Closed form for $\int_0^\infty\arctan\Bigl(\frac{2\pi}{x-\ln\,x+\ln(\frac\pi2)}\Bigr)\frac{dx}{x+1}$ [closed]

Computing a Related Contour Integral:

Define $$f(z)=\frac{i}{2}\frac{z-1}{1+az}\left(\frac{1}{z-\ln{z}+\ln\left(\frac{\pi}{2}\right)}+\frac{1}{z-\ln{z}+2\pi i+\ln\left(\frac{\pi}{2}\right)}\right)$$ and let $\gamma$ denote a keyhole contour deformed around $[0,\infty]$. Restricting the argument between $0$ and $2\pi$, it is not hard to see that $f(z)$ has poles at $z=-\dfrac{1}{a}$, $z=-W_{-1}\left(-\dfrac{\pi}{2}\right)=\dfrac{\pi i}{2}$, and $z=-W_0\left(-\dfrac{\pi}{2}\right)=-\dfrac{\pi i}{2}$. The residues at these poles are \begin{align} \operatorname*{Res}_{z=\frac{\pi i}{2}}f(z) &=\frac{i}{2}\frac{\frac{\pi i}{2}-1}{\frac{\pi i}{2}a+1}\frac{1}{1-\frac{2}{\pi i}}\\ \operatorname*{Res}_{z=-\frac{\pi i}{2}}f(z) &=\frac{i}{2}\frac{\frac{\pi i}{2}+1}{\frac{\pi i}{2}a-1}\frac{1}{1+\frac{2}{\pi i}}\\ \operatorname*{Res}_{z=-\frac{1}{a}}f(z) &=-\frac{i}{2}u'(a)\left(\frac{1}{u(a)+\ln\left(\frac{\pi}{2}\right)-\pi i}+\frac{1}{u(a)+\ln\left(\frac{\pi}{2}\right)+\pi i}\right)\\ \end{align} where $u(a)=\ln{a}-\dfrac{1}{a}$. By the residue theorem, \begin{align} \oint_{\gamma}f(z)\ dz &=2\pi i\sum_{z_k\in\left\{-a^{-1}, \pm\pi i/2\right\}}\operatorname*{Res}_{z=z_k}f(z)\\ &=\pi\left(\frac{2\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)}{\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}u'(a)-\frac{2\pi^2a}{\pi^2a^2+4}\right) \end{align}

Parameterisation of the Contour Integral:

We take the argument of $z$ to be $0$ above the branch cut, and $2\pi$ below the branch cut. Also, the contribution from the big arc is clearly $2\pi i\times\dfrac{i}{2}\times\dfrac{1}{a}\times (1+1)=-\dfrac{2\pi}{a}$. Taking all of these points into consideration, we eventually arrive at \begin{align} \oint_\gamma f(z)\ dz+\frac{2\pi}{a} &=\small\frac{i}{2}\int^\infty_0\frac{x-1}{1+ax}\left(-\frac{1}{x-\ln|x|-2\pi i+\ln\left(\frac{\pi}{2}\right)}+\frac{1}{x-\ln|x|+2\pi i+\ln\left(\frac{\pi}{2}\right)+\pi^2}\right)\ dx\\ &=2\pi\int^\infty_0\frac{x-1}{\left(x-\ln{x}+\ln\left(\frac{\pi} {2}\right)\right)^2+4\pi^2}\frac{dx}{1+ax}\\ \end{align}

Obtaining the Closed Form:

Integrating with respect to $a$, we obtain \begin{align} \small\int^\infty_0\frac{2\pi\left(1-\frac{1}{x}\right)\ln(1+ax)}{\left(x-\ln{x}+\ln\left(\frac{\pi}{2}\right)\right)^2+4\pi^2}\ dx &\small=\ \pi\int\left(\frac{2\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)}{\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}u'(a)-\frac{2\pi^2a}{\pi^2a^2+4}+\frac{2}{a}\right)\ da\\ &=\small\pi\left(\ln\left(\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2\right)-\ln\left(\pi^2a^2+4\right)+\ln{a^2}\right)+\text{const.}\\ &=\small\pi\ln\left(\frac{\left(\ln{a}-\frac{1}{a}+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}{\pi^2+\frac{4}{a^2}}\right)+\text{const.} \end{align} Letting $a\to 0$, we find that the constant term is $\pi\ln{4}$. Plugging in $a=1$ and integrating by parts, we finally arrive at the closed form. \begin{align} \int^\infty_0\frac{2\pi\left(1-\frac{1}{x}\right)\ln(1+x)}{\left(x-\ln{x}+\ln\left(\frac{\pi}{2}\right)\right)^2+4\pi^2}\ dx &=\int^\infty_0\arctan\left(\frac{2\pi}{x-\ln{x}+\ln\left(\frac{\pi}{2}\right)}\right)\frac{dx}{1+x}\\ &=\left.\pi\ln\left(\frac{\left(\ln{a}-\frac{1}{a}+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}{\frac{\pi^2}{4}+\frac{1}{a^2}}\right)\right|_{a=1}\\ &=\color{red}{\pi\ln\left(\frac{\ln^2\left(\frac{\pi}{2}\right)-2\ln\left(\frac{\pi}{2}\right)+1+\pi^2}{\frac{\pi^2}{4}+1}\right)} \end{align}

$$I=\pi\,\ln\left(\frac{1+\pi^2+\ln^2\left(\frac\pi2\right)-2\ln\left(\frac\pi2\right)}{1+\frac{\pi^2}4}\right)$$