Fourier transform of Singular Function $f(x)$ with $\frac{1}{x^2}$ as $x\to 0^{+}$ and $\frac{1}{x^2} +\frac{1}{x}$ as $x\to 0^{-}$ In Lighthill's `An Introduction to Fourier Analysis and Generalised Functions', the Fourier transform of a function $f(x)$ is defined as:
$$
\mathscr{F}[f](y) \ = \ \int_{-\infty}^{\infty}\ f(x) e^{-2\pi i x y} dx
$$
He uses the notion of generalised functions, like the Dirac Delta and the Heaviside function to assign Fourier transforms to functions that would otherwise yield divergent Fourier integrals (in the strict sense).
In the book, one of the results stated is that the function $\ f_{m}(x) = x^{-m}$ for integers $m\geq 1$ has the following Fourier transform:
$$
\mathscr{F}[f_{m}](y) = - i \pi \frac{(- 2 \pi i y)^{m-1}}{(m-1)!}\mathrm{sgn}(y)
$$
Which means for example that:
$$
f_{1}(x) = \frac{1}{x} \ \ \ \implies \ \ \ \mathscr{F}[f_{1}](y) \ = \ - i \pi \mathrm{sgn}(y) \\
f_{2}(x) = \frac{1}{x^2} \ \ \ \implies \ \ \ \mathscr{F}[f_{2}](y) \ = \ - 2 \pi^2 |y|
$$
My question is the following, what if the function $f(x)$ has a singularity at $x=0$ and the dependence of the function is different as $x \to 0^{+}$ or $x \to 0^{-}$? The function I have in mind is the following:
$$
f(x) \ = \ \frac{1}{x^2} + \frac{\Theta(-x)}{x} \ = \ \begin{cases} \ \frac{1}{x^2}\ \ \ \ \ \ \ \ \ \ \ , \ x>0\\
 \ \frac{1}{x^2} + \frac{1}{x}\ \ \ , \ x<0\\
\end{cases}
$$
This function is differently singular depending on which way you approach the singularity. Is there still a way to assign a Fourier transform to this function?
 A: Yes, it has a  meaning. By additivity, it suffices to consider the function $\Theta(-x)/x$. Let $F$ be its Fourier transform. Recall that multiplication by $x$ is equivalent to taking the derivative on Fourier transform side, up to a constant. Formally,
$$
\int_{\mathbb{R}} xf(x)e^{-2\pi i xy }\,dx =  -\frac{1}{2\pi i} \frac{d}{dy}\int_{\mathbb{R}} f(x)e^{-2\pi i xy }\,dx
$$
Thus, the Fourier transform of $\Theta(-x)$ is $-\dfrac{1}{2\pi i} F'(y)$. 
On the other hand, $\Theta(-x)$ is  $\dfrac12 - \dfrac12\operatorname{sgn} x$, the transform of which comes out as $\dfrac12 \delta_0 - \dfrac{1}{2 i\pi y}$. So, 
$$
-\dfrac{1}{2\pi i} F'(y) = \dfrac12 \delta_0 - \dfrac{1}{2i \pi y}
$$
hence 
$$
F(y) = -\frac{\pi i}{2} \operatorname{sgn} y + \log|y|  + C
$$
for some constant $C$. Figuring out $C$ is not as easy; it turns out to be Euler's constant $\gamma$. I refer to the notes Homogeneous Distributions and The Fourier Transform by Ethan Y. Jaffe: Theorem 2.3, take $s=1$ which corresponds to $k=0$. The formula given there simplifies to
$$
\widehat{x_+^{-1}} = \gamma + \frac{\pi i}{2} \operatorname{sgn}y + \log|y|
$$
where $x_+ = \max(x, 0)$. Your situation differs by reflecting $f$, which amounts to conjugating the Fourier transform.
A: First you need to define $\Theta(-x)/x$. It has to be defined as a functional, since it's not in the space of well-behaved test functions. The natural definition is
$$\left( \frac {\Theta(-x)} x, \phi \right) =
\int_{-\infty}^{-1} dx \frac {\phi(x)} x +
 \int_{-1}^0 dx \frac {\phi(x) - \phi(0)} x,$$
which is the distributional derivative of $\ln(-x)\Theta(-x)$. In principle, you can have other definitions, differing from this one by $\delta(x)$ and its derivatives, and those functionals will also be valid regularizations of the function $\Theta(-x)/x$.
Once you have the definition established, the answer is straightforward, because now you just need to apply the functional to $e^{-2\pi i y x}$ and carry out the integration of the ordinary functions:
$$\int_{-\infty}^{\infty}dx \frac {\Theta(-x)} x e^{-2\pi i y x} =
\ln(2 \pi |y|) - \frac {i \pi \operatorname{sgn} y} 2 + \gamma.$$
