Here are two integrals:
$$1:\hspace{2cm} \int_0^1\frac{\log x\arctan\left(\frac{\log x}{2\pi}\right)}{1+x^2}dx=\frac{\pi^2}{4}\log\left(\frac{192\pi^4}{\Gamma\left(\frac14\right)^8}\right)+\frac{\pi C}{2} $$
$$2:\hspace{1.7cm}\int_0^1\frac{\log x\arctan\left(\frac{\pi}{2\log x}\right)}{1+x^2}dx=\frac{\pi C}{2}+\frac{\pi^2}{8}-\frac{\pi^2}{24}\log\left(\frac{A^{36}}{2\pi^3}\right) $$
where $C$ is Catalan's constant and $A$ is Glaisher's constant.
My proof for both of them uses Feynman's trick on
$$G(a)=\int_0^1\frac{\log x\arctan\left(\frac{a}{\log x}\right)}{1+x^2}dx $$
but it requires to know some values of $\psi^{-2}(x)$, which I had to look up on Wikipedia.
So, given the "nice" closed forms, I hope there is a way to evaluate them without using Feynman technique.
For instance in $(1)$ I tried to expand the $\arctan$ as a series, and then use the fact that
$$\int_0^1\frac{\log^{2n}(x)}{1+x^2}dx=\frac{(-1)^n}{2}\left(\frac{\pi}{2}\right)^{2n+1}E_{2n} $$
to get a series with Euler numbers. But this is wrong, since $\left|\frac{\log x}{2\pi}\right|>1$ for $0<x<1$, and so the series is not convergent. In contrast, if one tries the same approach on $(2)$, where the series is convergent due to $\left|\frac{\pi}{2\log x}\right|<1$, one gets integrals of the form
$$\int_0^1\frac{\log^{-2n}(x)}{1+x^2}dx$$
which are not convergent at all.
So how would you do it?
Best Answer
Original Post
I've got the first one out, but I'm not sure if the solution meets the criteria since I use double integral which is in some sense, the same with Feynman's technique. Alas, I reckon I write it down here for the funzies.
lemma 1 $$ I\left( a \right) =\int_0^1{\ln \left( \Gamma \left( t+a \right) \right)}\mathrm{d}t=\int_a^{a+1}{\ln \left( \Gamma \left( t \right) \right)}\mathrm{d}t=a\ln \left( a \right) -a+\frac{1}{2}\ln \left( 2\pi \right) $$ proof : \begin{align*} I\left( 0 \right) =&\int_0^1{\ln \left( \Gamma \left( t \right) \right)}\mathrm{d}t=\int_0^1{\ln \left( \Gamma \left( 1-t \right) \right)}\mathrm{d}t \\ =&\frac{1}{2}\int_0^1{\ln \left( \Gamma \left( t \right) \Gamma \left( 1-t \right) \right)}\mathrm{d}t=\frac{1}{2}\int_0^1{\ln \left( \frac{\pi}{\sin \left( \pi t \right)} \right)}\mathrm{d}t=\frac{1}{2}\ln \left( 2\pi \right) \end{align*} Also $$ I'\left( a \right) =\frac{\mathrm{d}}{\mathrm{d}a}\int_a^{a+1}{\ln \left( \Gamma \left( t \right) \right)}\mathrm{d}t=\ln \left( \Gamma \left( a+1 \right) \right) -\ln \left( \Gamma \left( a \right) \right) =\ln \left( a \right) $$ Therefore $$ I\left( a \right) =\int_0^a{I'\left( t \right) \mathrm{d}t}-I\left( 0 \right) =\int_0^a{\ln \left( t \right) \mathrm{d}t}-I\left( 0 \right) =a\ln \left( a \right) -a+\frac{1}{2}\ln \left( 2\pi \right) $$
Integral 1 \begin{align*} I_1=&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{\ln \left( x \right)}{2\pi} \right)}\mathrm{d}x \\ =&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\left( \frac{\pi}{2}-\mathrm{arctan} \left( \frac{2\pi}{\ln \left( x \right)} \right) \right)}\mathrm{d}x \\ =&\frac{\pi}{2}\int_0^1{\frac{\ln \left( x \right)}{1+x^2}}\mathrm{d}x-\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{2\pi}{\ln \left( x \right)} \right)}\mathrm{d}x \\ =&\frac{\pi}{2}\mathbf{G}-\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{2\pi}{\ln \left( x \right)} \right)}\mathrm{d}x \\ =&\frac{\pi}{2}\mathbf{G}-2\pi \int_0^1{\int_0^1{\frac{1}{1+x^2}\left( \frac{\ln ^2\left( x \right)}{\ln ^2\left( x \right) +4\pi ^2t^2} \right)}}\mathrm{d}t\mathrm{d}x \\ =&\frac{\pi}{2}\mathbf{G}-\frac{\pi ^2}{2}+4\pi ^2\int_0^1{\int_0^1{\frac{t}{1+x^2}\left( \frac{2\pi t}{\ln ^2\left( x \right) +4\pi ^2t^2} \right)}}\mathrm{d}t\mathrm{d}x \\ =&\frac{\pi}{2}\mathbf{G}-\frac{\pi ^2}{2}+4\pi ^2\int_0^{\infty}{\int_0^1{\frac{te^{-x}}{1+e^{-2x}}\left( \frac{2\pi t}{x^2+4\pi ^2t^2} \right)}}\mathrm{d}t\mathrm{d}x \end{align*} The last step, $-\ln \left( x \right) \rightsquigarrow x$. Now, let's evaluate the integral expression solely by itself othewise this is getting out of hands. Anyway, \begin{align*} 4\pi ^2\int_0^1{\int_0^{\infty}{\frac{te^{-x}}{1+e^{-2x}}\left( \frac{2\pi t}{x^2+4\pi ^2t^2} \right)}}\mathrm{d}x\mathrm{d}t =&4\pi ^2\int_0^1{\int_0^{\infty}{t\left( \sum_{n=0}^{\infty}{\left( -1 \right) ^ne^{-\left( 2n+1 \right) x}} \right) \left( \int_0^{\infty}{\cos \left( xs \right) e^{-2\pi ts}\mathrm{d}s} \right)}}\mathrm{d}x\mathrm{d}t \\ =&4\pi ^2\int_0^1{\int_0^{\infty}{te^{-2\pi ts}\sum_{n=0}^{\infty}{\left( -1 \right) ^n}\int_0^{\infty}{\cos \left( xs \right) e^{-\left( 2n+1 \right) x}\mathrm{d}x}}}\mathrm{d}s\mathrm{d}t \\ =&4\pi ^2\int_0^1{\int_0^{\infty}{te^{-2\pi ts}\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n+1 \right)}{\left( 2n+1 \right) ^2+s^2}}}}\mathrm{d}s\mathrm{d}t \\ =&\pi ^3\int_0^1{\int_0^{\infty}{te^{-2\pi ts}\mathrm{sech} \left( \frac{\pi s}{2} \right) \mathrm{d}s}}\mathrm{d}t \\ \stackrel{(*)}{=}&4\pi ^2\int_0^1{t\int_0^1{\frac{u^{4t}}{u^2+1}\mathrm{d}u}}\mathrm{d}t \\ =&4\pi ^2\int_0^1{t\int_0^1{u^{4t}\sum_{n=0}^{\infty}{\left( -1 \right) ^nu^{2n}}\mathrm{d}u}}\mathrm{d}t \\ =&4\pi ^2\int_0^1{t\sum_{n=0}^{\infty}{\left( -1 \right) ^n}\int_0^1{u^{4t+2n}\mathrm{d}u}}\mathrm{d}t \\ =&4\pi ^2\int_0^1{t\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{4t+2n+1}}}\mathrm{d}t \\ =&\pi ^2\int_0^1{t\sum_{n=0}^{\infty}{\left( \frac{1}{n+t+1/4}-\frac{1}{n+t+3/4} \right)}}\mathrm{d}t \\ =&\pi ^2\int_0^1{t\left( \psi \left( t+\frac{3}{4} \right) -\psi \left( t+\frac{1}{4} \right) \right)}\mathrm{d}t \\ =&\left. \pi ^2t\ln \left( \frac{\Gamma \left( t+\frac{3}{4} \right)}{\Gamma \left( t+\frac{1}{4} \right)} \right) \right|_{0}^{1}+\pi ^2\left( I\left( \frac{1}{4} \right) -I\left( \frac{3}{4} \right) \right) \\ =&\frac{\pi ^2}{4}\ln \left( \frac{324\pi ^4}{\Gamma ^8\left( \frac{1}{4} \right)} \right) +\frac{\pi ^2}{4}\ln \left( \frac{16}{27} \right) +\frac{\pi ^2}{2} \end{align*} In (*), let $e^{-\frac{\pi s}{2}}=u$. So we have, at the end $$ I_1=\frac{\pi}{2}\mathbf{G}-\frac{\pi ^2}{2}+\frac{\pi ^2}{4}\ln \left( \frac{324\pi ^4}{\Gamma ^8\left( \frac{1}{4} \right)} \right) +\frac{\pi ^2}{4}\ln \left( \frac{16}{27} \right) +\frac{\pi ^2}{2}=\frac{\pi}{2}\mathbf{G}+\frac{\pi ^2}{4}\ln \left( \frac{192\pi ^4}{\Gamma ^8\left( \frac{1}{4} \right)} \right) $$
The General Case
Similar methedology shown above can be deployed for the development of the general case:
$$ \int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{\theta}{\ln \left( x \right)} \right)}\mathrm{d}x=\frac{\pi ^2}{4}\ln \left( \frac{\Gamma ^3\left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \Gamma \left( \frac{3}{4} \right) \mathrm{G}^4\left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}^4\left( \frac{3}{4} \right)}{\Gamma ^3\left( \frac{1}{4} \right) \Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}^4\left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}^4\left( \frac{1}{4} \right)} \right) $$
lemma 2 $$ \int_0^z{\ln \left( \Gamma \left( x \right) \right)}\mathrm{d}x=\frac{z}{2}\ln \left( 2\pi \right) +\frac{z\left( 1-z \right)}{2}+z\ln \left( \Gamma \left( z \right) \right) +\ln \left( \mathrm{G}\left( z+1 \right) \right) $$ proof : Omited, The way to derive this is to integrate the fourier expansion of log gamma funcion, which is VERY TEDIOUS IN EVERY ASPECT
lemma 3 \begin{align*} &\int_0^1{\ln \left( \Gamma \left( a+bt \right) \right)}\mathrm{d}t \\ =&\frac{1}{2}\ln \left( 2\pi \right) +\frac{1-b-2a}{2}+\frac{a}{b}\ln \left( \frac{\Gamma \left( a+b \right)}{\Gamma \left( a \right)} \right) +\ln \left( \Gamma \left( a+b \right) \right) -\frac{1}{b}\ln \left( \frac{\mathrm{G}\left( a+b+1 \right)}{\mathrm{G}\left( a+1 \right)} \right) \end{align*} proof : \begin{align*} &\int_0^1{\ln \left( \Gamma \left( a+bt \right) \right)}\mathrm{d}t=\frac{1}{b}\int_a^{a+b}{\ln \left( \Gamma \left( t \right) \right)}\mathrm{d}t=\frac{1}{b}\left( \int_0^{a+b}{\ln \left( \Gamma \left( t \right) \right)}\mathrm{d}t-\int_0^a{\ln \left( \Gamma \left( t \right) \right)}\mathrm{d}t \right) \\ =&\frac{1}{2}\ln \left( 2\pi \right) +\frac{1-b-2a}{2}+\frac{a}{b}\ln \left( \frac{\Gamma \left( a+b \right)}{\Gamma \left( a \right)} \right) +\ln \left( \Gamma \left( a+b \right) \right) -\frac{1}{b}\ln \left( \frac{\mathrm{G}\left( a+b+1 \right)}{\mathrm{G}\left( a+1 \right)} \right) \end{align*}
Proving the general case
\begin{align*} I\left( \theta \right) =&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{\theta}{\ln \left( x \right)} \right)}\mathrm{d}x \\ =&\int_0^1{\frac{\theta}{1+x^2}\int_0^1{\frac{\ln ^2\left( x \right)}{\ln ^2\left( x \right) +\theta ^2t^2}\mathrm{d}t}}\mathrm{d}x \\ =&\int_0^1{\frac{\theta}{1+x^2}\left( 1-\int_0^1{\frac{\theta ^2t^2}{\ln ^2\left( x \right) +\theta ^2t^2}\mathrm{d}t} \right)}\mathrm{d}x \\ =&\frac{\pi \theta}{4}-\int_0^1{\int_0^1{\frac{\theta ^2t}{1+x^2}\left( \frac{\theta t}{\ln ^2\left( x \right) +\theta ^2t^2} \right) \mathrm{d}t}}\mathrm{d}x \\ =&\frac{\pi \theta}{4}-\int_0^1{\int_0^1{\theta ^2t\left( \sum_{n=0}^{\infty}{\left( -1 \right) ^nx^{2n}} \right) \left( \int_0^{\infty}{\cos \left( \ln \left( x \right) s \right) e^{-\theta ts}\mathrm{d}s} \right)}}\mathrm{d}t\mathrm{d}x \\ =&\frac{\pi \theta}{4}-\int_0^{\infty}{\int_0^1{\theta ^2te^{-\theta ts}\sum_{n=0}^{\infty}{\left( -1 \right) ^n}\int_0^1{\cos \left( \ln \left( x \right) s \right) x^{2n}\mathrm{d}x}}}\mathrm{d}t\mathrm{d}s \\ =&\frac{\pi \theta}{4}-\int_0^{\infty}{\int_0^1{\theta ^2te^{-\theta ts}\sum_{n=0}^{\infty}{\left( -1 \right) ^n}\Re \left\{ \int_0^1{x^{2n+is}\mathrm{d}x} \right\}}}\mathrm{d}t\mathrm{d}s \\ =&\frac{\pi \theta}{4}-\int_0^{\infty}{\int_0^1{\theta ^2te^{-\theta ts}\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n+1 \right)}{\left( 2n+1 \right) ^2+s^2}}}}\mathrm{d}t\mathrm{d}s \\ =&\frac{\pi \theta}{4}-\int_0^1{\int_0^{\infty}{\theta ^2te^{-\theta ts}\frac{\pi}{4}\mathrm{sech} \left( \frac{\pi s}{2} \right) \mathrm{d}s}}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\frac{\theta ^2\pi}{2}\int_0^1{\int_0^{\infty}{t\frac{e^{-\left( \frac{\pi}{2}+\theta t \right) s}}{1+e^{-\pi s}}\mathrm{d}s}}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\frac{\theta ^2\pi}{2}\int_0^1{\int_0^{\infty}{te^{-\left( \frac{\pi}{2}+\theta t \right) s}\sum_{n=0}^{\infty}{\left( -1 \right) ^ne^{-n\pi s}}\mathrm{d}s}}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\frac{\theta ^2\pi}{2}\int_0^1{t\sum_{n=0}^{\infty}{\left( -1 \right) ^n}\int_0^{\infty}{e^{-\left( \frac{\pi}{2}\left( 2n+1 \right) +\theta t \right) s}\mathrm{d}s}}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\theta ^2\int_0^1{t\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n+1 \right) +\frac{2\theta t}{\pi}}}}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\frac{\theta ^2}{4}\int_0^1{t\sum_{n=0}^{\infty}{\left[ \frac{1}{n+\frac{1}{4}+\frac{\theta t}{2\pi}}-\frac{1}{n+\frac{3}{4}+\frac{\theta t}{2\pi}} \right]}}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\frac{\theta ^2}{4}\int_0^1{t\left( \psi \left( \frac{3}{4}+\frac{\theta t}{2\pi} \right) -\psi \left( \frac{1}{4}+\frac{\theta t}{2\pi} \right) \right)}\mathrm{d}t \\ =&\frac{\pi \theta}{4}-\frac{\pi \theta}{2}\int_0^1{t}\mathrm{d}\left[ \ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta t}{2\pi} \right) \right) -\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta t}{2\pi} \right) \right) \right] \end{align*} Here goes the part where my lose my sanity. By performing integraion by part we yield: \begin{align*} \frac{\pi \theta}{4}-\frac{\pi \theta}{2}\left[ \ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \right) -\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \right) -\int_0^1{\ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta t}{2\pi} \right) \right)}\mathrm{d}t+\int_0^1{\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta t}{2\pi} \right) \right)}\mathrm{d}t \right] \end{align*} Then you expand the remaining integral term by lemma 3, you have the following carnage of brain cells...which I persevered to simplify... \begin{align*} &\ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \right) -\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \right) +\int_0^1{\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta t}{2\pi} \right) \right)}\mathrm{d}t-\int_0^1{\ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta t}{2\pi} \right) \right)}\mathrm{d}t \\ =&\ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \right) -\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \right) +\frac{1}{2}\ln \left( 2\pi \right) +\frac{2\pi -\theta}{4\pi}-\frac{1}{4}+\frac{\pi}{2\theta}\ln \left( \frac{\Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{1}{4} \right)} \right) +\ln \left( \Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \right) -\frac{2\pi}{\theta}\ln \left( \frac{\mathrm{G}\left( \frac{5}{4}+\frac{\theta}{2\pi} \right)}{\mathrm{G}\left( \frac{5}{4} \right)} \right) -\frac{1}{2}\ln \left( 2\pi \right) -\frac{2\pi -\theta}{4\pi}+\frac{3}{4}-\frac{3\pi}{2\theta}\ln \left( \frac{\Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{3}{4} \right)} \right) -\ln \left( \Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \right) +\frac{2\pi}{\theta}\ln \left( \frac{\mathrm{G}\left( \frac{7}{4}+\frac{\theta}{2\pi} \right)}{\mathrm{G}\left( \frac{7}{4} \right)} \right) \\ =&\frac{1}{2}+\frac{\pi}{2\theta}\ln \left( \frac{\Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{1}{4} \right)} \right) -\frac{3\pi}{2\theta}\ln \left( \frac{\Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{3}{4} \right)} \right) -\frac{2\pi}{\theta}\ln \left( \frac{\mathrm{G}\left( \frac{5}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}\left( \frac{7}{4} \right)}{\mathrm{G}\left( \frac{7}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}\left( \frac{5}{4} \right)} \right) \end{align*} Which when we plug it back to our solution, followed by a lot of manipulation and cancellation, we have the a closed form purposed atop: \begin{align*} I\left( \theta \right) =&\frac{\pi \theta}{4}-\frac{\pi \theta}{2}\left( \frac{1}{2}+\frac{\pi}{2\theta}\ln \left( \frac{\Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{1}{4} \right)} \right) -\frac{3\pi}{2\theta}\ln \left( \frac{\Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{3}{4} \right)} \right) -\frac{2\pi}{\theta}\ln \left( \frac{\mathrm{G}\left( \frac{5}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}\left( \frac{7}{4} \right)}{\mathrm{G}\left( \frac{7}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}\left( \frac{5}{4} \right)} \right) \right) \\ =&-\frac{1}{4}\pi ^2\ln \left( \frac{\Gamma \left( \frac{1}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{1}{4} \right)} \right) +\frac{3}{4}\pi ^2\ln \left( \frac{\Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{3}{4} \right)} \right) +\pi ^2\ln \left( \frac{\mathrm{G}\left( \frac{5}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}\left( \frac{7}{4} \right)}{\mathrm{G}\left( \frac{7}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}\left( \frac{5}{4} \right)} \right) \\ =&\frac{\pi ^2}{4}\ln \left( \frac{\Gamma ^3\left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \Gamma \left( \frac{3}{4} \right) \mathrm{G}^4\left( \frac{1}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}^4\left( \frac{3}{4} \right)}{\Gamma ^3\left( \frac{1}{4} \right) \Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}^4\left( \frac{3}{4}+\frac{\theta}{2\pi} \right) \mathrm{G}^4\left( \frac{1}{4} \right)} \right) \end{align*}
Put it into a different from: $$ I\left( \theta \right) =\frac{\pi ^2}{4}\ln \left( \frac{\Gamma ^3\left( \frac{1}{4}+\frac{\theta}{2\pi} \right)}{\Gamma \left( \frac{3}{4}+\frac{\theta}{2\pi} \right)} \right) +\pi ^2\ln \left( \frac{\mathrm{G}\left( \frac{1}{4}+\frac{\theta}{2\pi} \right)}{\mathrm{G}\left( \frac{3}{4}+\frac{\theta}{2\pi} \right)} \right) +\frac{\pi}{2}\mathbf{G} $$
The result
we can plug in different value into the expression and simplify it to get the result OP claimed, like: \begin{align*} I\left( \frac{\pi}{2} \right) =&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{\pi}{2\ln \left( x \right)} \right)}\mathrm{d}x=\frac{\pi ^2}{4}\ln \left( \Gamma ^3\left( \frac{1}{2} \right) \right) +\pi ^2\ln \left( \mathrm{G}\left( \frac{1}{2} \right) \right) +\frac{\pi}{2}\mathbf{G} \\ =&\frac{\pi ^2}{4}\ln \left( \Gamma ^3\left( \frac{1}{2} \right) \mathrm{G}^4\left( \frac{1}{2} \right) \right) +\frac{1}{2\pi}\mathrm{Cl}_2\left( \frac{\pi}{2} \right) =\frac{\pi ^2}{24}\ln \left( \frac{2\pi ^3}{\mathbf{A}^{36}} \right) +\frac{\pi ^2}{8}+\frac{1}{2\pi}\mathbf{G} \\ I\left( \pi \right) =&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{\pi}{\ln \left( x \right)} \right)}\mathrm{d}x=\frac{\pi ^2}{2}\ln \left( \frac{\sqrt{2}}{\pi}\Gamma ^2\left( \frac{3}{4} \right) \right) +\pi \mathbf{G} \\ I\left( \frac{3}{2}\pi \right) =&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{3\pi}{2\ln \left( x \right)} \right)}\mathrm{d}x=\frac{\pi ^2}{24}\ln \left( \frac{2^7\mathbf{A}^{36}}{\pi ^9} \right) +\frac{\pi ^2}{8}+\frac{\pi}{2}\mathbf{G} \\ I\left( 2\pi \right) =&\int_0^1{\frac{\ln \left( x \right)}{1+x^2}\mathrm{arctan} \left( \frac{2\pi}{\ln \left( x \right)} \right)}\mathrm{d}x=\frac{\pi ^2}{4}\ln \left( \frac{\Gamma ^3\left( \frac{1}{4}+1 \right) \Gamma \left( \frac{3}{4} \right) \mathrm{G}^4\left( \frac{1}{4}+1 \right) \mathrm{G}^4\left( \frac{3}{4} \right)}{\Gamma ^3\left( \frac{1}{4} \right) \Gamma \left( \frac{3}{4}+1 \right) \mathrm{G}^4\left( \frac{3}{4}+1 \right) \mathrm{G}^4\left( \frac{1}{4} \right)} \right) \\ =&\frac{\pi ^2}{4}\ln \left( \frac{1}{48}\frac{\Gamma ^3\left( \frac{1}{4} \right) \Gamma \left( \frac{3}{4} \right) \Gamma ^4\left( \frac{1}{4} \right) \mathrm{G}^4\left( \frac{1}{4} \right) \mathrm{G}^4\left( \frac{3}{4} \right)}{\Gamma ^3\left( \frac{1}{4} \right) \Gamma \left( \frac{3}{4} \right) \Gamma ^4\left( \frac{3}{4} \right) \mathrm{G}^4\left( \frac{3}{4} \right) \mathrm{G}^4\left( \frac{1}{4} \right)} \right) =\frac{\pi ^2}{4}\ln \left( \frac{1}{192\pi ^4}\Gamma ^8\left( \frac{1}{4} \right) \right) \end{align*} There would be more result if my Mathematica worked...