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#### chisigma

##### Well-known member

- Feb 13, 2012

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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

$\chi$ $\sigma$

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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