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How to evaluate $$ \displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx $$ where $a\in[0,1]$ ?

I think of this problem as a generalization of the following proposition $$ \displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2 $$

My try

Put $$ I(a)=\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx $$ From the substitution $x \to \frac{\pi}{2}-x$ , we get $$ \displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx = \displaystyle\int_0^{\pi/2} \log \left| \cos^2 x - a \right|\,dx $$ Thus $$ \displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx = \displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - (1-a) \right|\,dx $$ which means $$I(a)=I(1-a) \tag{1}$$

On the other hand, \begin{align} 2I(a) &= \displaystyle\int_0^{\pi/2} \log \left| (\sin^2 x - a)(\cos^2 x -a) \right|\,dx \\ &= \displaystyle\int_0^{\pi/2} \log \left| a^2-a+\sin^2 x \cos^2 x \right|\,dx \\ &= \displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 (2x) \right|\,dx -\pi \log 2 \\ &= \frac{1}{2}\displaystyle\int_0^{\pi} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx -\pi \log 2 \\ &= \displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx -\pi \log 2 \\ &= \displaystyle\int_0^{\pi/2} \log \left| 1+4(a^2-a)-\sin^2 x \right|\,dx -\pi \log 2 \\ &= I((2a-1)^2) -\pi \log 2 \end{align} Thus $$ 2I(a)=I((2a-1)^2)-\pi \log 2 \tag{2} $$

Let $a=0$ we get the proposition mentioned above $\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2.$

But how to move on ? Can we solve the problem only by $(1)$ and $(2)$? Or what other properties should we use to evaluate that?

Looking forward to your new solutions as well.

Thank you in advance!

Added:

As pointed out in the comments, it seems like that the integral is identical to $-\pi\log 2$.

From $(1)$ and $(2)$ we can also find many numbers such that $I(a)=-\pi\log 2$.

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  • $\begingroup$ Breaking the integral into two with different interval so that the modulus gets converted , might help for solving the problem. $\endgroup$ – Bijayan Ray Mar 14 at 17:32
  • $\begingroup$ I'm fairly certain that the integral is identical to $-e$ for $a\in [0,1]$ $\endgroup$ – clathratus Mar 14 at 17:47
  • $\begingroup$ @BijayanRay I tried it but it was horrible :D $\endgroup$ – Advil Sell Mar 14 at 17:54
  • $\begingroup$ @clathratus It seems like that you meant it's identical to $-\pi \log 2$. $\endgroup$ – Zero Mar 14 at 18:14
  • $\begingroup$ Wait yeah its $-\pi\log2$ my bad. I just got my decimals mixed up I guess :) $\endgroup$ – clathratus Mar 14 at 18:16
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Using the symmetries and developing the $\sin^2$ term, we can express \begin{align} I&=\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \sin^2 x - a \right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1-2a}{2}-\frac{1}{2}\cos 2x\right|\,dx %&=-\frac{\pi}{2}\ln 2+\frac{1}{4}\int_0^{2\pi} \log \left| \left( 2a-1 \right)+\cos 2x\right|\,dx \end{align} By denoting $2a-1=\cos 2\alpha$, \begin{align} I&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1}{2}\left( \cos 2\alpha+\cos 2x\right)\right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x+\alpha \right)\cos \left( x-\alpha \right)\right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x+\alpha \right)\right|\,dx+\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x-\alpha \right)\right|\,dx \end{align} As the functions are periodic, the integration variables can be shifted, thus \begin{equation} I=\frac{1}{2}\int_0^{2\pi} \log \left| \cos \left( x \right)\right|\,dx \end{equation} Finally using the symmetries of the integrand, \begin{align} I&=2\int_0^{\pi/2} \log \left| \cos \left( x \right)\right|\,dx\\ &=2\int_0^{\pi/2} \log \left| \sin \left( x \right)\right|\,dx\\ &=-\pi\ln 2 \end{align} from the quoted result.

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  • $\begingroup$ This proof is very satisfying to read. Thank you very much! +1 $\endgroup$ – clathratus Mar 14 at 23:25

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