- #1
CAF123
Gold Member
- 2,948
- 88
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?
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?