integral arising in statistical mechanics The following integral arises in statistical mechanics of lattice models:
$
\displaystyle I = \int_{0}^{\pi/2}
\ln\left(\, 1 +
\sqrt{\, 1 - a^{2}\sin^{2}\left(\phi\right)\,}\,\right)\,
\mathrm{d} \phi\quad$ with $\quad a \leq   1
$.
By trial and error, the integral was shown to be nearly equal to
$
\displaystyle I =
{\pi \over 4}\ln\left(2\right) + {\pi \over 4}
\ln\left(1 + \sqrt{\, 1 - Ga^{2}\, }\right)
$.
where $G$ denotes Catalan constant.
Is there a method of obtaining an exact analytical solution or justifying the approximate answer given above ?.
 A: Similar to @Maxim's comment
$$I(a) = \int_{0}^{\pi/2}\log\left(\, 1 +\sqrt{\, 1 - a^{2}\sin^{2}\left(\phi\right)\,}\,\right)\,d\phi$$
$$I'(a)=- \int_{0}^{\pi/2} \frac{a \sin ^2(\phi)}{\sqrt{1-a^2 \sin ^2(\phi)} \left(1+\sqrt{1-a^2 \sin^2(\phi)}\right)}\,d\phi=\frac{\pi -2 K\left(a^2\right)}{2 a}$$
$$\int \frac{K\left(a^2\right)}{ a}\,da=-\frac{1}{4} G_{3,3}^{2,2}\left(-a^2|
\begin{array}{c}
 \frac{1}{2},\frac{1}{2},1 \\
 0,0,0
\end{array}
\right)$$ and $I(0)=\frac{1}{2} \pi  \log (2)$.
After simplifications, this led me (laboriously) to @Maxim's result.
$$I(a) = \frac {\pi } 2 \log( 2)- \frac {\pi a^2} {16}  {_4 F_3} {\left( 1, 1, \frac 3 2, \frac 3 2; 2, 2, 2; a^2 \right)}$$
The good approximation given in the post
$$ J(a) ={\pi \over 4}\log\left(2\right) + {\pi \over 4}
\log\left(1 + \sqrt{\, 1 - Ca^{2}\, }\right)$$
can easily be justified looking at the series expansions built around $a=0$
$$I(a)=\frac{\pi}{2}   \log (2)-\frac{\pi  }{16}a^2-\frac{9 \pi  }{512}a^4-\frac{25 \pi 
   }{3072}a^6+O\left(a^8\right)$$
$$J(a)=\frac{\pi}{2}   \log (2)-\frac{\pi  C}{16} a^2 -\frac{3\pi 
   C^2}{128} a^4 -\frac{5\pi  C^3}{384} a^6 +O\left(a^8\right)$$ Since $C\sim 0.915966 $, this makes the coefficients quite similar (their ratios are $C$, $\frac{4}{3} C^2$,  $\frac{8 }{5}C^3$).
If you need approximations, you could probably think about $[2n,2n]$ Padé approximants of the ${_4 F_3}$ hypergeometric function (built around $a=0$).
A: I take a different approach to get a simpler approximation formula.
In terms of convenience i use $x$ instead of $\phi$.
Let's use a simple approximation formula:
$$\ln(1+x)\approx \frac{\ln 2}{3}x(4-x)$$
Taking into account this formula, (i will skip routine calculations) $I(a)$ takes an approximate form:
$$I(a)\approx\frac{a^2-2}{12}\pi \ln 2+\frac{4}{3}\ln 2\int_{0}^{\pi/2}\sqrt{1 - a^{2}\sin^{2}x}\,dx$$
The integral(elliptic) still appears here. To overcome this obstacle we use the next approximation formula:
$$\sqrt{1 - x}\approx 1 - x^2 $$
With this approximation the integral can be computed in closed form and we get finally the following approximation for $I(a)$
$$I(a)\approx(6+a^2-3a^4)\frac{\pi}{12}\ln 2$$
The approximation error within $0<a<1$ is about $0.05$ on average.
Now, there arose a question about the derivative of $I(a)$ i.e. $I'(a)$
From a comment:
"the derivative at 'a' =1 has a physical significance."
But
$$I'(1)=\infty$$
It is difficult to understand where a physical significance is hidden here.
