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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?

Lachy
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  • Can you clarify your notation, to me that would be interpreted as Usually that notation means $\delta\left(x^{-1}1_{[-R,R]}\right)=0$ for all $R>0$ where here we are applying the distribution (i.e. linear functional) $\delta$ to the measurable function $x^{-1}1_{[-R,R]}$, however that's not the case since this would give the value of $x^{-1}1_{[-R,R]}$ at $0$ which is undefined. – Adam Hughes Feb 01 '17 at 19:15
  • The delta function is not a function. It is not at all clear what meaning that integral you wrote has, and unless we know it is impossible to know if it is zero or not. – Mariano Suárez-Álvarez Feb 01 '17 at 21:20

2 Answers2

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

LL 3.14
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  • I was wondering about using the $PV$ distribution and a regularization that is an even function. This is invalid, of course, since one cannot restrict regularizations to even functions only. But physicists and engineers often impose such symmetries, which would lead to a conclusion that is flawed. Not sure if this will make sense but wanted to hear your thoughts. – Mark Viola Dec 12 '22 at 16:20
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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.

Pierpaolo Vivo
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