Dirac delta function in reciprocal function

[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!

 

[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.

 

[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.

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.