Evaluation of $\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$ [closed]

Question

$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$$

$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x))}{\tan(2x)} \,dx = -\int_{0}^{\frac{\pi}{4}} \log(-\log(\tan x)) \cdot \log(\tan x) \cdot \tan(2x) \,dx$$

$$= -2 \int_{0}^{\frac{\pi}{4}} \frac{\log(-\log(\tan x)) \cdot \log(\tan x) \cdot \tan x}{1-\tan^2 x} \,dx \Bigg|_{\tan x=e^{-t}}$$

$$= 2 \int_{0}^{\infty} \frac{t \cdot \log(t \cdot e^{-2t})}{1 - e^{-4t}} \,dt = 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} t \cdot \log(t \cdot e^{-(4n+2)t}) \,dt$$

$$= 2 \sum_{n=0}^{\infty} L \cdot (t \cdot \log(t)) \cdot (4n + 2)$$

Answer 1

$$\begin{align}I=&\int_{0}^{\frac{\pi}{4}}\frac{\ln\left(\ln\left(\tan\left(x+\frac{\pi}{4}\right)\right)\right)\cdot\ln\left(\tan\left(x+\frac{\pi}{4}\right)\right)}{\tan\left(2x\right)}dx\\ =&-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\left(\ln\left(\tan\left(x\right)\right)\right)\cdot\ln\left(\tan\left(x\right)\right)\tan\left(2x\right)dx\\ =&\int_{1}^{+\infty}\ln\left(\ln\left(t\right)\right)\ln\left(t\right)\cdot\frac{2t}{t^{4}-1}dt\\ =&\int_{0}^{\infty}\ln\left(s\right)s\frac{2e^{2s}}{e^{4s}-1}ds\end{align}$$ Using the Feymann's trick: $$f(z):=\int_{0}^{\infty}x^{z}\frac{2e^{2s}}{e^{4s}-1}ds$$ $$f(z)=(2^{-z} - 2^{-2z - 1}) \zeta(1 + z) \Gamma(1 + z)$$ $$I=f'(1)$$ $$I=\frac{\pi^2}{48}\left(3 + \ln(2) + 3 \ln(\pi)-36 \ln(A)\right)\approx -0.375831$$

Where:

• $\zeta(z)$ is the zeta Riemann function
• $\Gamma(z)$ is the Gamma function
• $A$ is the Glaisher-Kinkelin Constant