Calculate the integral $\int_{-\infty}^\infty\frac{dy}{(1+y^2)(1+[x-y]^2)}$ 
Calculate the integral for $x\in\mathbb{R}$
  $$\int_{-\infty}^\infty\frac{dy}{(1+y^2)(1+[x-y]^2)}$$where $[\ ]$ is the floor function.

By using the fitting tool of MATLAB I'm almost certain that the answer is
$$\frac{2\pi}{(x-0.5)^2+4}$$compared with the result of an easier integral
$$\int_{-\infty}^{\infty}\frac{dy}{(1+y^2)(1+(x-y)^2)}=\frac{2\pi}{x^2+4}$$
Noting that the wanting integral is $f(x)*f([x])$ if $f(x)=1/(1+x^2)$, I tried to calculate $\mathcal{F}^{-1}(\mathcal{F}(f(x))\cdot\mathcal{F}(f([x])))$ by convolution theorem. But things didn't get any simpler.
 A: The proposed formula is based on approach
$$[t]\approx t-0.5,$$
so it isn't exact.
There is another way, based on exact formula
$$[x-y]=-k\quad\text{for}\quad y\in(x+k-1,x+k).$$
Then
$$J(x)=\int_{-\infty}^\infty\dfrac{dy}{(y^2+1)([y-x]^2+1}=\sum_{k=-\infty}^\infty\int_{x+k-1}^{x+k}\dfrac{dy}{(y^2+1)(k^2+1)}$$
$$=\sum_{k=-\infty}^\infty\dfrac{\arctan(x+k)-\arctan(x+k-1)}{k^2+1}$$
$$=\sum_{k=-\infty}^\infty\dfrac{\arctan(x+k)}{k^2+1} -\sum_{k=-\infty}^\infty\dfrac{\arctan(x+k-1)}{k^2+1}$$
$$=\sum_{k=-\infty}^\infty\left(\dfrac1{k^2+1}-\dfrac1{(k+1)^2+1}\right)\arctan(x+k)$$
$$=\sum_{k=-\infty}^\infty\left(\dfrac1{k^2+1}-\dfrac1{(k+1)^2+1}\right)\arctan(k+[x]+\{x\})$$
$$=\sum_{k=-\infty}^\infty\left(\dfrac1{(k-[x])^2+1}-\dfrac1{(k-[x]+1)^2+1}\right)\arctan(k+\{x\}).$$
Calculations with Wolfram Alpha give $$J(3.7)\approx0.446,\quad\text{while}\quad\frac{2\pi}{(3.7-0.5)^2+4}\approx0.441.$$
A: By using:
$$\forall a\in\mathbb{R},\qquad \frac{1}{1+a^2} = \int_{0}^{+\infty}\sin(t) e^{-|a|t}\,dt $$
we may compute first:
$$ \int_{-\infty}^{+\infty} e^{-|y|t} e^{-|\lfloor x-y \rfloor|s}\,dy = \int_{-\infty}^{+\infty}e^{-|x-y|t} e^{-|\lfloor y\rfloor|s}\,dy=\sum_{n\in\mathbb{Z}}\int_{n}^{n+1}e^{-|n|s} e^{-|x-y|t}\,dy$$
then integrate the resulting expression multiplied by $\sin(s)\sin(t)$ over $(0,+\infty)^2$.
