# Infinity divided by infinity and dirac delta?

A dirac delta produces something that's infinitely long, and it could also be seen as infinitely thin. Why do we define the surface of a dirac delta to be 1. If length$\times$width $= \infty \cdot \frac{1}{\infty} = \infty/\infty$ is not by definition $1$. See: Whats infinity divided by infinity?

• Indeed, what if one integrates $3$ times the delta function? $\displaystyle\int_{-\infty}^\infty 3\delta(x)\,dx=3$, so in this case $\infty\cdot\dfrac1\infty=3$. ${}\qquad{}$ – Michael Hardy Apr 12 '13 at 16:19
• If it would be one - which it isn't - Can't we rewrite that one like this: $3 \cdot \infty \cdot \frac{1}{\infty} = 3 \cdot 1 = 3$? – user1095332 Apr 12 '13 at 16:21
• The point is that if you multiply something approaching $\infty$ by something approaching $0$, the product can approach any number or $\pm\infty$, depending on what the two things are that are approaching those limits. – Michael Hardy Apr 12 '13 at 16:29

One way to think about this might be to consider approximations to the delta function. Consider $$\varphi(x) = \frac{1}{2\pi}e^{-x^2/2},\text{ and }\varphi_a(x) = a\varphi(ax)\qquad (a>0).$$ Then, as with the delta function, we have $$\int_{-\infty}^\infty \varphi_a(x)\,dx=1,$$ and $\varphi_a(x)\to 0$ as $a\to\infty$, unless $x=0$, in which case $\varphi_a(x)\to\infty$ as $a\to\infty$. Just as the convolution of $f$ with $\delta$ is $f$, so also the convolution of $f$ with $\varphi$ approaches $f$ as $a\to\infty$. And $$\int_{-\infty}^\infty f(x)\varphi_a(x)\,dx \to f(0)\text{ as }a\to\infty.$$ So we're multiplying something approaching $\infty$ by something approaching $0$, and getting $1$. But if you consider $3\varphi_a$ instead of $\varphi_a$, then we're multiplying something approaching $\infty$ by something approaching $0$, and getting $3$.
So $0\cdot\infty$ is still indeterminate.
I believe the best way to understand the behavior of the Dirac delta function is to start with: $$\delta(a, x) = \left\{ \begin{array}{ll}\frac x{a^2} + \frac 1a & -a\le x < 0 \\ \frac {-x}{a^2} + \frac 1a & 0\le x < a \\ 0 & \mbox{otherwise}\end{array}\right.$$
This function defines a triangle with base $2a$ and height $\frac 1a$ centered at $0$. In particular, it is easy to check that $\displaystyle{\int_{-\infty}^{\infty} \delta (a, x) dx} = 1$ for any value of $a$. In particular, if $a\to 0$, $\delta(a, x)$ tends to a "function" $\delta(x)$ which is $0$ everywhere, except at $x=0$, where its value is $\infty$. But the most interesting thing is that, by construction, $\displaystyle{\lim_{a\to 0}\int_{-\infty}^{\infty} \delta (a, x) dx} = \displaystyle{\int_{-\infty}^{\infty} \delta (x) dx} = 1$.