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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?

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  • $\begingroup$ The singularity of $1/z$ is not integrable (the improper integral doesn't exist). If the singularity is integrable, the Cauchy p.v. is the same as the improper integral. If there is a continuous antiderivative $F$ on $[a,0)$ and on $(0,b]$, then $$\operatorname{v.\!p.} \int_a^b f(z)dz = \lim_{h \downarrow 0} (F(b)-F(h)+F(-h)-F(a)).$$ If $F$ is also an antiderivative of $f$ in the upper half-plane and is continuous from above at $\pm h$, then you can evaluate $F(-h)-F(h)$ as the integral of $f$ over the semicirle. There are other possible regularizations, such as the Hadamard finite part. $\endgroup$ – Maxim Mar 7 at 17:15
  • $\begingroup$ @Maxim the singularity at z=0 of 1/z is integrable over a symmetric domain though yes? $\endgroup$ – CAF Mar 8 at 19:16
  • $\begingroup$ Suppose that $a < 0 < b$ and $f(z)$ has a singularity at zero. If we want the improper integral to be additive, then $$\int_a^b f(z) dz = \lim_{h \downarrow 0} \int_a^{-h} f(z) dz + \lim_{h \downarrow 0} \int_h^b f(z) dz.$$ Then the improper integral of $1/z$ doesn't exist, while the p.v. integral exists. It doesn't matter whether $a = -b$ or not. $\endgroup$ – Maxim Mar 8 at 19:48

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