In my project, I am stuck with the following integral. Please help. $$ I = \int_{x = a}^b \int_{y = a}^b \left\{ \frac xy\right\} \left\{\frac >yx\right\}~dx~dy$$ $0 < a < b < 1, b < 2a$, where $\lbrace \rbrace$ is the fractional part function.
I found an approximation of the integral calculated on the square $(0,\alpha)^2, \alpha < 1$ on your forum. I tried using the known values of the integral on the squares $(0,a)^2, (0,b)^2$ in calculation of the integral in discussion. Let $f(x,y)=\left\{ \frac xy\right\} \left\{\frac yx\right\}$, then $$\int_{x = a}^b \int_{y = a}^b f(x,y)~dx~dy = \int_{x = 0}^b \int_{y = 0}^b f(x,y)~dx~dy- \int_{x = 0}^a \int_{y = 0}^a f(x,y)~dx~dy - 2 \int_{x = 0}^b \int_{y = 0}^a f(x,y)~dx~dy$$
I get stuck trying to follow the idea used in calculation of the integrals on the squares $(0,a)^2$ and $(0,b)^2$ in calculation of the last one. Even taking $b = 2a$ was not useful.
I hope I'll have the chance to receive a response.
Thank you in advance.
Al
Split the region $R=(a,b)^2$ into two: $R_1:=\{(x,y)\in R:x>y\},R_2:=\{(x,y)\in R:x<y\}$.
$$\begin{align*}I_{R_1}&=\int_{x = a}^b \int_{y = a}^x \frac yx\left(\frac xy-1\right)~dy~dx\\& =\int_{x = a}^b \int_{y = a}^x\left(1-\frac yx\right)~dy~dx\\& =\frac{b^2+3a^2-4ab}4+\frac{a^2}2\ln\left(\frac ba\right) \end{align*}$$
$$I_{R_2}=\int_{y=a}^b \int_{x = a}^y \frac xy\left(\frac yx-1\right)~dx~dy=I_{R_1}$$Thus $I=I_{R_1}+I_{R_2}=\frac{b^2+3a^2-4ab}2+a^2\ln\left(\frac ba\right)$.