Modulo Integration?

First of all, the graph $y=\mod{(f(x),g(x))}$ would be the graph $y=f(x)+ng(x)$ , bounded by the x-axis and the function $g(x)$ where $n$ ranges over all the integers, positive and negative and $0$ , and that there will not be more than $1$ value for $y$ for a value of $x$ . here is a simulation showing what I mean.

You can try proving that ^ yourself.

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So, for the graph of $\mod{(x,x^{2})}$ , it would be the graph $y=x+nx^{2}$ . bounded by the x-axis and the function $x^{2}$ .

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Lets focus on $\int _{ 0 }^{ 1 }{ \mod{(x,{ x }^{ 2 })} } dx$

The points at which $y=x-nx^{2}$ touch the x-axis can be easily derived to be $x=\frac{1}{n}$ , for all positive integer $n$ more than $0$ . Note that for now it is $y=x-nx^{2}$ instead of $y=x+nx^{2}$ because I am using positive values of $n$ .

Therefore, $\int _{ 0 }^{ 1 }{ \mod{(x,{ x }^{ 2 })} } dx=\sum _{ n=1 }^{ \infty }{ \int _{ \frac { 1 }{ n+1 } }^{ \frac { 1 }{ n } }{ x-n{ x }^{ 2 } } } dx=\sum _{ n=1 }^{ \infty }{ \frac { 3n+1 }{ 6{ n }^{ 2 }{ (n+1) }^{ 3 } } }$

We would leave that mess for later.

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Now lets focus on $\int _{ -1 }^{ 0 }{ \mod{(x,{ x }^{ 2 })} } dx$ .

The points at which $y=x+nx^{2}$ touch the x-axis can be easily derived to be $-\frac{1}{n}$ , for all positive integers $n$ more than $1$ . However, we will be leaving out the graph for when $n=1$ || $x+x^{2}$ because it does not interfere in the of $x=-1$ to $x=1$ in the graph $y=\mod{(f(x),g(x))}$ .

Therefore, $\int _{ -1 }^{ 0 }{ \mod{(x,{ x }^{ 2 })} }=\sum _{ n=2 }^{ \infty }{ \int _{ -\frac { 1 }{ n-1 } }^{ -\frac { 1 }{ n } }{ x+n{ x }^{ 2 } } } dx=\sum _{ n=2 }^{ \infty }{ \frac { 3n-1 }{ 6{ n }^{ 2 }{ (n-1) }^{ 3 } } }$

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Summing them together:

$\int _{ -1 }^{ 0 }{ \mod{(x,{ x }^{ 2 })} } dx+\int _{ 0 }^{ 1 }{ \mod{(x,{ x }^{ 2 })} } dx = \int _{ -1 }^{ 1 }{ \mod{(x,{ x }^{ 2 })} } dx$ $=\sum _{ n=1 }^{ \infty }{ \frac { 3n+1 }{ 6{ n }^{ 2 }{ (n+1) }^{ 3 } } } +\sum _{ n=2 }^{ \infty }{ \frac { 3n-1 }{ 6{ n }^{ 2 }{ (n-1) }^{ 3 } } }$ $=\sum _{ n=2 }^{ \infty }{ \frac { 3n-1 }{ 6{ n }^{ 2 }{ (n-1) }^{ 3 } } } +\sum _{ n=2 }^{ \infty }{ \frac { 3(n-1)+1 }{ 6{ (n-1) }^{ 2 }{ (n) }^{ 3 } } } =\sum _{ n=2 }^{ \infty }{ \left( \frac { 1 }{ 3{ (n-1) }^{ 3 } } -\frac { 1 }{ 3{ n }^{ 3 } } \right) }$

And look! A telescoping series!

Evaluating that gives $\int _{ -1 }^{ 1 }{ \mod{(x,{ x }^{ 2 })} } dx=\boxed{\frac{1}{3}}$

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Additional:

$\int _{ -1 }^{ 0 }{ \mod{(x,{ x }^{ 2 })} } dx=\frac { 1 }{ 6 } \sum _{ n=2 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 2 } } \right) } -\frac { 1 }{ 6 } \sum _{ n=1 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 2 } } \right) } +\frac { 1 }{ 3 } \sum _{ n=1 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 3 } } \right) } =\frac { 1 }{ 6 } \left( \frac { { \pi }^{ 2 } }{ 6 } -1 \right) -\frac { { \pi }^{ 2 } }{ 36 } +\frac { \zeta (3) }{ 3 } =\frac { \zeta (3) }{ 3 } -\frac { 1 }{ 6 }$

$\int _{ 0 }^{ 1 }{ \mod{(x,{ x }^{ 2 })} } dx=\frac { 1 }{ 6 } \sum _{ n=1 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 2 } } \right) } -\frac { 1 }{ 6 } \sum _{ n=2 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 2 } } \right) } -\frac { 1 }{ 3 } \sum _{ n=2 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 3 } } \right) } =-\frac { 1 }{ 6 } \left( \frac { { \pi }^{ 2 } }{ 6 } -1 \right) +\frac { { \pi }^{ 2 } }{ 36 } -\frac { \zeta (3) }{ 3 } +\frac { 1 }{ 3 } =\frac { 1 }{ 2 } -\frac { \zeta (3) }{ 3 }$

This is what we need to find -

$\displaystyle A = \int_{-1}^{1}{mod(x,{x}^{2}) \quad dx}$

Now, as it is obvious that $mod(a,b) = c \rightarrow a - c = kb \rightarrow c = a - kb$

where $k = \left \lfloor \frac{a}{b} \right \rfloor$

Hence, $mod(x,{x}^{2}) = x - \left \lfloor \frac{x}{{x}^{2}} \right \rfloor {x}^{2} = x - \left \lfloor \frac{1}{x} \right \rfloor {x}^{2}$

$\displaystyle A = \int_{-1}^{1}{x - \left \lfloor \frac{1}{x} \right \rfloor {x}^{2} \quad dx}$

$\displaystyle A = 0 - \int_{-1}^{1}{\left \lfloor \frac{1}{x} \right \rfloor {x}^{2} \quad dx}$

Substituting $u = \frac{1}{x}$

$\displaystyle A = \int_{-1}^{1}{\left \lfloor u \right \rfloor \frac{1}{{u}^{4}} \quad du}$

$\displaystyle A = 0 \int_{0}^{1}{\frac{1}{{u}^{4}} du} + (-1)\int_{-1}^{0}{\frac{1}{{u}^{4}} du} = \boxed{\frac{1}{3}}$