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A false approach to an integral...

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...

eb191d59c14248f1e362dd5eb1d3102c.jpg


Merry Christmas from Serbia


$\chi$ $\sigma$
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
In theory it could be evaluated using the residue theorem. But you would need to deform the contour around the branch points at $z=1 - \sqrt{2}$ and $z= \sqrt{2}-1$.
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
What do you mean by fail ? , is it the case the we cannot apply the transformation ? or the integral is difficult to solve using that transformation ?
As RV indicated the square root produces two branch points for the polynomial so in case they are inside \(\displaystyle |z|=1\) we have to deform the contour around them. Looking at the complexity of the answer it might be solvable by this contour but more challenging than using elliptic integrals .
 

chisigma

Well-known member
Feb 13, 2012
1,704
As RV said the problem is the fact that f(z) has two brantch points inside the unit circle and that means that, unless You choose more or less complicated paths excluding them, the direct use of the residue theorem is impossible...



View attachment 1799


Merry Christmas from Serbia


$\chi$ $\sigma$