When and How Does an Integral Have a Cauchy Principal Value?

[CAF123]

Under what conditions does an integral have a cauchy principal value and how is it related to an integral having an integrable singularity?

E.g $$p.v \int_{-\delta}^{\delta} \frac{dz}{z} = 0$$ If I evaluate the integral along a semi circle in the complex plane I'll get $i \pi$. So the cauchy principal value seems to be the real part of this calculation which is zero.

Similarly, $$\int_{-\delta}^{\delta} dz \frac{\ln(z)}{z} = i \pi \ln (\delta) + \pi^2/2, $$ but I don't think it's correct to say that the cauchy principle value is pi^2/2 in this case because the imaginary part depends on delta still, i.e. the singularity was not integrable. Are these statements correct?

 

[aheight]

Basically the principal value exists when a symmetrical limit about the singular point exists. Consider the integrals:

$$\text{p.v.}\int_{-1}^1 \frac{1}{z}dz$$
$$\text{p.v.} \int_{-1}^1 \frac{1}{z^2}dz$$
$$\text{p.v.} \int_{-1}^1 \frac{1}{z e^z}dz$$

Just looking at the plots, intuitively you suspect the first one is zero, the second one does not exist and the third if it exists is probably negative right? I'll do the first one:

$$\int_{-1}^1 \frac{1}{z}dz=\lim_{\delta\to 0} \biggr\{\log(z)\biggr|_{-1}^{-\delta}+\log(z)\biggr|_{\delta}^1\biggr\}$$
$$=\lim_{\delta\to 0}\biggr\{\log(-\delta)-\log(-1)+\log(1)-\log(\delta)\biggr\}$$
Now using the branch $\log(z)=\ln|z|+i\theta,\;\; -\pi<\theta\leq \pi$, we have:
$$=\lim_{\delta\to 0}\biggr\{(\ln|\delta|+\pi i)-(\ln|1|+\pi i)+0-\ln|\delta|\biggr\}$$
$$=\lim_{\delta\to 0}\biggr\{\ln|\delta|-\ln|\delta|\biggr\}=0$$

Do the second one similarly and the third one you can do in Mathematica with special functions and should get a little less than -2.

 

[CAF123]

Thanks, what does the fact that $$\int_{-\delta}^{\delta} \frac{dz}{z-i\epsilon} = i \pi $$ (i.e delta independent) mean compared to say, $$\int_{-\delta}^{\delta} \frac{dz \ln(z-i\epsilon)}{z-i\epsilon} = i \pi \ln(\delta) + \pi^2/2$$ (i.e delta dependent)

In the first integral, the singularity at z=0 is not integrable yet the principal value exists while in the second case the singularity is not integrable and the principal value doesn't exist. As delta goes to zero in the first result, the rhs is well behaved while in the second it is diverging due to the $\ln(\delta)$.

 

[aheight]

Thanks, what does the fact that $$\int_{-\delta}^{\delta} \frac{dz}{z-i\epsilon} = i \pi $$ (i.e delta independent) mean compared to say, $$\int_{-\delta}^{\delta} \frac{dz \ln(z-i\epsilon)}{z-i\epsilon} = i \pi \ln(\delta) + \pi^2/2$$ (i.e delta dependent)

In the first integral, the singularity at z=0 is not integrable yet the principal value exists while in the second case the singularity is not integrable and the principal value doesn't exist. As delta goes to zero in the first result, the rhs is well behaved while in the second it is diverging due to the $\ln(\delta)$.

I don't understand what you mean by "singularity is not integrable". Also, perhaps you made a typo: The singularity of both is not zero but rather at $i\epsilon$.

Also, I think these are equivalent to yours and a little easier to analyze:
$$\int_{-1}^1 \frac{1}{z} dz$$
$$\int_{-1}^1 \frac{\log(z)}{z}dz$$
And as written the integrals diverge. However the Cauchy Principal Value exists for the first and not the second. Also, little confusing what you're doing with the $\delta$ above: In order to take the Principal Value, we take a symmetrical limit around the singularity, zero for these two, with parameters $(-\delta,\delta)$ and then let $\delta\to 0$. We showed the first integral converges in the prindipal sense above and can do a similar analysis for the second. For the second one, we end up with an expression:
$$\lim_{\delta\to 0}\biggr\{\frac{\ln^2|\delta|+2\pi i \ln|\delta|-\pi^2}{2}-\frac{\ln^2|\delta|}{2}\biggr\}$$
$$\lim_{\delta\to 0}\biggr\{\pi i \ln|\delta|-\frac{\pi^2}{2}\biggr\}\to \infty$$