I understand the Dirac Delta is the limit of a normal distribution when the variance of the normal distribution tends to 0: $$ \delta(x) = \lim_{v\to 0}\frac{e^{-x^2/2v}}{\sqrt{2\pi v}} $$ Then what is the limit $$ \lim_{v\to 0} \frac{e^{-x^2/2v}}{\sqrt{2\pi v^n}} $$ for some integer $n$? Can we expand the normal density function, or an integral with the normal integrator, into perhaps a power series of the variance v, which is assumed to be small?

By the way, I typed the first limit into Mathematica9, but didn't really get the Dirac Delta.


The statement $$\delta(x) = \lim_{v\to 0}\frac{e^{-x^2/2v}}{\sqrt{2\pi v}}$$ means that for every $C^\infty$ smooth function $\varphi$ with compact support we have $$\varphi(0) = \lim_{v\to 0} \int_{-\infty}^\infty \frac{e^{-x^2/2v}}{\sqrt{2\pi v}} \varphi(x)\,dx \tag{1}$$

(I would not expect computer algebra systems to correctly handle various modes of function convergence, especially convergence in the sense of distributions.)

Let's replace $v$ by $v^n$ under the square root in (1); I will consider real $n$, not just integers. The resulting limit can be written as $$ \lim_{v\to 0} v^{(1-n)/2} \int_{-\infty}^\infty \frac{e^{-x^2/2v}}{\sqrt{2\pi v}} \varphi(x)\,dx \tag{2}$$ where the integral converges to $\varphi(0)$, as previously professed. There are two cases:

  • $n<1$. Then the limit is zero. So, the functions $\lim_{v\to 0} \frac{e^{-x^2/2v}}{\sqrt{2\pi v^n}}$ converge to zero in the sense of distributions.
  • $n>1$. The limit does not exist, which means $\lim_{v\to 0} \frac{e^{-x^2/2v}}{\sqrt{2\pi v^n}}$ does not exist, even in the sense of distributions.

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