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In the question Geometry problem involving infinite number of circles I showed that the answer could be obtained by the sum

$$ \sum_{k=0}^{\infty}\int_{B_{k}} {4 \over \,\left\vert\,1 + \left(\,x + y{\rm i}\,\right)\,\right\vert^{\,4}\,} \,{\rm d}x\,{\rm d}y\,, $$ where $ B_k = \left\{\, z \in \mathbb{C}:\ \left\vert\,z - \left[\frac{1}{2} + \left(k + {1 \over 2}\right) {\rm i}\right]\,\right\vert\ \leq\ {1 \over 2}\,\right\} $ is the ball centered at ${1 \over 2} + \left(k + {1 \over 2}\right){\rm i}$ and radius ${1 \over 2}$.

Eventually I solved the problem in a different manner, and from my answer we can deduce that

$$\int_{B_{k}}{4 \over \left\vert\,1 + (x + iy)\,\right\vert^{\,4}} \,{\rm d}x\,{\rm d}y={16 \pi \over \left(\,4k^{2} + 4k + 9\,\right)^{2}}.$$

Can we somehow get this result directly ?. If so, I could perhaps shorten my answer quite a bit.

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J. J.
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  • Just out of curiosity, what was the different manner you used to solve the initial geometry problem? – Asier Calbet Aug 20 '14 at 13:40
  • @Assaultous2: You can read my answer there. – J. J. Aug 20 '14 at 13:41
  • oh, ok thanks haha – Asier Calbet Aug 20 '14 at 13:41
  • I can no longer find the question - can you post a link please? – Asier Calbet Aug 20 '14 at 13:45
  • @Assaultous2: It is linked in the question: http://math.stackexchange.com/questions/903834/geometry-problem-involving-infinite-number-of-circles – J. J. Aug 20 '14 at 13:47
  • @DanielFischer: This is basically how I found the value of the integral: By calculating the radius. I would like to get the area directly without calculating the radius first, however. Of course if the shortest way to compute the integral is via doing the Möbius transform and calculating the area of the circle there, then there isn't much point in this question. – J. J. Aug 20 '14 at 14:01
  • Yeah, I saw that too when I looked closer at your answer. – Daniel Fischer Aug 20 '14 at 14:02

1 Answers1

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You can compute the integral directly using complex coordinates.

Let $z = x + iy$ and $\bar{z} = x-iy$ and notice $$dx \wedge dy = \frac{d\bar{z} \wedge dz}{2i}$$ The integral $I_k$ over the disk $B_k$ becomes

$$\int_{B_k} \frac{4}{|1 + (x+iy)|^4} dx \wedge dy = -2i \int_{B_k} \frac{1}{(1+z)^2(1+\bar{z})^2} d\bar{z} \wedge dz\\ = 2i \int_{B_k} d \left( \frac{1}{(1+\bar{z})(1+z)^2} dz \right) = 2i \int_{\partial B_k} \frac{dz}{(1 + \bar{z})(1+z)^2} $$ Let $\omega_k = \frac12 + i\left(k + \frac12\right)$. For $z \in \partial B_k$, parametrize $z$ as $\;\omega_k + \frac12 \rho\;$ for $\rho \in S^1$. We have

$$I_k = i \int_{S^1} \frac{d\rho}{(1 + \bar{\omega}_k + \frac{1}{2\rho})(1 + \omega_k + \frac12\rho)^2} = \frac{i}{1+\bar{\omega}_k} \int_{S^1} \frac{\rho d\rho}{ \left(\rho + \frac{1}{2(1+\bar{\omega}_k)}\right)(1 + \omega_k + \frac12\rho)^2} $$ In the integral at RHS, there is only one pole inside the unit circle $S^1$. Namely, $\rho = -\frac{1}{2(1+\bar{\omega}_k)}$. As a result, the integral is equal to $2 \pi i$ times corresponding residue at $-\frac{1}{2(1+\bar{\omega}_k)}$.

$$\begin{align} I_k &= (2\pi i) \frac{i}{1+\bar{\omega}_k}\frac{-\frac{1}{2(1+\bar{\omega}_k)}}{\left(1 + \omega_k - \frac{1}{4(1 + \bar{\omega}_k)}\right)^2}\\ &= \frac{\pi}{\left(|1 + \omega_k|^2 - \frac14\right)^2} = \frac{\pi}{\left( \left( \frac32\right)^2 + \left(k + \frac12\right)^2 - \frac14 \right)^2} = \frac{16\pi}{(4k^2+4k+9)^2} \end{align} $$

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