Find the Fourier transform of
$$f(x) = \frac{1}{x^2+6x+10} $$
By definition:
$$\mathcal{F}\{f(x)\} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-iwx}f(x)\mbox{d}x, w\in\mathbb{R} $$
So, if $w>0$ we can look at it as follows with $C:=[-R,R]\cup\{z : |z| = R,0<\mbox{arg}z<\pi\} =: \gamma _C$:
$$\oint_C e^{-iwz}f(z)\mbox{dz} $$
We have $z^2+6z+10 = (z-(-3+i))(z-(-3-i))$, on the upper half plane we have singularity at $-3+i$, therefore:
$$\oint_C e^{-iwz}f(z) = 2i\pi \frac{e^{-iw(-3+i)}}{2i} = \pi e^{(1+3i)w}\Longrightarrow \mathcal{F}\{f(x)\} = \sqrt{\frac{\pi}{2}}e^{(1+3i)w} $$
because as $R\to\infty$ the integral over $\gamma _C$ vanishes.
Wolfram Alpha says as $w>0$ we have the transform as $\sqrt{\frac{\pi}{2}}e^{-(1+3i)w}$, where does the minus come into play?
What do we do when $w<0$? $|e^{-iwz}|$ is no longer bounded, the same trick doesn't work, unless I'm missing something. Can we switch to similar contour on the lower half plane?
Idea: $w<0\Longrightarrow -w>0$. $\int_C e^{iwz}f(z)\mbox{d}z$, but as I mentioned, it won't work.
We define $$\hat{f}(\omega)=\mathcal F\{f(x)\}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i \omega x}\,\mathrm dx $$ and $$ f(x)=\frac{1}{x^2+6x+10}=\frac{1}{(x+3)^2+1} $$ It's easy to find that $$\mathcal F\{\mathrm e^{-|x|}\}=\sqrt{\frac{2}{\pi}}\frac{1}{1+\omega^2}$$ and using the duality property $\mathcal F\{\hat f(x)\}=f(-\omega)$, where $\hat f$ denotes the Fourier transform of $f$, we have
$$\mathcal F\left\{\frac{1}{1+x^2}\right\}=\sqrt{\frac{\pi}{2}}\mathrm e^{-|\omega|}$$
and using the shift property $$\mathcal F\{f(x-a)\}=\mathrm e^{-i\omega a}\mathcal F\{f(x)\}$$ for $a=-3$ we have $$ \mathcal F\left\{\frac{1}{(x+3)^2+1}\right\}=\mathrm e^{i \omega 3}\sqrt{\frac{\pi}{2}}\mathrm e^{-|\omega|} $$
WolframAlpha defines the Fourier transform as $\mathcal F\{f(t)\}=\displaystyle \hat{f}(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(t) e^{+i \omega t}\, dt $