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Let ${\mathbb{R}^2}^+=\{(x,y)\in \mathbb{R}^2\mid x>0, y>0\}$.

Let $P:{\mathbb{R}^2}^+\to \mathbb{R}$ be the function with $P(a,b)=$ $\text{The perimeter of ellipse}\;\; \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Is $P$ a real analytic function? Is there a real analytic extention of of $P$ to whole $\mathbb{R}^2$? Are the Taylor coefficients computed in some reference? Is there a holomorphic function $\tilde{P}$** defined on $\mathbb{C}^2$ whose restriction to ${\mathbb{R}^2}^+$ is equal the ellipse perimeter function.

Note: In the case of positive answer to the latter question, it would be interesting to think to the complex interpretation $\tilde{P}(a,b)$ where are $a,b$ are two complex numbers associated to "certain"?? complex object. Some complex analogy of "Ellipse Perimeter"?

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This function was studied since Newton's time; I suppose the power series was obtained by Newton himself. It is analytic, but not in the whole $(a,b)$ space since it has singularities, in particular when $a=b=0$. See, for example this file. The answer is an explicit function, which is a special case of the hypergeometric function.

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  • $\begingroup$ I just find this link which ensure us the function is globally holomorphic(note that we have even power of $e$. So I do not see why it is singualr at a=b? so it is seemms that the last part of my question(the note part) is remained to discussAm I mistaken? .mathsisfun.com/geometry/ellipse-perimeter.html $\endgroup$
    – Ali Taghavi
    Mar 17, 2021 at 17:43
  • $\begingroup$ But lets look at it againit is not holomorphic at a=? May beit is? $\endgroup$
    – Ali Taghavi
    Mar 17, 2021 at 17:45
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    $\begingroup$ We also have an evident singularity when $a=b=0$. $\endgroup$
    – Alexandre Eremenko
    Mar 18, 2021 at 21:49
  • $\begingroup$ But the origin can not be an isolated complex singularity accoeding to Hartogs theorem. $\endgroup$
    – Ali Taghavi
    Mar 19, 2021 at 14:44
  • $\begingroup$ $(0,0)$ is a singularity in the REAL plane. There are many more complex singularities. It is an explicit function after all, and you can see all its real and complex singularities. I do not know what "Hartogs theorem" you are referring to but complex singularities CAN be isolated, for example $(0,0)$ for $x^2=x^3$. $\endgroup$
    – Alexandre Eremenko
    Mar 19, 2021 at 18:20

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