How to evaluate $\int_0^{\frac{\pi}{2}} \ln\left(x^2+\ln^2(a \cos x)\right) dx , \forall a>0$?

Question

I tried to find generalization for the integral for $a>0$ $$\Omega\left(a\right)=\int_0^{\frac{\pi}{2}} \ln\left(x^2+\ln^2(a \cos x)\right) dx$$ here we can see the value of $\Omega(1)$ ,I tried to solve and found that firstly $$ \Omega\left(a\right)=\Re\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln \ln \left(\frac{a}{2} \left(e^{2i x}+1\right) \right)$$ and I used $$ \ln \ln x=\lim_{(y,z)\to (0,0)} \frac{\partial}{\partial z} \frac{\partial^z}{\partial y^z} x^y$$ therefore (the line which I think is true for only $0<a\le1$) $$ \Omega\left(a\right)=\Re \lim_{(y,z)\to (0,0)} \frac{\partial}{\partial z} \frac{\partial^z}{\partial y^z} \left(\frac{a}{2} \right)^y \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(e^{2ix}+1 \right)^y dx$$ and from here we have $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(e^{2ix}+1 \right)^y dx=\pi ,\forall y>0$$ So $$ \Omega\left(a\right)=\Re \lim_{(y,z)\to (0,0)} \frac{\partial}{\partial z} \frac{\partial^z}{\partial y^z} \left(\frac{a}{2} \right)^y \pi=\Re \pi \ln \ln \frac{a}{2} $$ but numerically I find that this formula is true for only $0<a\le1$ and then $$\Omega\left(a\right)=\pi \ln \ln \frac{2}{a} , 0<a\le 1 $$ and for special values (for $a> 1$) I got that $$ \Omega\left(\frac{3}{2}\right)=\pi \ln \ln \frac{64}{27} \\ \Omega\left(\frac{5}{2}\right)=\pi \ln \ln \frac{3125}{1024} \\ \Omega\left(2\right)=0$$ So is it possible to find $\Omega\left(a\right)$ for all positive $a$ ?

Addition

from comments a new special value $\Omega(4)=\pi \ln \ln 4$

Answer 1

For $a>1$, $a \ne 2$, I will show that $$\Omega(a) = \int_0^{\pi/2} \ln\left(x^2+\ln^2(a \cos x)\right) \, \mathrm dx = \pi \ln \left(\frac{a}{2-a} \, \ln \frac{2}{a} \right) + C,$$ where $C$ is an unknown constant.

To show that $C$ must be zero, we can take the limit on both sides of the above equation as $ a \to 1^{+}$ and use the known value $\Omega(1) = \pi \ln \left(\ln 2 \right) $.

(If we take the limit of $\pi \ln \left(\frac{a}{2-a} \, \ln \frac{2}{a} \right)$ as $a \to 2$, we get $0$ (which is the correct value for $a=2$).)

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Assume that $0 < a< 1$, $a \ne 1/2$.

Also assume that $\ln$ is the principal branch of the logarithm.

Differentiating under the integral sign, we have $$\Omega'(a) = \frac{2}{a} \int_{0}^{\pi/2} \frac{\ln(a \cos x)}{\ln^{2}(a \cos(x))+x^{2}} \, \mathrm dx. $$

Let's integrate the function $$f(z) = \frac{1}{\ln(a-iz)} \frac{1}{a^{2}+z^{2}} $$ around a contour consisting of the real axis and a large semicircle in the upper half-plane.

(This is possible because the branch cut for $\ln(a-iz)$ is the lower half-plane.)

Since $0 < a< 1$, there is a simple pole in the upper half-plane at $z=i(1-a)$.

There is also a simple pole at $z=ia$.

Therefore, since the integral along the semicircle vanishes in the limit, we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{1}{\ln(a-ix)} \frac{1}{1+x^{2}} \, \mathrm dx &= 2 \pi i \left(\operatorname*{Res}_{z=ia}f(z) + \operatorname*{Res}_{z=i(1-a)}f(z) \right) \\ &= \pi \left(\frac{1}{a \ln(2a)} - \frac{2 }{2a-1}\right). \end{align}$$

Equating the real parts on both sides of the above equation, we have $$\int_{-\infty}^{\infty} \frac{\frac{1}{2} \ln(a^{2}+x^{2})}{\frac{1}{4} \ln^{2}(a^{2}+x^{2})+\arctan^{2}(\frac{x}{a})} \frac{1}{a^{2}+x^{2}} \, \mathrm dx = \pi \left(\frac{1}{a \ln(2a)} - \frac{2 }{2a-1}\right).$$

And then making the substitution $x= a \tan(u)$, we have

$$ \begin{align} \frac{1}{a} \int_{-\pi/2}^{\pi/2} \frac{\frac{1}{2} \ln(a^{2} \sec^{2} u)}{\frac{1}{4} \ln^{2}(a^{2} \sec^{2}u)+u^{2}} \, \mathrm du &= - \frac{2}{a} \int_{0}^{\pi/2} \frac{\ln \left(\frac{1}{a} \cos u\right)}{\ln^{2}\left(\frac{1}{a} \cos u\right)+u^{2}} \mathrm du \\ &= \pi \left(\frac{1}{a \ln(2a)} - \frac{2 }{2a-1}\right). \end{align}$$

Therefore, for $a>1$, $a \ne 2$, $$-2a \int_{0}^{\pi/2} \frac{\ln \left(a \cos u\right)}{\ln^{2}\left(a \cos u\right)+u^{2}} \mathrm du = \pi \left(\frac{a}{\ln (\frac{2}{a})} - \frac{2a}{2-a}\right), $$

from which it follows that $$\Omega'(a) = - \frac{\pi}{a^{2}}\left(\frac{a}{\ln (\frac{2}{a})} - \frac{2a}{2-a}\right) = \frac{\pi}{a} \left( \frac{2}{2-a} - \frac{1}{\ln(\frac{2}{a})} \right). $$

Integrating back with respect to $a$, we get $$ \Omega(a) = \pi \ln(a) - \pi \ln(2-a) + \pi \ln \left(\ln \frac{2}{a} \right) +C = \pi \ln \left(\frac{a}{2-a} \, \ln \frac{2}{a} \right) +C$$ for $a>1$, $a \ne 2$.

Answer 2

With no proof.

A bit of data analysis suggests that, for large values of $a$ $$\Omega(a) \sim \pi \log \left(\log\left(\frac{a}{2}+\log(a)\right)\right)$$ $$\left( \begin{array}{ccc} a & \text{approximation} & \text{integration}\\ 10^1 & 2.1590407 & 2.1960629 \\ 10^2 & 4.3552732 & 4.3487727 \\ 10^3 & 5.7463125 & 5.7456735 \\ 10^4 & 6.7302431 & 6.7301927 \\ 10^5 & 7.4813795 & 7.4813755 \\ 10^6 & 8.0874648 & 8.0874645 \\ 10^7 & 8.5953549 & 8.5953549 \\ 10^8 & 9.0324528 & 9.0324528 \\ \end{array} \right)$$

The correlation between the two vectors is $0.9999923$.