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.

  • 5
    $\begingroup$ I may be wrong but this looks related to Gauss mean value theorem. $\endgroup$ Jun 22, 2013 at 10:47
  • $\begingroup$ @O.L: Thank you so much for this hint! I just got the answer using Gauss mean value theorem. (+1) $\endgroup$ Jun 22, 2013 at 11:07
  • $\begingroup$ @O.L: Do you have idea about the other integral, I posted some time ago - math.stackexchange.com/questions/426325/… It seems to be very tough. $\endgroup$ Jun 22, 2013 at 11:11
  • $\begingroup$ You can find a similar technique for evaluating this integral here. I am also considering a real-analytic method, though not successful so far. $\endgroup$ Aug 4, 2013 at 16:05

3 Answers 3


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:

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


Since the other thread was marked duplicate, I will post my solution here as well.

The integral is: $$2\Re\int_0^{\pi/2} \ln\ln\left(\frac{1+e^{2ix}}{2}\right)\,dx=2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx$$ Consider $$f(x)=\ln(\ln(1+x)-\ln2)$$ Around $x=0$, the taylor expansion can be written as: $$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+....$$ Replace $x$ with $e^{2ix}$. Notice that integrating the powers of $e^{2ix}$ would result in either zero or a purely imaginary number and since the derivatives of $f(x)$ at $0$ are real, we need to consider only the constant term i.e $f(0)$. Since $f(0)=\ln(-\ln 2)=\ln\ln 2+i\pi$, hence, $$2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx=2\int_0^{\pi/2} \ln\ln 2\,dx=\boxed{\pi\ln\ln 2}$$



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


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