I want to calculate the series expansion of the function $E(1-x)$, where $E$ is the complete elliptic integral of second kind defined as
$$ E(x)=\int_0^{\pi/2} d\theta\, \sqrt{1-x^2\sin^2\theta} $$
Mathematica result gives the following:
$E(1-x)$." />
From this, we can consider that $E(1-x)$ can be written in the form of
$$ E(1-x)=f(x)+g(x)\log x $$ where $f$ and $g$ are Taylor expandable functions.
My questions are the followings:
- Is my expecetation correct?
- If so, how can I find the functions $f$ and $g$? Is it possible to find out the general form Taylor coefficients of $f$ and $g$? Or at least few terms? I found Approximation for elliptic integral of second kind, which calculates the coefficients of the smallest order, but it seems that this method is a bit complicated to achieve a higher terms.
- If not, what is the exact series expansion of the function $E(1-x)$?
I tried binomial expansion to $\sqrt{1-x^2 \sin^2\theta}$, which makes us possible to calculate the coefficients term-by-term, then tried to calculate the coefficients of $f$ and $g$, which does not work very well. The point which makes complication is that the function $\log x$ is not Taylor expandable, but $\log(1-x)$ is.
I also tried the following: assuming that $E(1-x)=f(x)+g(x)\log x$, we have $\log x = (E(1-x)-f(x))/g(x)$, and taking differential gives $1/x=[(E(1-x)-f(x))/g(x)]'$, removing the bothering $\log x$ function. However, this approach does not give any meaningful result also.
