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I'm interested in showing that:

$$ \frac{d}{dt}P \; \int_{-\infty}^{\infty} \frac{\phi(x)}{x-t}dt = P \int_{-\infty}^{\infty}\frac{\phi(x)-\phi(t)}{(x-t)^2}dt $$

where $\phi(x)$ is a test function (goes to zero at $-\infty$ and $\infty$ fast enough that we don't have to worry about x not going to zero fast enough)

The problem is that when I attempt this:

$$ \frac{d}{dt}P \; \int_{-\infty}^{\infty} \frac{\phi(x)}{x-t}dt = \lim_{\epsilon \rightarrow 0} \left\{ -\frac{\phi(t+\epsilon)}{\epsilon}-\frac{\phi(t-\epsilon)}{\epsilon} +\int_{-\infty}^{t-\epsilon}\frac{\phi(x)}{(x-t)^2} + \int_{t+\epsilon}^{\infty}\frac{\phi(x)}{(x-t)^2} \right\} $$

But I don't know where to go from here... and I'm not sure I'm on the right track. The $\phi(t)$ term doesn't seem to want to pop out.

Could this be some sort of dirac delta identity I'm missing?

Thanks!

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Differentiation under the integral sign still works for Cauchy principal values. –  J. M. Nov 4 '11 at 3:11
    
ok, so where does the $\phi(t)$ come from? And I'm integrating over a singularity... –  Andrew Spott Nov 4 '11 at 3:14
    
how do you end up with P∫∞−∞dx 1/(x−t)^2=2/ϵ? I got 1/(ϵ+t)+1/(ϵ-t) –  user108693 Nov 14 '13 at 3:40
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2 Answers 2

up vote 6 down vote accepted

I will assume you are trying to show $$\frac{d}{d t} \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)}{x-t} = \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)-\phi(t)}{(x-t)^2}$$ Using the definition of principal value and the Leibniz integral rule you will find $$\frac{d}{d t} \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)}{x-t} = -\lim_{\epsilon\rightarrow 0}\frac{2}{\epsilon} \phi(t) + \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)}{(x-t)^2}$$ Then notice that $$\mathrm{P} \int_{-\infty}^\infty dx \ \frac{1}{(x-t)^2} = \frac{2}{\epsilon}$$ The result follows immediately.

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One can begin with $$\frac{d}{dt}P \; \int_{-\infty}^\infty \frac{\phi(x)-\phi(t)}{x-t}dx.$$

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