When I try to review the solution process of solving the 2-D Ising model, I get interested in a integral, which is considered to be impossible to get a closed-form.
The integral is $I=\int_0^\pi\log\left(1+\sqrt{(1-\kappa^2\sin^2x)}\right)dx$
If we split the integral interval into two parts, then
$I=2\int_0^\frac{\pi}{2}\log\left(1+\sqrt{(1-\kappa^2\sin^2x)}\right)dx$
Taking advantage of the Taylor expansion of logarithmic function,namely
$\log(1+x)=\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}x^n}{n}$
the integral turns out to be
$I=2\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}}{n}\int_0^\frac{\pi}{2}\left(\sqrt{1-\kappa^2\sin^2x}\right)^n~dx$
When $n=1$ ,the first item is nothing but the complete elliptic integral of the second kind:
$E(\kappa)=\int_0^\frac{\pi}{2}\sqrt{1-\kappa^2\sin^2x}~dx$
So, I wonder whether or not can we rewrite the integral $I$ according to the complete elliptic integral of the second kind $E(\kappa)$ ? Once we succeed, the result will be much more beautiful!
Thanks!
Note that the following procedure is wrong (but I don't know the reason):
$I=\int_0^\pi\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)dx$
$=\dfrac{1}{2}\int_0^{2\pi}\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)dx$ $\left(\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)~\text{is a periodic function of period}~\pi\right)$
$=\dfrac{1}{2}\left[x\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)\right]_0^{2\pi}-\dfrac{1}{2}\int_0^{2\pi}x~d\left(\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)\right)$
$=\dfrac{\pi\ln2}{2}+\dfrac{1}{2}\int_0^{2\pi}\dfrac{\kappa^2x\sin x\cos x}{\sqrt{1-\kappa^2\sin^2x}\left(1+\sqrt{1-\kappa^2\sin^2x}\right)}dx$
$=\dfrac{\pi\ln2}{2}+\dfrac{1}{2}\int_{2\pi}^0\dfrac{\kappa^2(2\pi-x)\sin(2\pi-x)\cos(2\pi-x)}{\sqrt{1-\kappa^2\sin^2(2\pi-x)}\left(1+\sqrt{1-\kappa^2\sin^2(2\pi-x)}\right)}d(2\pi-x)$
$=\dfrac{\pi\ln2}{2}+\dfrac{1}{2}\int_0^{2\pi}\dfrac{\kappa^2(2\pi-x)\sin x\cos x}{\sqrt{1-\kappa^2\sin^2x}\left(1+\sqrt{1-\kappa^2\sin^2x}\right)}dx$
$\therefore2I=\pi\ln2+\int_0^{2\pi}\dfrac{2\kappa^2\pi\sin x\cos x}{\sqrt{1-\kappa^2\sin^2x}\left(1+\sqrt{1-\kappa^2\sin^2x}\right)}dx$
$I=\dfrac{\pi\ln2}{2}-\left[\pi\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)\right]_0^{2\pi}$
$I=\dfrac{\pi\ln2}{2}$
In fact the following procedure is really correct:
$I=\int_0^\pi\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)dx=2\int_0^\frac{\pi}{2}\ln\left(1+\sqrt{1-\kappa^2\sin^2x}\right)dx$
$\dfrac{dI}{d\kappa}=-\int_0^\frac{\pi}{2}\dfrac{2\kappa\sin^2x}{\sqrt{1-\kappa^2\sin^2x}\left(1+\sqrt{1-\kappa^2\sin^2x}\right)}dx$
$=-\int_0^\frac{\pi}{2}\dfrac{2\kappa\sin^2x\left(1-\sqrt{1-\kappa^2\sin^2x}\right)}{\sqrt{1-\kappa^2\sin^2x}\left(1+\sqrt{1-\kappa^2\sin^2x}\right)\left(1-\sqrt{1-\kappa^2\sin^2x}\right)}dx$
$=\int_0^\frac{\pi}{2}\dfrac{2\kappa\sin^2x\left(\sqrt{1-\kappa^2\sin^2x}-1\right)}{(1-(1-\kappa^2\sin^2x))\sqrt{1-\kappa^2\sin^2x}}dx$
$=\int_0^\frac{\pi}{2}\dfrac{2\kappa\sin^2x\left(\sqrt{1-\kappa^2\sin^2x}-1\right)}{\kappa^2\sin^2x\sqrt{1-\kappa^2\sin^2x}}dx$
$=\int_0^\frac{\pi}{2}\dfrac{2}{\kappa}dx-\int_0^\frac{\pi}{2}\dfrac{2}{\kappa\sqrt{1-\kappa^2\sin^2x}}dx$
When $0\leq\kappa\leq1$ ,
$\int_0^\frac{\pi}{2}\dfrac{2}{\kappa}dx-\int_0^\frac{\pi}{2}\dfrac{2}{\kappa\sqrt{1-\kappa^2\sin^2x}}dx$
$=\int_0^\frac{\pi}{2}\dfrac{2}{\kappa}dx-\int_0^\frac{\pi}{2}\dfrac{2}{\kappa}\sum\limits_{n=0}^\infty\dfrac{(2n)!\kappa^{2n}\sin^{2n}x}{4^n(n!)^2}dx$
$=-\int_0^\frac{\pi}{2}\sum\limits_{n=1}^\infty\dfrac{(2n)!\kappa^{2n-1}\sin^{2n}x}{2^{2n-1}(n!)^2}dx$
For $\int\sin^{2n}x~dx$ , where $n$ is any natural number,
$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
$\therefore-\int_0^\frac{\pi}{2}\sum\limits_{n=1}^\infty\dfrac{(2n)!\kappa^{2n-1}\sin^{2n}x}{2^{2n-1}(n!)^2}dx$
$=-\left[\sum\limits_{n=1}^\infty\dfrac{((2n)!)^2\kappa^{2n-1}x}{2^{4n-1}(n!)^4}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2\kappa^{2n-1}((k-1)!)^2\sin^{2k-1}x\cos x}{2^{4n-2k+1}(n!)^4(2k-1)!}\right]_0^\frac{\pi}{2}$
$=-\sum\limits_{n=1}^\infty\dfrac{((2n)!)^2\pi\kappa^{2n-1}}{16^n(n!)^4}$
$\therefore I=-\sum\limits_{n=1}^\infty\dfrac{(2n)!(2n-1)!\pi\kappa^{2n}}{16^n(n!)^4}+c$
$\because I(0)=\int_0^\pi\ln2~dx=[x\ln2]_0^\pi=\pi\ln2$
$\therefore c=\pi\ln2$
Hence $I=\pi\ln2-\sum\limits_{n=1}^\infty\dfrac{(2n)!(2n-1)!\pi\kappa^{2n}}{16^n(n!)^4}$