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Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$

I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar. Anyway I thought it's quite a trivial integral so I'm gonna solve it quickly, but I am having some hard time to finish it. I went on with Feynman's trick:

$$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{dx}{a+x+ax^2}$$ $$=\frac1a\int_0^\infty \frac{dx}{\left(x+\frac{1}{2a}\right)^2+1-\frac{1}{4a^2}}=\frac{1}{a}\frac{1}{\sqrt{1-\frac{1}{4a^2}}}\arctan\left(\frac{x+\frac{1}{2a}}{\sqrt{1-\frac{1}{4a^2}}}\right)\bigg|_0^\infty$$$$=\frac{\pi}{\sqrt{4a^2-1}}-\frac{2}{\sqrt{4a^2-1}}\arctan\left(\frac{1}{\sqrt{4a^2-1}}\right)=\frac{2\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}$$ We can prove easily via the substitution $x\to \frac{1}{x}$ that $I(0)=0$ so we have that: $$I=I(1)-I(0)=2\int_0^1 \frac{\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}da$$ Now I thought about two substitutions: $$ \overset{a=\frac12\cosh x}=\int_{\operatorname{arccosh}(0)}^{\operatorname{arccosh}(2)} \arctan(\sinh x)dx$$ $$\overset{a=\frac12\sec x}=\int_{\operatorname{arcsec}(0)}^{\frac{\pi}{3}}\frac{x}{\cos x}dx$$ But in both cases the lower bound is annoying and I think I am missing something here (maybe obvious). So I would love to get some help in order to finish this.


Edit: We can apply once again Feynman's trick. First consider: $$I(t)=\int_0^1 \frac{2\arctan(t\sqrt{4a^2-1})}{\sqrt{4a^2-1}}da\Rightarrow I'(t)=2\int_0^1 \frac{1}{1+t^2(4a^2-1)}da$$ $$=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2at}{\sqrt{1-t^2}}\right)\bigg|_0^1=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)$$ So once again we have $I(0)=0$, so $I=I(1)-I(0)$. $$\Rightarrow I=\int_0^1\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)dt\overset{t=\sin x}=\int_0^\frac{\pi}{2}\frac{\arctan(2\tan x)}{\sin x}dx$$ At this point Mathematica can evaluate the integral to be: $$I=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$ I didn't try the last integral yet, but I am thinking of Feynman again $\ddot \smile$.


Edit 2: Found that I already was on it some time ago, and actually posted it here, which means I have solved it before using Feynman's trick, but right now I can't remember how I did it.

So given the circumstances I am positive that it can be solved starting with my approach, but if you have any other ways then feel free to share it.

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    $\begingroup$ The integral is complex for $a\in(-1/2,1/2)$. $\endgroup$ – John Wayland Bales Apr 28 at 18:31
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    $\begingroup$ And after $n$ applications of Feynman, you come back to the original integral. :) $\endgroup$ – John Wayland Bales Apr 28 at 18:45
  • $\begingroup$ Well, it's not hard to reduce the problem to finding $$\int_0^{\frac\pi2}\log\left(1+\frac12\sin x\right)\mathrm dx$$ I remember a post on AoPS about a similair integral, namely with a variable instead of the $\frac12$ but I cannot find the post right now. $\endgroup$ – mrtaurho Apr 28 at 18:55
  • $\begingroup$ @Zacky No, that's not all; but that's the part I cannot solve by myself ^^' $\endgroup$ – mrtaurho Apr 28 at 18:58
  • $\begingroup$ @Zacky Precisely! I think there is a possible way of solving this integral using Clausen's Function but I am not sure where to start. $\endgroup$ – mrtaurho Apr 28 at 18:59
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Start by letting $x\mapsto\tan x$ we obtain $$\int_0^\infty\frac{\log(1+x+x^2)}{1+x^2}\mathrm dx\stackrel{x\mapsto\tan x}=\int_0^\frac\pi2\log(1+\tan x+\tan^2x)\mathrm dx=\int_0^\frac\pi2\log\left(\frac{1+\sin x\cos x}{\cos^2x}\right)\mathrm dx$$ Splitting the logarithm we are left with a standard integral, solvable by differentiating the Beta Function for instance, and another one which I already referred to within the comments. To be precise we get \begin{align*} \int_0^\frac\pi2\log\left(\frac{1+\sin x\cos x}{\cos^2x}\right)\mathrm dx&=\pi\log 2+\int_0^\frac\pi2\log(1+\sin x\cos x)\mathrm dx\\ &=\pi\log 2+2\int_0^\frac\pi4\log\left(1+\frac12\sin2x\right)\mathrm dx\\ &=\pi\log 2+\int_0^\frac\pi2\log\left(1+\frac12\sin x\right)\mathrm dx\\ &=\frac\pi2\log2+\int_0^\frac\pi2\log\left(2+\sin x\right)\mathrm dx \end{align*} The latter integral $-$ even a more general case $-$ is examined within this AoPS thread. An expression is deduced by the user gustin33. I won't copy his derivation here since his own solution is impressive enough. For the given case he obtained $$\int_0^\frac\pi2\log\left(2+\sin x\right)\mathrm dx=\frac{4G}3+\frac\pi3\log(2+\sqrt3)-\frac\pi2\log2 $$ Which overall yields to the result.

$$\therefore~\int_0^\infty\frac{\log(1+x+x^2)}{1+x^2}\mathrm dx~=~\frac{4G}3+\frac\pi3\log(2+\sqrt3)$$

The crucial point of the linked post is the identity $$\int_0^\frac\pi2\log(a+\sin x)\mathrm dx=2\operatorname{Ti}_2(a+\sqrt{a^2-1})-\frac\pi2(\log2+\cosh^{-1}a)$$ For $a=2$ the result follows. I will see if I can find another proof for this identity; otherwise I will just leave this here.


EDIT I

Maybe I am on the right track now! Using the integral representation for the Dilogarithm used in this post and reexpressing the Inverse Tangent Integral in terms of the Dilogarithm aswell we obtain $$\small \begin{align*} \operatorname{Ti}_2(a+\sqrt{a^2-1})&=\frac1{2i}\left[\operatorname{Li}_2(ia+i\sqrt{a^2-1})-\operatorname{Li}_2(-ia+-i\sqrt{a^2-1})\right]\\ &=\frac1{2i}\left[\int_0^1\frac{ia+i\sqrt{a^2-1}}{(ia+i\sqrt{a^2-1})t-1}\log t\mathrm dt-\int_0^1\frac{-ia+-i\sqrt{a^2-1}}{(-ia+-i\sqrt{a^2-1})t-1}\log t\mathrm dt\right]\\ &=\frac{a+\sqrt{a^2-1}}2\int_0^1\left[\frac1{(-1)+i(a+\sqrt{a^2-1})t}+\frac1{(-1)-i(a+\sqrt{a^2-1})t}\right]\log t\mathrm dt\\ &=-(a+\sqrt{a^2-1})\int_0^1\frac{\log t}{1+(a+\sqrt{a^2-1})^2t^2}\mathrm dt \end{align*} $$ Mabye this integral is useful for someone. I will try to find something from which it is useful to me too.


EDIT II

The integral can also be reduced to finding $$\int_0^1\frac{\arctan t}{t^2+t+1}\frac{1-t^2}{1+t^2}\mathrm dt$$ I am almost certain I have seen this one before aswell. I will search for it.

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  • $\begingroup$ Feynman's trick doesn't work fine for $\displaystyle \int_0^1 \dfrac{\ln(1+ax^2)}{1+x^2}\,dx$ because of $1$ $\endgroup$ – FDP Apr 28 at 20:39
  • $\begingroup$ @FDP Were you trying the same approach as I posted in my answer? $\endgroup$ – LeBlanc Apr 29 at 9:36
  • $\begingroup$ @mrtaurho in the meantime I found another approach, and now I'm gonna try to find another way for: $$\int_0^\frac\pi2\log\left(2+\sin x\right)\mathrm dx=\frac{4G}3+\frac\pi3\log(2+\sqrt3)-\frac\pi2\log2$$ I highly believe this can be used: math.stackexchange.com/a/3044892/515527 $\endgroup$ – LeBlanc Apr 29 at 9:47
  • $\begingroup$ @Zacky: Yes but now i try to find out something else. Reverse engineering, 2G/3 is obtainable by an integral, something related to $\dfrac{\pi}{12}$ and $\pi\ln(2+\sqrt{3})$ is related to $\dfrac{\pi}{12}$ as well. $\endgroup$ – FDP Apr 29 at 9:54
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    $\begingroup$ Thanks ! It's nice but a proof without using Feynman's trick would be nicer for me. $\endgroup$ – FDP Apr 29 at 10:13
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Solution 1.

By splitting the integral at $1$ and letting $x\to \frac{1}{x}$ in the second part, we get:$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\int_0^1 \frac{\ln(1+x+x^2)+\ln\left(1+\frac{1}{x}+\frac{1}{x^2}\right)}{1+x^2}dx$$ $$=2\int_0^1 \frac{\ln(1+x+x^2)}{1+x^2}dx-2\int_0^1 \frac{\ln x}{1+x^2}dx$$ Via the substitution $x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$ and using this, we obtain: $$I=2\int_0^1\frac{\ln\left(\frac{3+t^2}{(1+t)^2}\right)}{1+t^2}dt+2G=2\int_0^1 \frac{\ln(3+t^2)}{1+t^2}dt-4\int_0^1\frac{\ln(1+t)}{1+t^2}+2G$$ The second one is a well known Putnam integral, and for the first one we can try to use Feynman's trick. $$I=2J-\frac{\pi}{2}\ln 2+2G, \quad J=\int_0^1 \frac{\ln(3+x^2)}{1+x^2}dx$$

$$J(a)=\int_0^1 \frac{\ln(2+a(1+x^2))}{1+x^2}dx\Rightarrow J'(a)=\frac1a\int_0^1 \frac{dx}{\frac{a+2}{a}+x^2}dx$$ $$=\frac1a\sqrt{\frac{a}{a+2}}\arctan\left(x\sqrt{\frac{a}{a+2}}\right)\bigg|_0^1=\frac{1}{\sqrt{a(a+2)}}\arctan\left(\sqrt{\frac{a}{a+2}}\right)$$ We are looking to find $J=J(1)$, but we also have: $J(0)=\frac{\pi}{4}\ln 2$ so: $$J=J(1)-J(0)+J(0)=\underbrace{\int_0^1 J'(a)da}_{=K}+\frac{\pi}{4}\ln 2 $$ Now letting $\sqrt{\frac{a+2}{a}}=x\Rightarrow \frac{1}{\sqrt{a(a+2)}}da=-a dx=-\frac{2}{x^2-1}dx\,$ gives us: $$K=\int_0^1 \frac{1}{\sqrt{a(a+2)}}\arctan\left(\sqrt{\frac{a}{a+2}}\right)da=2\int_\sqrt 3^\infty \frac{\arctan \left(\frac{1}{x}\right)}{x^2-1}dx$$ $$=\frac{\pi}{2}\ln(2+\sqrt 3)-2\int_{\sqrt 3}^\infty \frac{\arctan x}{x^2-1}dx $$ $$H=2\int_{\sqrt 3}^\infty \frac{\arctan x}{x^2-1}dx\overset{x=\tan t}=-2\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{t}{\cos(2t)}dt\overset{\large 2t=x+\frac{\pi}{2}}=\int_{\frac{\pi}{6}}^\frac{\pi}{2} \frac{\frac{\pi}{4}+\frac{x}{2}}{\sin x}dx$$ $$=\frac{\pi}{4}\ln\left(\tan\frac{x}{2}\right)\bigg|_\frac{\pi}{6}^\frac{\pi}{2}+\frac12 \int_0^\frac{\pi}{2}\frac{x}{\sin x}dx-\frac12\int_0^\frac{\pi}{6}\frac{x}{\sin x}dx$$ The last two integrals are linked in this post and using their values we get: $$H=\frac{\pi}{4}\ln(2+\sqrt 3)+G+\frac{\pi}{12}\ln(2+\sqrt 3)-\frac23G=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac13G}$$ $$\Rightarrow \boxed{K=\frac{\pi}{6}\ln(2+\sqrt 3)-\frac13G}\Rightarrow \boxed{J=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac{\pi}{4}\ln 2-\frac13G}$$ $$\Rightarrow I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$


Solution 2.

We can start by considering: $$A=\int_0^\frac{\pi}{2} \ln(2+\sin x)dx,\quad B=\int_0^\frac{\pi}{2}\ln(2-\sin x)dx$$ Like in mrtaurho's approach we have: $$I=\frac{\pi}{2}\ln 2 +A=\frac{\pi}{2}\ln 2+\frac12\left((A+B)+(A-B)\right)\tag 1$$ A solution for $A-B\,$ can be found here. $$A-B=\int_0^\frac{\pi}{2}\ln\left(\frac{2+\sin x}{2-\sin x}\right)dx=-\frac{\pi}{3}\ln(2+\sqrt 3) +\frac{8}{3}G\tag2$$ And for $A+B$ we can directly use this result. $$A+B=\int_0^\frac{\pi}{2} \ln(4-\sin^2 x)=\int_0^\frac{\pi}{2} \ln(4\cos^2x +3\sin^2 x)dx$$$$=\pi \ln 2 +\int_0^\frac{\pi}{2} \ln\left(\cos^2 x+\frac34 \sin^2 x\right)dx=\pi\ln\left(1+\frac{\sqrt 3}{2}\right)\tag3$$ Now plugging $(2)$ and $(3)$ into $(1)$ yields the result.

$$\boxed{I=\frac{\pi}{2}\ln 2+\frac12\left(\pi\ln(2+\sqrt 3)-\pi \ln 2-\frac{\pi}{3}\ln(2+\sqrt 3)+\frac83G\right)=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$

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\begin{align}I&=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx\\ &=\int_0^1 \frac{\ln(1+x+x^2)}{1+x^2}dx+\int_1^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx\\ \end{align} In the latter integral perform the change of variable $y=\dfrac{1}{x}$

\begin{align}I&=2\int_0^1 \frac{\ln(1+x+x^2)}{1+x^2}dx+2\text{G} \end{align} Perform the change of variable $y=\dfrac{1-x}{1+x}$, \begin{align}I&=2\int_0^1 \frac{\ln(3+x^2)}{1+x^2}dx-4\int_0^1 \frac{\ln(1+x)}{1+x^2}dx+2\text{G}\\ &=\frac{\pi}{2} \ln 3+2\int_0^1 \frac{\ln\left(1+\frac{x^2}{3}\right)}{1+x^2}dx-4\int_0^1 \frac{\ln(1+x)}{1+x^2}dx+2\text{G}\\ \end{align}

Define $F$ on $[0;1]$ by, \begin{align}F(a)=\int_0^1 \frac{\ln(1+a^2x^2)}{1+x^2}dx\end{align} Observe that, $\displaystyle F(0)=0,F\left(\frac{1}{\sqrt{3}}\right)=\int_0^1 \frac{\ln\left(1+\frac{x^2}{3}\right)}{1+x^2}dx$.

\begin{align}F^\prime (a)&=\int_0^1 \frac{2a x^2}{(1+x^2)(1+a^2x^2)}dx\\ &=2\left[a\left(\frac{\arctan x}{a^2-1}-\frac{\arctan(ax)}{a(a^2-1)}\right)\right]_0^1\\ &=\frac{\pi a}{2(a^2-1)}-\frac{2\arctan a}{a^2-1} \end{align} Therefore, \begin{align}F\left(\frac{1}{\sqrt{3}}\right)&=\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{3}}}\frac{ a}{a^2-1}\,da+2\int_0^{\frac{1}{\sqrt{3}}}\frac{\arctan a}{1-a^2}\,da\\ &=\frac{\pi}{4}\Big[\ln(1-a^2)\Big]_0^{\frac{1}{\sqrt{3}}}+2\int_0^{\frac{1}{\sqrt{3}}}\frac{\arctan a}{1-a^2}\,da\\ &=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)+2\int_0^{\frac{1}{\sqrt{3}}}\frac{\arctan a}{1-a^2}\,da\\ \end{align} Perform the change of variable $y=\dfrac{1-a}{1+a}$, \begin{align}F\left(\frac{1}{\sqrt{3}}\right)&=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)+\int_{2-\sqrt{3}}^1\frac{\arctan\left(\frac{1-a}{1+a}\right)}{a}\,da\\ &=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)+\frac{\pi}{4}\int_{2-\sqrt{3}}^1\frac{1}{a}\,da-\left(\int_0^1\frac{\arctan a}{a}\,da-\int_0^{2-\sqrt{3}}\frac{\arctan a}{a}\,da\right)\\ &=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)-\frac{\pi}{4}\ln\left(2-\sqrt{3}\right)-\text{G}+\int_0^{2-\sqrt{3}}\frac{\arctan a}{a}\,da\\ &=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)-\frac{\pi}{4}\ln\left(2-\sqrt{3}\right)-\text{G}+\Big[\arctan a\ln a\Big]_0^{2-\sqrt{3}}-\int_0^{2-\sqrt{3}}\frac{\ln a}{1+a^2}\,da\\ &=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)-\frac{\pi}{6}\ln\left(2-\sqrt{3}\right)-\text{G}-\int_0^{2-\sqrt{3}}\frac{\ln a}{1+a^2}\,da\\ \end{align} Perform the change of variable $\displaystyle a=\tan u$, \begin{align}F\left(\frac{1}{\sqrt{3}}\right)&=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)-\frac{\pi}{6}\ln\left(2-\sqrt{3}\right)-\text{G}-\int_0^{\frac{\pi}{12}}\ln(\tan u)\,du\end{align} The last integral value is $-\dfrac{2}{3}\text{G}$

(see https://math.stackexchange.com/a/987972/186817 )

Therefore, \begin{align}F\left(\frac{1}{\sqrt{3}}\right)&=\frac{\pi}{4}\ln\left(\frac{2}{3}\right)-\frac{\pi}{6}\ln\left(2-\sqrt{3}\right)-\dfrac{1}{3}\text{G}\end{align}

it is well known that, \begin{align}\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx=\frac{1}{8}\pi\ln 2\end{align} Therefore, \begin{align}I&=\frac{\pi}{2}\ln 3+\frac{\pi}{2}\ln\left(\frac{2}{3}\right)-\frac{\pi}{3}\ln\left(2-\sqrt{3}\right)-\dfrac{2}{3}\text{G}-\frac{\pi}{2}\ln 2+2\text{G}\\ &=\dfrac{4}{3}\text{G}-\frac{\pi}{3}\ln\left(2-\sqrt{3}\right)\\ &=\boxed{\dfrac{4}{3}\text{G}+\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)} \end{align} NB:

Perform the change of variable $y=\dfrac{1-x}{1+x}$, \begin{align}K&=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\\ &=\int_0^1\frac{\ln\left(\frac{2}{1+x}\right)}{1+x^2}\,dx\\ &=\int_0^1\frac{\ln 2}{1+x^2}\,dx-K\\ &=\frac{1}{4}\pi\ln 2-K \end{align} Therefore, \begin{align}K&=\frac{1}{8}\pi\ln 2\end{align}

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So finally I found a way to deal with it. Credits to Cornel Ioan Valean because when I saw his approach I realised how easily I could've solve the integral.

Here's a way to continue my approach. Let's take the following integral: $$\sf I(a)=\int_0^\frac{\pi}{2}\frac{\arctan(a\tan x)}{\sin x}dx\Rightarrow I'(a)=\int_0^\frac{\pi}{2}\frac{\sec x}{1+a^2\tan^2 x}dx$$ $$\sf =\int_0^\frac{\pi}{2}\frac{\cos x}{\cos^2 x+a^2\sin^2 x}dx\overset{\sin x=y}=\int_0^1 \frac{dy}{1+(a^2-1)y^2}=\frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}$$ Now at this point I kept taking $\sf I(0)=0$ as a reference in order to obtain the integral that we're looking for, which is $\sf I(2)$ and the outcome was clearly: $$\sf I=I(2)-I(0)=\int_0^2 \frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da$$ And well here it's were the trouble began because I kept trying substitutions like: $\sf a=\sec x$ and it didn't work with the lower bound.

Anyway a trick to avoid this is just not being greedy to take $\sf I(0)=0$ and move on with $\sf I(1)$, namely: $$\rm I=\underbrace{I(2)-I(1)}_{=J}+I(1), \quad I(1)=\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx$$ Now we're good to go since there is no $\operatorname{arcsec }0$ that bothers us. $$\rm J=\int_1^2 \frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da\overset{a=\sec x}=\int_0^\frac{\pi}{3}\frac{x}{\cos x}dx\overset{x=\frac{\pi}{2}-t}=\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{\frac{\pi}{2}-t}{\sin t}dt$$ $$\rm=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2} \frac{1}{\sin t}dt- \int_0^\frac{\pi}{2} \frac{t}{\sin t}dt+\int_0^\frac{\pi}{6} \frac{t}{\sin t}dt$$ $$\sf \Rightarrow I=J+I(1)=\frac{\pi}{2}\ln\left(\tan \frac{x}{2}\right)\bigg|_\frac{\pi}{6}^\frac{\pi}{2}+\int_0^\frac{\pi}{6} \frac{t}{\sin t}dt$$ And finally, using the result from here, we get: $$\sf I=\frac{\pi}{2}\ln(2+\sqrt 3)-\frac{\pi}{6}\ln(2+\sqrt 3)+\frac43G=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$ I should keep a reminder to myself in order to not be greedy, like taking the easiest way at first sight, $\sf I(0)$ instead of $\sf I(1)$ in our case $\ddot \smile$.


But if you're me and you find yourself still stuck at: $$\sf I=\int_0^2\frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da=\int_0^2\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da$$ Then no worries, I learnt recently from Yaghoub Sharifi the trick to deal with that case (see here).

Basically we would have to split the integral as: $$\sf I=\int_0^1\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da+\int_1^2\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da$$ The second integral is our old friend from above, and for the first case we need to use the complex definition of $\sf \arccos z$, namely $\sf -i\ln\left(z+\sqrt{z^2-1}\right)$. $$\sf \Rightarrow \frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}=\frac{-\ln\left(\frac{1-\sqrt{1-a^2}}{a}\right)}{\sqrt{1-a^2}}$$ And now via the substitution $a=\sin y$ everything goes smoothly.

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