? I guess I am confused because in the original equation I posted I just had x. In the integral Yves posted ∫ from 0 to x but then uses t to divide n. $\endgroup$
    – ajt
    Mar 8, 2016 at 5:29
  • $\begingroup$ @ajt: $t$ is the dummy variable required for integration. $\endgroup$
    – user65203
    Mar 8, 2016 at 8:07
  • $\begingroup$ @YvesDaoust Thank you for everything. I have only taken up to AP calculus at this point so I must admit I don't fully understand what is happening here. I don't want to drag this on but for the last equation you give =π26−H(2)m+(n−mx)2−n2(m+1)2. Would I do this for 10 mod x at x = 10 as: pi**2/6 - Hm2 + (10 - floor(10/10)*10)**2 - 100/4 ? Also how do I get the sum of harmonic numbers of degree 2? Is it 3/2 or something else? mathworld.wolfram.com/HarmonicNumber.html $\endgroup$
    – ajt
    Mar 9, 2016 at 1:04
1
$\begingroup$

http://m.wolframalpha.com/input/?i=x%3D10+%28mod+y%29&x=0&y=0

See the graph here, the equation $x\equiv 10$ (mod $y$) has infinite number of lines between $[0,1]$. The roots $x\equiv 0\equiv 10$ (mod $y$) can be found at all $\frac{p}{q}$ where $p$ is a factor of 10 and $q\in\Bbb{N}$. For examples, $\frac{10}{11}$, $\frac{5}{19}$. So it could be hard to find out the value.

But for $y\in[1,10]$, we know the roots of $x\equiv0$ (mod $y$) are located at $1$, $\frac{10}{9}$, $\frac{10}{8}$, $\frac{10}{7}$, $\frac{10}{6}$, $\frac{10}{5}=2$, $\frac{10}{4}$, $\frac{10}{3}$, $\frac{10}{2}=5$ and $10$.

So we can compute the integral by computing the sum of area of triangles. (At least between $[0,1]$),

$\int_1^{10}{x(y)dy}=\sum_{k=1}^{9} [\frac{1}{2}x(y_k)*(y_{k+1}-y_k)]$

where $y_k$ is the $k^{th}$ root starting from $y_1=1$. And $x(y_k)$ is taken to be equal to $\lim_{y \to y_k^-}x(y_k)=y_k$

(I take this value because otherwise $x(y_k)$ is counted as $0$)

By computing the sum, you can find that the integral is equal to approximately $17.5116...$, exactly $\frac{2224115}{127008}$.

On the graph, notice that $y=x$ is an asymptote for the function. So when the value of $y$ is small. (Maybe between $[0,1]$). You can approximate the integral by observing the triangles take up about half of the area. So $\int_0^{y_0}x(y)dy\approx \frac{1}{4}y_0^2$ for small $y_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.