integral arising in statistical mechanics

Question

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 ?.

Answer 1

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 $a=0$).

Answer 2

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.