A false approach to an integral....

[chisigma]

In...

http://mathhelpboards.com/questions-other-sites-52/unsolved-analysis-number-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...

https://www.physicsforums.com/attachments/1799._xfImportMerry Christmas from Serbia


$\chi$ $\sigma$

 

[polygamma]

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 $z= \sqrt{2}-1$.

 

[alyafey22]

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]

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 1799Merry Christmas from Serbia


$\chi$ $\sigma$

 

[CellCoree]

Hello,

Thank you for bringing this integral to my attention. After examining the proposed approach, it is clear that it is a false one. While it may seem feasible to use the substitution $z = e^{ix}$ and the Euler's relation $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$, there are a few issues with this approach.

Firstly, when we substitute $z = e^{ix}$, we are essentially transforming the integral from being over the real interval $[0, 2\pi]$ to being over the unit circle $\gamma$. This transformation is not always valid and can lead to incorrect results.

Secondly, the use of the Euler's relation to simplify the integral leads to a different function being integrated, which is not equivalent to the original integral. This can also lead to incorrect results.

Finally, even if we do manage to correctly transform and simplify the integral, using the residue theorem to solve it may not be the most appropriate method. The residue theorem is typically used for integrals over closed contours, but in this case, we are integrating over a real interval.

In conclusion, it is important to carefully consider the validity and appropriateness of any approach when solving integrals. It is also important to understand the limitations and assumptions of different methods, and to choose the most suitable one for the given problem.

Happy holidays from a fellow scientist in the US.