Integral with delta distribution kernel at singularity Let $\delta(x)$ be the Dirac delta distribution and suppose $f$ is a function with a singularity at $x = a$, that is, there exists a function $g$ which is nonsingular everywhere such that
$$
f(x) = \frac{g(x)}{x-a}.
$$ 
What is the value of 
$$
\int_{-\infty}^{\infty}f(x)\delta(x-a)dx,
$$
if the integral exists?
 A: From one definition of the Dirac Delta distribution (the 'sifting property'):
$\int_{-\infty}^{\infty}f(x)\delta(x-a)dx = f(a)$
which is singular by your definition in this case, so the integral is singular and in that sense does not exist.
A: NOTATION.  The symbol $1_{x\in A}$ denotes the function of $x$ that equals $1$ if $x\in A$ and $0$ otherwise.
I'll make a concrete example to show that the question is ill-posed. Set $a=0$ and consider $f(x)=\tfrac1x 1_{|x|\le 100}$.  Consider now the set $A:=[-1, -\tfrac12]\cup[\tfrac12, 1]$ and define a sequence of functions 
$$\delta_1(x):=1_{x\in A},\quad \delta_n(x):=n\delta_1(nx).$$ 
Notice that $\delta_n\to \delta$ in distributional sense. If the left-hand side integral of the following were well-defined, we should have
$$
\int_{\mathbb R} f(x)\delta_n(x)\, dx =\lim_{n\to \infty} \int_{\mathbb R} f(x)\delta_n(x)\, dx = \lim_{n\to\infty} \int_{\frac1{2n}\le |x|\le\frac1n} \frac{1}{x}\, dx=0,$$
because the integrand functions in the right-hand side are odd and bounded.
Now redo the same procedure with the set $B=[\tfrac12, \tfrac32]$, that is, define 
$$\tilde\delta_1(x):=1_{x\in B},\quad \tilde\delta_n(x):=n\tilde\delta_1(nx).$$ 
This sequence converges to $\delta$ too, however, we now have 
$$
\lim_{n\to\infty} \int_{\mathbb R} f(x)\tilde\delta_n(x)\, dx = \infty.$$
CONCLUSION. The question is ill-posed beyond repair.
