A Challenging Integral $\int_0^{\frac{\pi}{2}}\log \left( x^2+\log^2(\cos x)\right)dx$

Question

I encountered a strange integral with a strange result.

$$\int_0^{\frac{\pi}{2}}\log \left( x^2+\log^2(\cos x)\right)dx = \pi \log \left(\log (2) \right)$$

Believe it or not, the result agrees numerically.

How can we prove this result?

Please help me. I feel very curious to know how this can proved.

Answer 1

Performine the change of variables: $z = e^{ix}$, then , $x =\frac{1}{i}\log(z)$. The integral takes the form:

$ I = \Re \int_{|z|=1 \arg(z)=0}^{|z|=1 \arg(z)=\frac{\pi}{2}} \log \big(-(\log(z))^2 +(\log(\frac{z^2+1}{2z}))^2\big ) \frac{dz}{iz} $

The real part is added, since the logarithm of the cosine is singular at $x = \frac{\pi}{2}$ and can pick up an imaginary part. Expressing the difference of squares as a product we obtain:

$ = \Re\int_{|z|=1 \arg(z)=0}^{|z|=1\arg(z)=\frac{\pi}{2}}\big ( (\log(\log(\frac{z^2+1}{2})) + (\log(\log(\frac{z^{-2}+1}{2})) \frac{dz}{iz}$.

The second part of the integral can be brought to the form of the first part by the transformation $z\rightarrow z^{-1}$ , thus

$ I = \Re\int_{|z|=1 \arg(z)=0}^{|z|=1\arg(z)=\pi}\big ( (\log(\log(\frac{z^2+1}{2})) \frac{dz}{iz}$.

The integrand is invariant under the transformation $z\rightarrow -z$, thus: $ I = \Re\frac{1}{2}\oint_{|z|=1 } \big ( (\log(\log(\frac{z^2+1}{2})) \frac{dz}{iz}$.

The numerator has no poles in the unit disc, thus using the residue theorem:

$I = \Re \frac{2\pi i}{2 i}\log(\log(-2)) = \pi \log(\log(2))$.

Answer 2

This is another answer from antas kaushik to solve this problem. Amazing method using the Feynman trick