Integral evaluation with delta Dirac

Question

I am having doubts about the following integral:$$\int \limits _{0}^{10} \int \limits _{0}^{10} \frac{x^2y^2}{(x^2+y^2)^{5/2}}\ \delta(x)\ \mathrm{d}x\mathrm{d}y$$ If we apply the definition of the Delta Dirac function we should get: $$\int \limits _{0}^{10}\int \limits _{0}^{10} \frac{x^2y^2}{(x^2+y^2)^{5/2}}\ \delta(x)\ \mathrm{d}x\mathrm{d}y=\int \limits _{0}^{10} 0\ \mathrm{d}y=0$$ Nevertheless, when one plots the 3D function: $z(x,y)=\frac{x^2y^2}{(x^2+y^2)^{5/2}}$ and intersect it with the plane $x=0$, what he gets is:

1) $z(0,y)=0$ for $y\neq 0$

2) $z(0,0)=\lim \limits _{x,y\rightarrow0}z(x,y)=- \infty$

so basically $$z(0,y)=-\delta\ (y)$$ and therefore: $$\int \limits _{0}^{10}\int \limits _{0}^{10} \frac{x^2y^2}{(x^2+y^2)^{5/2}}\ \delta(x)\ \mathrm{d}x\mathrm{d}y=-\int \limits _{0}^{10} \delta\ (y)\ \mathrm{d}y=-H(0)$$ N.B. If I choose any negative number as the lower integration extremum for the variable $y$ the final result is obviously $-H(a)=1\neq 0$. I do not know which of the two approaches is right, or if they are both wrong and in this case, what is the correct one?

Answer 1

Let us re-name the inner integral in terms of a mapping $$ g:[0,10]\to\mathbb R\qquad y\mapsto \int_0^{10}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\delta(x)\,dx=\int_0^{10}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\delta(dx) $$ with $\delta$ being the Dirac measure (which is not all too relevant for this question, I just wanted to clarify that whatever we do with the so-called "delta function" is rigorously defined when one considers measure theory). Anyways, we get $g(y)=0$ for all $y\in[0,10]$ (actually for all $y\in\mathbb R$) as is readily verified. Then $$ \int_0^{10} \int_0^{10}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\delta(dx) \,dy=\int_0^{10} g(y)\,dy=0 $$ as you correctly claimed in the beginning.

Regarding the arguments you further make there are two problems:

1. $\lim_{(x,y)\to (0,0)} z(x,y)$ does not exist as $$ \lim_{n\to\infty}z\Big(\frac1n,0\Big)=\lim_{n\to\infty}0=0 $$ but $$ \lim_{n\to\infty} z\Big(\frac1n,\frac1n\Big)=\lim_{n\to\infty} \frac{n}{2^{5/2}}=\infty $$ so the path on which you approach the origin does matter for the limit, which would not happen if the limit in question existed in the usual sense. (One would say that $z$ is discontinuous at the origin or that $z$ cannot be extended continuously into the origin).

2. The idea that $\delta(0)=\infty$, $\delta(x)=0$ if $x\neq 0$ and $\int_{\mathbb R}\delta(x)\,dx=1$ ís nothing more than a heuristic characterization simply to get an intuition what the dirac measure does - because no mapping on the real numbers satisfies the three above conditions at once. Therefore, one should refrain from using this for anything technical beyond the standard property $\int_{\mathbb R} f(x)\delta(x)\,dx=f(0)$ for $f$ continuous.