I'm interested in the following integral $$ f(a,b) = \int dx \int dy e^{i a x } e^{i b y} \frac{1}{x ( x + y ) } $$ The integral is over the reals. We use the property $$ \int dx e^{i a x} \frac{1}{x} = i \pi s(a) , \qquad s(a) = \text{sign}(a) . $$ Let's the do the $y$ integral first followed by the $x$ one. Then, \begin{align} f(a,b) &= \int dx e^{i a x } \frac{1}{x} \int dy e^{i b y} \frac{1}{x + y} \\ &= \int dx e^{i a x } \frac{1}{x} \int dy e^{i b ( y - x ) } \frac{1}{y} \\ &= \int dx e^{i ( a - b ) x } \frac{1}{x} \int dy e^{i b y } \frac{1}{y} \\ &= ( i \pi )^2 s(a-b) s(b) . \end{align} Instead, let us do the $x$ integral first following by the $y$ one. To do this, we write $$ \frac{1}{x(x+y)} = \frac{1}{xy} - \frac{1}{y(y+x)} $$ Then, \begin{align} f(a,b) &= \int dy e^{i b y} \int dx e^{i a x } \left[ \frac{1}{xy} - \frac{1}{y(y+x)} \right] \\ &= \int dy e^{i b y} \int dx e^{i a x } \frac{1}{xy} - \int dy e^{i b y} \int dx e^{i a x } \frac{1}{y(y+x)} \\ &= \int dy e^{i b y} \frac{1}{y} \int dx e^{i a x } \frac{1}{x} - \int dy e^{i b y} \frac{1}{y} \int dx e^{i a x } \frac{1}{y + x } \\ &= ( i \pi )^2 s(a)s(b) - \int dy e^{i ( b - a ) y} \frac{1}{y} \int dx e^{i a x } \frac{1}{x} \\ &= ( i \pi )^2 s(a)s(b) - (i \pi )^2 s(b-a) s(a) \end{align} The two answers aren't equal - rather they differ by a constant $$ s(a-b) s(b) \color{red}{+ 1} = s(a)s(b) - s(b-a) s(a) $$ What's going on?
PS - I know I'm being pretty lax about many things so some guidance will be helpful.
I can now show that $\frac{1}{x(x+y)} + \frac{1}{(x+y)y} - \frac{1}{xy}$ does not vanish under integration but has a "residue" of $\pi^2$ at origin.
Let $\delta,\epsilon>0$ and integrate each term over the rectangle $R=[-\delta,\delta]\times[-\epsilon,\epsilon].$
For $\frac{1}{xy}$ this is trivial: $$ \iint_{R} \frac{1}{xy} \ dx \ dy = \left( \int_{-\delta}^{\delta} \frac{1}{x} \ dx \right) \left( \int_{-\epsilon}^{\epsilon} \frac{1}{y} \ dy \right) = 0. $$ Note that we are here using the principal value integral.
For $\frac{1}{x(x+y)}$ we get $$ \iint_{R} \frac{1}{x(x+y)} \ dx \ dy = \int_{-\delta}^{\delta} \frac{1}{x} \left( \int_{-\epsilon}^{\epsilon} \frac{1}{x+y} \ dy \right) \ dx = \int_{-\delta}^{\delta} \frac{1}{x} (\ln|x+\epsilon| - \ln|x-\epsilon|) \ dx \\ = \int_{-\delta}^{\delta} \frac{1}{x} \ln|x+\epsilon| \ dx - \int_{-\delta}^{\delta} \frac{1}{x} \ln|x-\epsilon| \ dx . $$ Now, setting $x=\epsilon t$ we get $$ \int_{-\delta}^{\delta} \frac{1}{x} \ln|x+\epsilon| \ dx = \int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{\epsilon t} \ln|\epsilon t+\epsilon| \ \epsilon \ dt = \int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{t} (\ln|t+1| + \ln\epsilon) \ dt \\ = \int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{t} \ln|t+1| \ dt + \int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{t} \ dt \ \ln\epsilon = \int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{t} \ln|t+1| \ dt $$ since the integral for the $\ln\epsilon$ term vanishes.
Likewise, setting $x=-\epsilon t$ we get $$ \int_{-\delta}^{\delta} \frac{1}{x} \ln|x-\epsilon| \ dx = \int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{t} \ln|t+1| \ dt. $$
Thus, $$ \iint_{R} \frac{1}{x(x+y)} \ dx \ dy = 2\int_{-\delta/\epsilon}^{\delta/\epsilon} \frac{1}{t} \ln|t+1| \ dt. $$
For $\frac{1}{(x+y)y}$ we similarly get $$ \iint_{R} \frac{1}{(x+y)y} \ dx \ dy = 2\int_{-\epsilon/\delta}^{\epsilon/\delta} \frac{1}{t} \ln|t+1| \ dt. $$
Thus, $$ \iint_R \left( \frac{1}{x(x+y)} + \frac{1}{(x+y)y} - \frac{1}{xy} \right) \ dx \ dy = 2 \left( \int_{-\epsilon/\delta}^{\epsilon/\delta}+\int_{-\delta/\epsilon}^{\delta/\epsilon} \right) \frac{1}{t} \ln|t+1| \ dt $$ which is of the form $$ I := 2 \left( \int_{-\alpha}^{\alpha}+\int_{-1/\alpha}^{1/\alpha} \right) \frac{1}{t} \ln|t+1| \ dt. $$
Using substitution $t=1/s$ we get $$ \int_{-1/\alpha}^{1/\alpha} \frac{1}{t} \ln|t+1| \ dt = \lim_{\eta\to 0}\left(\int_{-1/\alpha}^{-\eta}+\int_{\eta}^{1/\alpha}\right) \frac{1}{t} \ln|t+1| \ dt = \lim_{\eta\to 0}\left(\int_{-\alpha}^{-1/\eta}+\int_{1\eta}^{\alpha}\right) \frac{1}{1/s} \ln\left|\frac{s+1}{s}\right| \ \frac{-ds}{s^2} = -\lim_{\eta\to 0} \left(\int_{-\alpha}^{-1/\eta}+\int_{1/\eta}^{\alpha}\right) \frac{1}{s} \ln|s+1| \ ds + \lim_{\eta\to 0} \left(\int_{-\alpha}^{-1/\eta}+\int_{1/\eta}^{\alpha}\right) \frac{1}{s} \ln|s| \ ds = \lim_{\eta\to 0} \left(\int_{-1/\eta}^{-\alpha}+\int_{\alpha}^{1/\eta}\right) \frac{1}{s} \ln|s+1| \ ds $$ since the second limit vanishes.
Thus, $$ I = 2 \lim_{\eta\to 0} \left( \int_{-1/\eta}^{-\alpha} + \int_{-\alpha}^{\alpha} + \int_{\alpha}^{1/\eta} \right) \frac{1}{s} \ln|s+1| \ ds = 2 \lim_{\eta\to 0} \int_{-1/\eta}^{1\eta} \frac{1}{s} \ln|s+1| \ ds $$ which is not dependent on $\alpha.$
Taking $\alpha=1$ we get $$ I = 2 \left( \int_{-1}^{1}+\int_{-1}^{1} \right) \frac{1}{t} \ln|t+1| \ dt = 4 \int_{-1}^{1} \frac{1}{t} \ln|t+1| \ dt = 4 \int_{-1}^{1} \frac{1}{t} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}t^k}{k} \ dt \\ = 4 \sum_{k=1}^{\infty} (-1)^{k+1} \int_{-1}^{1} \frac{t^{k-1}}{k} \ dt = 4 \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1^{k}-(-1)^{k}}{k^2} = 4 \sum_{k=1\\ \text{$k$ odd}}^{\infty} (-1)^{k+1} \frac{2}{k^2} = 8 \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = 8 \cdot \frac{\pi^2}{8} = \pi^2. $$
Thus, over every rectangle $[-\delta,\delta]\times[-\epsilon,\epsilon]$ the integral of $\frac{1}{x(x+y)} + \frac{1}{(x+y)y} - \frac{1}{xy}$ has value $\pi^2.$ This means that the sum at least has a term $\pi^2 \ \delta(x,y).$ There could also be derivatives of $\delta(x,y)$ but the Fourier transform in the question shows that this is not the case.