How to find this integral $I=\int_{0}^{+\infty}\frac{\{t\}(\{t\}-1)}{1+t^2}dx$?

The function $$f(t):={1\over2}\{t\}(\{t\}-1)$$ is periodic with period $1$ and is $\ ={1\over2}(t^2-t)$ for $0\leq t\leq1$, whence continuous. It can be developed into a Fourier series as follows: $$f(t)=-{1\over12}+\sum_{k=1}^\infty{1\over 2\pi^2 k^2}\>\cos(2k\pi t)\ .$$ Therefore we obtain $$\eqalign{I&:=\int_0^\infty{f(t)\over 1+t^2}\ dt=-{1\over12}\int_0^\infty{1\over 1+t^2}\ dt +\sum_{k=1}^\infty{1\over 2\pi^2 k^2}\int_0^\infty{\cos(2k\pi t)\over 1+t^2}\ dt\cr &\ =-{\pi\over24}+\sum_{k=1}^\infty{1\over 4\pi k^2}e^{-2\pi k}\quad .\cr}$$ Here we have used the well known integral $$\int_0^\infty{\cos(c\>t)\over 1+t^2}\ dt={\pi\over2}e^{-c}\qquad(c\geq0)\ .$$ The resulting sum cannot be expressed in elementary terms, but it is extremely well convergent. Taking the first six terms we obtain the numerical approximation $$I\doteq -0.13075101809261383373\ .$$