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

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?

• 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 1 '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 1 '17 at 21:20

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.