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The goal is to prove, using Mathematica, that for $\mathbb{D}$ the unit disk and $u,v \in \mathbb{D}$, $u \neq v$, $$\frac{1}{\pi} \int_{\mathbb{D}} \frac{\mathrm{d}^2z}{(z-u)\overline{(z-v)}} = \ln(1-u\bar{v}) - \ln|u-v|^2.$$

It looks like the techniques suggested in previous posts do not work. For example, the expression :

Integrate[
  (r/((u - r Cos[θ] - I r Sin[θ]) (v - r Cos[θ]+ I r Sin[θ]))), 
  {r, 0, 1}, {θ, 0, 2 π}, 
  Assumptions -> {0 < u < v < 1}]

or the same with more restrictive assumptions, returns 0. This is contradictory with what NIntegrate returns when given specific values for $u,v$. (The values returned by NIntegrate do match the above formula).

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  • $\begingroup$ {0 < u, u < 0.5, 0.5 < v, v < 1} can be replaced by {0<u<v<1}. $\endgroup$ – David G. Stork Jan 23 '18 at 23:25
  • $\begingroup$ Sure, this is what I wrote in the first place, but it returned the value $0$. Then I tried to write it differently, but nothing helped. $\endgroup$ – Graoully Dubois Jan 24 '18 at 15:45
  • $\begingroup$ ...thereby showing you should leave the condition in its shorter, more elegant form. $\endgroup$ – David G. Stork Jan 24 '18 at 17:07
  • $\begingroup$ Done. Thank you. Do you know why it doesn't work ? Any suggestion on how to make it find the correct answer ? $\endgroup$ – Graoully Dubois Jan 24 '18 at 17:19
  • $\begingroup$ I would write out everything in terms of reals and integrate {x,-1,1},{y, -Sqrt[1-x^2],Sqrt[1-x^2]} as a first try. $\endgroup$ – David G. Stork Jan 24 '18 at 17:44

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