# Behavior of the logarithm function in the complex domain ...

#### chisigma

##### Well-known member
In...

http://mathhelpboards.com/challenge-questions-puzzles-28/logarithm-integral-5273.html#post24028

... it has beenshown in an elementary way that is ...

$\displaystyle\int_{0}^{\infty}\frac{\ln x}{x^{2}+a^{2}}\ dx = \frac{\pi\ \ln a}{2\ a}\ (1)$

The integral (1) isvery useful to shed light on the behavior of the logarithm functionin the complex field. Let's suppose we want to solve the integralusing the methods of complex analysis, which means to integrate thefunction $\displaystyle f(z) = \frac{\ln z}{z^{2} + a^{2}}$ along theclosed path shown in figure... ... looking for the limit when the great circle tends to infinity and the small circle tends to zero. According to the residue theorem is...

$\displaystyle\int_{C} \frac{\ln z}{z^{2} + a^{2}}\ d z = 2\ \pi\ i\ \sum_{k}r_{k}\ (2)$

... where r are the residues of the poles of f (z) that fall inside the path. The polesof f (z) are z = a i and z = - a i , so that the problem lies in theexact definition of the logarithm in the complex field. According tomost of the 'sacred texts' the logarithm of a complex number z is given by $\displaystyle \ln z = \ln |z| + i\ \theta$ , where in anycase is $- \pi < \theta \le \pi$. We call this alternative as'a)'. According to the alternative a) it will be ...

$\displaystyle r_{1}= \lim_{z \rightarrow a\ i} \frac{\ln z}{z + a\ i} = \frac{i\ \frac{\pi}{2}}{2\ a\ i} = \frac{\pi}{4\ a}$

$\displaystyle r_{2}= \lim_{z \rightarrow - a\ i} \frac{\ln z}{z - a\ i} = \frac{- i\ \frac{\pi}{2}}{- 2\ a\ i} =\frac{\pi}{4\ a}\ (3)$

... so that is...

$\displaystyle\int_{C} \frac{\ln z}{z^{2} + a^{2}}\ d z = 2\ \pi\ i\ (r_{1} +r_{2}) = i\ \frac{\pi^{2}}{a}\ (4)$

On the other hand, always in agreement with the alternative a) is ...

$\displaystyle \int_{C} \frac{\ln z}{z^{2} + a^{2}}\ d z = \int_{r}^{R} \frac{\ln x}{x^{2} + a^{2}} + i\ R\ \{ \int_{0}^{\pi} \frac{e^{i\ \theta}\ (\ln R + i\ \theta)}{a^{2} + R^{2}\ e^{2\ i\ \theta}}\ d \theta + \int_{- \pi}^{0} \frac{e^{i\ \theta}\ (\ln R + i\ \theta)}{a^{2} + R^{2}\ e^{2\ i\ \theta}}\ d \theta\ \} - \int_{r}^{R} \frac{\ln x}{x^{2} + a^{2}} - i\ r\ \{ \int_{0}^{\pi} \frac{e^{i\ \theta}\ (\ln r + i\ \theta)}{a^{2} + r^{2}\ e^{2\ i\ \theta}}\ d \theta + \int_{- \pi}^{0} \frac{e^{i\ \theta}\ (\ln R + i\ \theta)}{a^{2} + R^{2}\ e^{2\ i\ \theta}}\ d \theta\ \}\ (5)$

Now it is not difficult to see that for r tending to zero and R tends to infinity the integral (5) tends to zero, so that comparing (4) and (5) we arrive at the 'identity' $\displaystyle i\ \frac{\pi^{2}}{a}=0$! ...

If this is so is evident that either the theorem of residues, or in the alternative a) is wrong. Since giving up the residue theorem would be painful, let's consider a different alternative: the logarithm of a complex number z is given by $\displaystyle \ln z = \ln |z| + i\ \theta$ , where is $0 < \theta \le 2\ \pi$. We call this alternative as 'b)'. According to the alternative b) it will be ...

$\displaystyle r_{1}= \lim_{z \rightarrow a\ i} \frac{\ln z}{z + a\ i} = \frac{i\ \frac{\pi}{2}}{2\ a\ i} = \frac{\pi}{4\ a}$

$\displaystyle r_{2}= \lim_{z \rightarrow - a\ i} \frac{\ln z}{z - a\ i} = \frac{i\ \frac{3\ \pi}{2}}{- 2\ a\ i} = - \frac{3\ \pi}{4\ a}\ (6)$

... so that is...

$\displaystyle\int_{C} \frac{\ln z}{z^{2} + a^{2}}\ d z = 2\ \pi\ i\ (r_{1} +r_{2}) = - i\ \frac{\pi^{2}}{a}\ (7)$

On the other hand, always in agreement with the alternative a) is ...

$\displaystyle \int_{C} \frac{\ln z}{z^{2} + a^{2}}\ d z = \int_{r}^{R} \frac{\ln x}{x^{2} + a^{2}} + i\ R\ \int_{0}^{2\ \pi} \frac{e^{i\ \theta}\ (\ln R + i\ \theta)}{a^{2} + R^{2}\ e^{2\ i\ \theta}}\ d \theta - \int_{r}^{R} \frac{\ln x}{x^{2} + a^{2}} - 2\ \pi \ i\ \int_{r}^{R} \frac{d x}{a^{2} + x^{2}}\ dx - i\ r\ \int_{0}^{\pi} \frac{e^{i\ \theta}\ (\ln r + i\ \theta)}{a^{2} + r^{2}\ e^{2\ i\ \theta}}\ d \theta\ (8)$

Now it is not difficult to see that for r tending to zero and R tends to infinity the integral (8) tends to $\displaystyle - i\ \frac{\pi^{2}}{a}$ so that comparing (7) and (8) we arrive at $\displaystyle - i\ \frac{\pi^{2}}{a}=- i\ \frac{\pi^{2}}{a}$, something that fills us with happiness! ...

Add consideration on which of the alternatives a) or b) is 'preferable' is clearly superfluous ...

Since this is a note, comments and clarifications have to be made on the commentary ...

Kind regards

$\chi$ $\sigma$

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#### chisigma

##### Well-known member
It is significant that this note has been taken in response to the post...

http://mathhelpboards.com/analysis-50/complex-logarithm-11275.html#post52548

...and that new fuel has been provided by the post ...

http://mathhelpboards.com/analysis-50/contour-integral-method-query-11457.html#post53618

...produced by the same user ... congratulations to Fermat! ...

In the previous post we examined two different alternatives to thedefinition of the logarithm in the complex field: alternative a),accepted by most of the literature, and alternative b) minority in the literature but strongly supported by the writer. The strict answer to the questions posed by Fermat requires the use of powerful complex analysis theorem known as the principle of argument, which states that, given an f (z) meromorphic in a region included in a closed path $\gamma$, is...

$\displaystyle \int_{\gamma} \frac{f^{\ '}(z)}{f(z)}\ d z = 2\ \pi\ i\ (n - m)\ (1)$

... where n is the number of zeros and m the number of poles of f (z) inside $\gamma$, each of which taken in agreement with its multiplicity.The proof of the theorem is quite easy if you start with the essential requirements ...

a) f (z) must be in the form $\displaystyle f(z)= \frac{p(z)}{q(z)}$ , where p (z) is a polynomial of degree n and q (z) be a polynomial of degree m ...

b) function ln [f (z)] is analytic everywhere except the zeros and poles of f (z) and its derivative is $\displaystyle \frac{d}{d z} \ln [f(z)] = \frac{f^{\ '} (z)}{f(z)}$ ...

c) given two complex numbers a and b, both different from 0, apply the base relations $\displaystyle \ln (a\ b) = \ln a + \ln b$ and $\displaystyle \ln \frac{a}{b}= \ln a - \ln b$...

It should be noted that the property c) is compatible with the alternative b) and not with the alternative a). Given the hypotheses a), b) and c), the proof proceeds without any problems. First is ...

$\displaystyle f(z) = \frac{a_{0}\ (z - a_{1})\ (z-a_{2})...(z-a_{n})}{b_{0}\ (z - b_{1})\ (z - b_{2})...(z-b_{m})}\ (2)$

From (2) and the properties b) and c) we obtain...

$\displaystyle \frac{d}{d z} [\ln f(z)] = \frac{f^{\ '}(z)}{f(z)} = \sum_{k=1}^{n} \frac{1}{z-a_{k}} - \sum_{k=1}^{m} \frac{1}{z-b_{k}}\ (3)$

... and by (3), using the residue theorem, we arrive at (1). Very interesting is the simplest case of all: f(z) = z and $\gamma$ is the unit circle. In that case we arrive to the result...

$\displaystyle \int_{|z=1|} \frac{d z}{z} = 2\ \pi\ i\ (4)$

... the correctness of which we all agree ...

Last not useless observation to make is that it is forced into a crucial choice: if the alternative a) is wrong, then the alternative b) is good and good is also the principle of the argument ... if the alternative a) is good then the alternative b) is wrong and wrong is also the principle of the argument. Now is the chance to point out that the principle of argument in recent decades has made possible to arrive at valuable results in the field of complex analysis, so ...

Kind regards

$\chi$ $\sigma$

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