# Tough integrals: $\int_0^1\frac{\log x\arctan\left(\frac{\log x}{2\pi}\right)}{1+x^2}dx$

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 $$0, and so the series is not convergent. In contrast, if one tries the same approach on $$(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?

• $$2\pi^2\int_0^\infty t \,\tan ^{-1}(t) \,\text{sech}(2 \pi t)\,dt$$ could be easier Aug 24, 2023 at 10:43
• Would you like to name the source from where you found these awesome integrals? Aug 26, 2023 at 3:30
• Using some simple substitutions, I reduced $(1)$ into this one $$4\pi^2 \int_0^\infty \frac{e^{-2\pi t} \arctan{t}}{1+e^{-4\pi t}} \, \mathrm{d}t$$ But If I use geometric series for denominator that I think It would be difficult to evaluate. Aug 26, 2023 at 3:34
• @LuckyChouhan the source is me. I derived the first one from other simpler integrals, and then decided to try the general case.
– Zima
Aug 26, 2023 at 6:21
• Can you give a little more insight regarding how you solve the integral originally? Aug 27, 2023 at 17:28

# 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...

• This is a nice proof, yes the steps are similar to what I did with feynman but you used several different tricks and the result is amazing! I suppose for the second one the same exact steps you did for the first one would work as well, what do you think?
– Zima
Sep 15, 2023 at 15:50
• @Zima I've worked out the general case, typing... Sep 15, 2023 at 16:06
• This is exceptional work!
– Zima
Sep 15, 2023 at 17:09
• @Zima Never saw that notaion before but it's essentially what I have gone thruogh for the past 8 hours... Sep 15, 2023 at 17:17
• @User Cheers!!! Sep 17, 2023 at 10:30

You could use a series expansion $$\tan ^{-1}\left(\frac{\log (x)}{2 \pi }\right)\log (x)=\frac 1{2\pi}\Bigg(\sum_{n=1}^\infty a_n (x-1)^{2n}-\sum_{n=1}^\infty b_n (x-1)^{2n+1}\Bigg)$$ where all coefficients are positive (but very slowly decreasing) leading to the "simple" $$I=\sum_{n=1}^\infty \frac{a_n}{2(2 n+1)\pi}\,\, _3F_2\left(\frac{1}{2},1,1;n+1,n+\frac{3}{2};-1\right)+$$ $$\sum_{n=1}^\infty \frac{b_n}{ 4(n+1)\pi}\,\, _3F_2\left(\frac{1}{2},1,1;n+\frac{3}{2},n+2;-1\right)$$

• But how to find $a_n$ and $b_n$?
– Zima
Aug 26, 2023 at 7:01
• @Zima. The problem is not to get them but to find their general formulation. Cheers :-) Aug 26, 2023 at 7:15