- Thread starter
- #1
chisigma
Well-known member
- Feb 13, 2012
- 1,704
In...
http://mathhelpboards.com/questions...mber-theory-other-sites-7479-3.html#post38136
... it has been found the value the integral...
$\displaystyle \int_{0}^{2\ \pi} \sqrt{1 + \sin^{2} x}\ dx\ (1)$
At first it seems feasible to set $z = e^{i\ x}$ and the Euler's relation $\displaystyle \sin x = \frac{e^{i\ x} - e^{- i\ x}}{2\ i}$ so that the integral becomes...
$\displaystyle \int_{0}^{2\ \pi} \sqrt{1 + \sin^{2} x}\ dx = \int_{\gamma} \frac{\sqrt{1 + (\frac{z - z
^{-1}}{2\ i})^{2}}}{i\ z}\ dz\ (2)$
... being $\gamma$ the unit circle and finally solve (2) with the residue theorem. Thi approach however fails and it is requested to explain why...

Merry Christmas from Serbia
$\chi$ $\sigma$
http://mathhelpboards.com/questions...mber-theory-other-sites-7479-3.html#post38136
... it has been found the value the integral...
$\displaystyle \int_{0}^{2\ \pi} \sqrt{1 + \sin^{2} x}\ dx\ (1)$
At first it seems feasible to set $z = e^{i\ x}$ and the Euler's relation $\displaystyle \sin x = \frac{e^{i\ x} - e^{- i\ x}}{2\ i}$ so that the integral becomes...
$\displaystyle \int_{0}^{2\ \pi} \sqrt{1 + \sin^{2} x}\ dx = \int_{\gamma} \frac{\sqrt{1 + (\frac{z - z
^{-1}}{2\ i})^{2}}}{i\ z}\ dz\ (2)$
... being $\gamma$ the unit circle and finally solve (2) with the residue theorem. Thi approach however fails and it is requested to explain why...

Merry Christmas from Serbia
$\chi$ $\sigma$