Hot Integral - 9

Since subtracting an integer doesn't change fractional part, $\{f(x)\}=\{x\}$ . Then

$\displaystyle\int_0^{\infty}\frac{\{f(x)\}}{e^{xy}}dx=\sum_{n=0}^{\infty}\int_0^1\frac{x}{e^{y(x+n)}}dx=\sum_{n=0}^{\infty}\left(-\frac{x}{ye^{y(x+n)}}-\frac{1}{y^2e^{y(x+n)}}\right)\bigg|_0^1$

$\displaystyle=\frac{1}{y^2}\sum_{n=0}^{\infty}\left(-\frac{1}{e^{y(n+1)}}+\frac{1}{e^{yn}}-\frac{y}{e^{y(n+1)}}\right)$ .

Note the first two parts of each term telescope, so we simplify to

$\displaystyle\frac{1}{y^2}\left(\frac{1}{e^0}-\frac{y}{e^y}\sum_{n=0}^{\infty}e^{-yn}\right)=\frac{1}{y^2}\left(1-\frac{y}{e^y}\frac{e^y}{e^y-1}\right)=\frac{1}{y^2}\left(1-\frac{y}{e^y-1}\right)$ .

Then $A=1,B=2,C=1,D=1,E=1,F=1,$ so the answer is $\boxed{7}$ .