I am terribly confused by complex integrals involving $|z|$. Please help me.
Let's say I want to evaluate
$$
\int_{a<|x| But on $\Gamma$, we have
$$
\int_\Gamma {dz \over |z|} = \int_0^\pi {b i e^{i \theta} \over b}d\theta = -2
$$
Similarly
$$
\int_\gamma {dz \over |z|} = 2
$$
Thus I conclude
$$
\left(
\int_{-b}^{-a} + \int_{a}^{b}
\right) {dz \over |z|} = 0
$$
And of course I know this is wrong -- the integral is 2 $\ln(b/a)$. What am I doing wrong? Thank you for your help!
integral of $1 \over |z|$ over a circle
1
$\begingroup$
complex-analysis
complex-integration
cauchy-integral-formula
-
0So in the first integral it is $\;x\in\Bbb C\;$ ? – 2017-01-24
-
1No, the first integral is on the real line. – 2017-01-24
-
1Is the integral you want to start with the ordinary real integral $\int_a^b \frac{1}{|x|}\,dx$? I have a bit of trouble understanding your notation. If it _is_ the ordinary real integral, your first step should be to eliminate the absolute-value sign, if necessary by splitting into cases. You can't begin usefully considering contour integrals until you have a _holomorphic_ function to integrate, and things that involve $|z|$ generally aren't. – 2017-01-24
-
3You can't use Cauchy here, because $1/|z|$ is not holomorphic anywhere in $\mathbb C$. – 2017-01-24
-
0If you want a real integral, why are you unnecessarily messing with complex ones? And also: in the real case, $\;a<|x|
2 Answers
2
Trying to make some sense of your question: if $\;0 $$\left.\int_a^b\frac{dx}x=\log x\right|_a^b=\log \frac ba$$ If $\;b<0\;$ it is something similar. If $\;a\le 0
2
Now I see what I am doing wrong. $1/|z|$ is not holomorphic because $$ {1 \over |z|} = {1 \over \sqrt{x^2 + y^2}} + 0 \times i = u(x,y) + i v(x,y) $$ Since $$ {\partial u(x,y) \over \partial x} \neq {\partial v(x,y) \over \partial y} $$ the Cauchy-Riemann equations are not satisfied and the Cauchy theorem is not applicable. Thank you all for your help!
-
0A more general result: With the Cauchy-Riemann equations you can also show that a holomorphic function which is real (on some open subset in $\mathbb{C}$), must be constant. Since $1/|z|$ is real a _not_ constant, we can (again) conclude that $1/|z|$ is _not_ holomorphic. – 2018-09-15