Given $f(y)= \frac{y}{y^2+2y+2}$
calulate $ \hat f(\omega)=\int_\mathbb{R} f(y) \exp(-iy\omega)$
I know that $\mathcal F\{\exp(-|x|)\}=\frac{1}{1^2+\omega^2}$, via Transformation $y=x-1$
$\frac{y}{y^2+2y+2} = \frac{x-1}{x^2+1}= \frac{x}{x^2+1}+\frac{-1}{x^2+1} = \frac{d}{dx} \frac{1}{2} \ln(x^2+1)+ \frac{-1}{x^2+1} $ , but I can't integrate the last term. I am very thankful for hints.
Greetings.
$$f(y)=\frac{y}{y^{2}+2y+2}=\frac{y}{(y+1-i)(y+1+i)}=\frac{1+i}{2(y+1-i)}+\frac{1-i}{2((y+1+i))}$$ Then $$\mathcal{F}[f(y)](\omega)=\int_{\mathbb{R}}f(y)e^{-i\omega{y}}dy=$$ $$=\frac{1+i}{2}\int_{\mathbb{R}}\frac{e^{-i\omega{y}}}{(y+1-i)}dy+\frac{1-i}{2}\int_{\mathbb{R}}\frac{e^{-i\omega{y}}}{(y+1+i)}dy$$ The first integrand has a pole at $y=-1+i$, so we intend an infinite semicircular contour in the upper half plane and the second integrand has a pole at $y=-1-i$, so we intend an infinite semicircular contour in the lower half plane The residue at the pole of the first integran is $$e^{(i+1)\omega}$$ The residue at the pole of the second integran is $$e^{(i-1)\omega}$$ Thus $$\int_{\mathbb{R}}\frac{e^{-i\omega{y}}}{(y+1-i)}dy=2\pi{i}e^{(i+1)\omega}$$ and $$\int_{\mathbb{R}}\frac{e^{-i\omega{y}}}{(y+1-i)}dy=2\pi{i}e^{(i-1)\omega}$$ So $$\mathcal{F}[f(y)](\omega)=\pi(i-1)e^{(i+1)\omega}+\pi(i+1)e^{(i-1)\omega}=2\pi{e^{i\omega}}\Big(i\cosh{\omega}-\sinh\omega\Big)$$