Integration of Weird Functions - 4

Converting the Integral into a summation:-

$\displaystyle \frac{1}{8}\sum_{r=0}^{7}\sum_{m=0}^{7}\int_{r}^{r+1}(y-r)^{m}dy$

Substitute $y-r = z$

$\displaystyle \frac{1}{8}\sum_{r=0}^{7}\sum_{m=0}^{7}\int_{0}^{1}y^{m}dy = \frac{1}{8}\sum_{r=0}^{7}\sum_{m=0}^{7}\frac{1}{m+1} = \left(\frac{1}{1}+\frac{1}{2}+......+\frac{1}{8}\right)$

In general for $\displaystyle \int_{0}^{n}\int_{0}^{n}\{y\}^{\lfloor x \rfloor}dxdy = nH_{n}$

Where $H_{n}$ denotes the $nth$ harmonic number .

Consider $I_n$ as follows:

$\begin{aligned} I_n & = \int_0^n \int_0^n \{y\}^{\lfloor x \rfloor} dx\ dy \\ & = \int_0^n \sum_{k=0}^{n-1} \{y\}^k dy \\ & = n \sum_{k=0}^{n-1} \int_0^1 y^k dy \\ & = nH_n & \small \blue{\text{where }H_n \text{ denotes the }n \text{th harmonic number.}} \\ \implies \frac {I_n}n & = H_n \end{aligned}$

Therefore $\dfrac {I_8}8 = H_8 = \dfrac {781}{280}$ , $\implies A+B = 781+280 = \boxed{1041}$ .