Is it true that the integral of $\delta(x)/x$ between symmetrical limits is zero?

Question

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?

Answer 1

I think your interesting question is almost on the same footing as asking what the value of the following integral is: $$ I=\int_{\mathbb{R}}dx\frac{x}{1+x^2}\ . $$ Of course, you have an odd integrand, but that doesn't per se mean that $I=0$ - the integral in fact is truly undefined (it is an odd raw moment of the Cauchy distribution, see discussion at https://en.wikipedia.org/wiki/Cauchy_distribution#Higher_moments). However, you can assign it a meaningful (finite) value by considering Cauchy's principal value. In the same way, your integral is per se undefined, as the function $1/x$ is not defined (hence it is not continuous) at the value $x=0$ which would be singled out by the Dirac delta. However, you can assign it the value $0$, for example by i) regularizing the delta function with its nascent version, and ii) excluding the non-integrable singularity $\sim 1/x$ at the origin $$ \int_{-\infty}^{\infty}\frac{\delta(x)}{x}dx\rightarrow\lim_{a\to 0}\lim_{\delta\to 0}\int_{\mathcal{D}_\delta}dx\ \frac{\exp \left(-\frac{x^2}{a^2}\right)}{\sqrt{\pi } a x}=0\ , $$ where $$ \mathcal{D}_\delta=(-\infty,-\delta)\cup(\delta,\infty). $$ In summary, you are assigning a 'meaningful meaning' to an object in principle undefined - exactly as you would do for the integral $I$ above.

Answer 2

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 $$ "∫_{-\infty}^\infty \frac{\delta_0(x)}{x}\,\mathrm d x" = 0. $$