TITLE: Is it true that the integral of $\delta(x)/x$ between symmetrical limits is zero?
QUESTION [5 upvotes]: My professor is claiming that the following is true:
$$\int_{-\infty}^{\infty}\frac{\delta(x)}{x}dx=0,$$
where $\delta(x)$ is the Dirac delta "function", as he calls it. I think the integral diverges from the definition of the delta "function", but his rational is that the solution must be zero because the integrand is "an odd function of $x$".
I think that if he is correct then the definition of the delta distribution is basically meaningless.I know it is true that $\delta(x)=\delta(-x)$, but I think that the reason his explanation that the integrand is an odd function fails because because the delta distribution isn't a function at all, and (presumably) distributions don't have this usual integration property. 
Would somebody please confirm or deny, and if possible explain why a distribution doesn't have to integrate to zero in the same way as a function when it is odd-valued?

REPLY [3 votes]: As written in the other answers, "$\delta_0(x)/x$" does not mean a priori anything (since there is no defintion for the product of two singular distributions in general). In a general procedure, one can only define the multiplication of $\delta_0$ with functions $g$ continuous at $0$ as the distribution acting on smooth functions $\varphi$ as
$$
\langle g\,\delta_0,\varphi\rangle = g(0)\,\delta_0.
$$
However, one can look in the particular case of "$\delta_0(x)/x$" what is the distribution $T$ that has the closest properties to what we would expect. A way is to take $T$ as a solution of the equation
$$
x\,T(x) = \delta_0(x)
$$
However there are several solutions to this equation (since if $T$ is solution, then $T + c\, \delta_0$ is also solution). An additional constraint can be to require the solution to be homogeneous. In this case, we are left with only one solution (see e.g. here for more details):
$$
T = -\delta_0',
$$
which is a well defined distribution defined by $\langle T,\varphi\rangle = \varphi'(0)$. Remarking that this defines a linear functional not only on smooth  compactly supported functions, but also on $C^1$ functions, we can take $\varphi  = 1$, which yields
$$\boxed{
\langle T,1\rangle = 0}
$$
And this is the rigorous version of the result claimed by your professor. Indeed, with less rigorous notations, we could define the (possibly misleading) notation $"\frac{\delta_0(x)}{x}" := T = - \delta_0'$ and then the (possibly misleading) notation $"∫_{-\infty}^\infty T(x)φ(x)\,\mathrm d x" := \langle T,φ\rangle$. With these notations, $\langle T,1\rangle = 0$ becomes
$$
"∫_{-\infty}^\infty \frac{\delta_0(x)}{x}\,\mathrm d x" = 0.
$$
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