# Dirac delta function in reciprocal function

1. Oct 22, 2011

### jackychenp

From dirac, if A=B, then $\frac{A}{x}=\frac{B}{x}+c\delta(x)$ (1) How this formula is derived?

Since $\frac{dlnx}{dx} = \frac{1}{x}-i\pi\delta(x)$
We can get $\frac{A}{x} = A\frac{dlnx}{dx}+Ai\pi\delta(x)$
$\frac{B}{x} = B\frac{dlnx}{dx}+Bi\pi\delta(x)$
So if A=B, $\frac{A}{x}=\frac{B}{x}.$

Another argument is if we integrate the equation (1) from -a to a, a->$\infty$ and assume in a small region $[-\varepsilon, \varepsilon ]$, $\int_{-\varepsilon}^{\varepsilon}\frac{1}{x}dx=0,$ so we can get $\int_{-a}^{a}\frac{1}{x}dx=0, but \int_{-a}^{a}c\delta(x)dx=c,$ so the left side of equation (1) doesn't equal to the right side. Please correct me if I am wrong!

2. Oct 23, 2011

### Mute

I'm not sure what you're getting at. It seems like you're talking about the identity

$$\lim_{\epsilon \rightarrow 0} \frac{1}{x\pm i\epsilon} = \mathcal P \frac{1}{x} \mp i\pi \delta(x),$$
where $\mathcal P$ denotes a principle value.

Beyond that, though, I'm not sure what your question is.

3. Oct 23, 2011

### jackychenp

Hi Mute,

In Dirac's book, he demonstrates that if A=B, one can't infer A/x=B/x, but only A/x=B/x+cδ(x). I cannot get the latter result. And if we integrate the latter equation, it doesn't look correct.