INTEGRATING ROUNDING FUNCTIONS-(II)

23 03 2009

SQUARE FRACTIONAL PART DEFINITE INTEGRAL:

For any $\displaystyle x\geq 0$:

$\displaystyle I_1= \int_0^x \left\{{x}\right\}^2 dx = \int_0^{\lfloor x \rfloor} \left\{{x}\right\}^2 dx+ \int_{{\lfloor x \rfloor}}^ {x} \left\{{x}\right\}^2 dx$

The function $\displaystyle \left\{{x}\right\}^2$ has a periodical behaviour, so the limits can be reduced to the first period:

$\displaystyle I_1= \lfloor x \rfloor \int_0^{1} {\left\{{x}\right\}^2 dx}+ \int_0^{\left\{{x}\right\}}{ \left\{{x}\right\}^2 dx} = \lfloor x \rfloor \int_0^1 x^2 dx+ \int_0^{\left\{{x}\right\}} {x^2 dx}$

$\displaystyle I_1= \frac{ \lfloor x \rfloor + \left\{{x}\right\}^3}{3}$

$\displaystyle I_1= \int_0^x \left\{{x}\right\}^2 dx= \frac{x+ \left\{{x}\right\}^{3}-\left\{{x}\right\}}{3}$

POWER FRACTIONAL PART DEFINITE INTEGRAL:

The same formula can be extended to any given positive integer power:

$\displaystyle I_2= \int_0^x \left\{{x}\right\}^n dx= \frac{x+ \left\{{x}\right\}^{n+1}-\left\{{x}\right\}}{n+1}$

$\displaystyle I_2= \frac{ \lfloor x \rfloor + \left\{{x}\right\}^{n+1}}{n+1}$