Prove that:
$\displaystyle\int_{\pi/4}^{\pi/2} \ln (\ln(\tan x))dx =\frac{\pi}{2}\ln \left( \frac{\sqrt{2\pi} \Gamma \left(\dfrac{3}{4} \right)}{\Gamma \left(\dfrac{1}{4} \right)}\right)$
I know that the Vardi's Integral can be evaluated in terms of derivatives of Hurwitz Zeta Function. I would like to see a method which uses differentiation under the integral sign.
I'm going to sketch a way to transform the integral into a sum. The sum looks difficult, but it converges and numerically checks with the stated result.
Begin by substituting $u=\log{\tan{x}}$. Then
$$du = \frac{1}{\tan{x}} \sec^2{x} \, dx = \left ( \frac{1}{\tan{x}} + \tan{x} \right ) dx = (e^{-u} + e^u) \, dx $$
and
$$\begin{align} \int_{\pi/4}^{\pi/2} dx \: \ln (\ln(\tan x)) &= \int_0^{\infty} du \: \frac{\log{u}}{e^u + e^{-u}} \\ &= \int_0^{\infty} du \: \frac{e^{-u} \log{u}}{1+e^{-2 u}} \\ &= \int_0^{\infty} du \: e^{-u} \log{u} \sum_{k=0}^{\infty} (-1)^k e^{-2 k u} \end{align}$$
Reverse the order of sum and integral, which is justified by Fubini's Theorem (both sum and integral are absolutely convergent). Then we may write
$$\begin{align} \int_{\pi/4}^{\pi/2} dx \: \ln (\ln(\tan x)) &= \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} du \: e^{-(2 k+1) u} \log{u} \\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \int_0^{\infty} du \: e^{-u} \log{u} - \sum_{k=0}^{\infty} (-1)^k \frac{\log{(2 k+1)}}{2 k+1} \int_0^{\infty} du \: e^{-u} \\ &= -\frac{\pi}{4} \gamma + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\log{(2 k+1)}}{2 k+1} \\ \end{align} $$
where $\gamma$ is the Euler-Mascheroni constant. The sum on the right-hand sign is known:
$$ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\log{(2 k+1)}}{2 k+1} = \frac{\pi}{4} \gamma + \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} $$
Use the fact that
$$\Gamma{\left ( \frac{3}{4} \right )} \Gamma{\left ( \frac{1}{4} \right )} = \sqrt{2} \pi$$
to deduce that
$$ \int_{\pi/4}^{\pi/2} dx \: \ln (\ln(\tan x)) = \frac{\pi}{2} \log{\left [\sqrt{2 \pi} \frac{\Gamma{\left ( \frac{3}{4} \right )}}{\Gamma{\left ( \frac{1}{4} \right )}}\right ]} $$
As a matter of fact, there is a very nice paper by Iaroslav Blagouchine in which the Vardi's integral, as well as numerous integrals akin to it, are treated in details (more correctly they should be called Malmsten's integrals).
$$ I := \int_{\pi/4}^{\pi/2} \log \log \tan x \,dx $$
is, after substitution $x \mapsto \arctan e^{x}$, equal to
$$ \frac{1}{2} \int_{0}^{\infty} \frac{\log x}{\cosh x} \,dx. $$
Further, this can be expressed as
$$ \frac{\pi}{2} \int_{0}^{\infty} \frac{\log \pi x}{\cosh \pi x} \,dx = \frac{\pi}{2} \log \sqrt{\pi} \int_{-\infty}^{\infty} \frac{dx}{\cosh \pi x} + \frac{\pi}{8} \int_{-\infty}^{\infty} \frac{\log x^{2}}{\cosh \pi x} \,dx. $$
Here, $\int_{-\infty}^{\infty} \frac{dx}{\cosh \pi x} = 1$, and according to the formula
$$ \int_{-\infty}^{\infty} \frac{\log(\alpha^{2}+x^{2})}{\cosh \pi x} \,dx = 4 \log \left(\sqrt{2} \; \frac{\Gamma\left(\frac{3}{4} + \frac{\alpha}{2}\right)}{\Gamma\left(\frac{1}{4} + \frac{\alpha}{2}\right)}\right) $$
for $\alpha \geq 0$, it follows that
$$ \frac{\pi}{8} \int_{-\infty}^{\infty} \frac{\log x^{2}}{\cosh \pi x} \,dx = \frac{\pi}{2} \log \left(\sqrt{2} \, \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right). $$
Therefore,
$$ I = \frac{\pi}{2} \log \left(\sqrt{2\pi} \; \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right). $$
In the formula
$$\int_{0}^{1} \frac{\log \log \left(\frac{1}{x}\right)}{1+2\cos \alpha \pi \cdot x+x^{2}} \,dx = \frac{\pi}{2\sin \alpha \pi} \left(\alpha \log 2\pi + \log \frac{\Gamma\left(\frac{1}{2}+\frac{\alpha}{2}\right)}{\Gamma\left(\frac{1}{2}-\frac{\alpha}{2}\right)}\right)$$
set $\alpha = \frac{1}{2}$:
$$\int_{0}^{1} \frac{\log \log \left(\frac{1}{x}\right)}{1+x^{2}} \,dx = \frac{\pi}{2} \left(\log \sqrt{2\pi} + \log \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$
Through the substitution $x\mapsto \cot x$, the aforementioned equation is obtained