- #1
wasia
- 52
- 0
I am slightly confused about the definition of principle value. If we have an integral
[tex]\int 1/z,[/tex]
where the integration from [tex]-\infty[/tex] to [tex]\infty[/tex] is implied, then by Cauchy integral theorem we know that the principle value
[tex]P \int 1/z=i\pi.[/tex]
However, I would like to write down this principle value explicitly. My best shot is
[tex] \lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty}\int_{-R}^{-\epsilon}1/z+\int_{\epsilon}^{R}1/z.[/tex]
Assuming that this is correct (is it?) I can (can I?) calculate the integrals first and take limits afterwards. I get
[tex]\lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty} \ln\left(-\frac{\epsilon}{\epsilon}\right) + \ln\left(-\frac{R}{R}\right)=2\ln(-1)=0.[/tex]
Can you tell me what am I doing wrong?
[tex]\int 1/z,[/tex]
where the integration from [tex]-\infty[/tex] to [tex]\infty[/tex] is implied, then by Cauchy integral theorem we know that the principle value
[tex]P \int 1/z=i\pi.[/tex]
However, I would like to write down this principle value explicitly. My best shot is
[tex] \lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty}\int_{-R}^{-\epsilon}1/z+\int_{\epsilon}^{R}1/z.[/tex]
Assuming that this is correct (is it?) I can (can I?) calculate the integrals first and take limits afterwards. I get
[tex]\lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty} \ln\left(-\frac{\epsilon}{\epsilon}\right) + \ln\left(-\frac{R}{R}\right)=2\ln(-1)=0.[/tex]
Can you tell me what am I doing wrong?