A curious equation containing an integral $\int_0^{\pi/4}\arctan\left(\tan^x\theta\right)d\theta=\frac{\ln2\cdot\ln x}{16}$

I came across an interesting problem that I do not know how to solve:

Find $x>0$ such that $$\int_0^{\pi/4}\arctan\left(\tan^x\theta\right)d\theta=\frac{\ln2\cdot\ln x}{16}.$$

Could you suggest how to approach it?

Solution 1:

The only positive answer is $$x=3+\sqrt{8}.$$ The corresponding integral (in a slightly different form) is given as formula (17) and its proof is given by Theorem 10 in the paper "A restricted Epstein zeta function and the evaluation of some definite integrals"$^{[1]}$$^{[2]}$, H. Muzaffar, K. S. Williams, Acta Arith. 104 (2002), 23-66, DOI:10.4064/aa104-1-2.

The same integral also appears as formula 120 in section 4.1.6, page 169, in the book "Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas"$^{[3]}$$^{[4]}$, Yuri A. Brychkov, Chapman and Hall/CRC, ISBN 978-1584889564.

Solution 2:

It would be amazing if this had a closed form solution. My approach was to use Newton's method, with a simple trapezoidal rule for the integral. For all initial guesses I tried I obtained the numerical result $x \approx 5.8284271247461$. Lookup using W|A or the ISC reveals that the most likely candidate is the algebraic number $$ x = 3 + 2\sqrt{2} \approx \color{red}{5.8284271247461}90. $$ If we can prove that there is a unique solution then it will suffice to prove that $$\int_0^{\pi/4}\arctan\left(\tan^{3+2\sqrt{2}}\theta\right)d\theta=\frac{\ln2\cdot\ln (3+2\sqrt{2})}{16}.$$