TITLE: Taking the derivative under a principal value integral
QUESTION [11 upvotes]: 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!

REPLY [8 votes]: 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.