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How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx=\frac43G+\frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?

I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
Attempt
$$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx\\=\int_0^\infty\frac{\operatorname{arcsinh}(2x)}{1+x^2}dx\\ =2\int_0^\infty\frac{x\cosh x}{4+\sinh^2x}dx\\ =2\int_0^\infty\frac{x\cosh x}{3+\cosh^2x}dx\\ =2\int_0^\infty\sum_{n=0}^\infty x(-3)^n\cosh^{-2n-1}(x)dx$$ I failed to integrate $x\cosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.

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I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let

$$ J(t) = \int_{0}^{\infty} \frac{\operatorname{arsinh}(tx)}{1+x^2} \, \mathrm{d}x. $$

Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=\sqrt{u^{-2}-1}$, we obtain

\begin{align*} J'(t) = \int_{0}^{\infty} \frac{x}{(1+x^2)\sqrt{1+t^2x^2}} \, \mathrm{d}x = \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u. \end{align*}

So it follows that

\begin{align*} J(2) &= \int_{0}^{2} \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t \\ &= \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t + \int_{1}^{2} \int_{0}^{1} \frac{1}{\sqrt{1 + (t^2-1)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t. \end{align*}

The inner integral is easily computed, yielding

\begin{align*} J(2) &= \int_{0}^{1} \frac{\operatorname{artanh}\left( \sqrt{1 - t^2} \right)}{\sqrt{1 - t^2}} \, \mathrm{d}t + \int_{1}^{2} \frac{\arctan\left(\sqrt{t^2-1}\right)}{\sqrt{t^2 - 1}} \, \mathrm{d}t. \end{align*}

Now we substitute $t = \operatorname{sech} \varphi$ for the first integral and $t = \sec \theta$ for the second integral. This yields

\begin{align*} J(2) &= \int_{0}^{\infty} \frac{\varphi}{\cosh\varphi} \, \mathrm{d}\varphi + \int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta. \end{align*}

These integrals can be computed as follows:

  • Using $ \operatorname{sech}\varphi = \frac{2e^{-\varphi}}{1 + e^{-2\varphi}} = 2 \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)\varphi} $, we obtain

    $$ \int_{0}^{\infty} \frac{\varphi}{\cosh\varphi} \, \mathrm{d}\varphi = 2 \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} \varphi e^{-(2n+1)\varphi} \, \mathrm{d}\varphi = 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} = 2G. $$

  • Taking integration by parts,

    \begin{align*} \int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta &= \left[ - \theta \log \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right) \right]_{0}^{\frac{\pi}{3}} + \int_{0}^{\frac{\pi}{3}} \log \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right) \, \mathrm{d}\theta \\ &= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) + 2 \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \log \left( \tan \theta \right) \, \mathrm{d}\theta. \end{align*}

    This can be computed by using the Fourier series $\log \left( \tan \theta \right) = - 2 \sum_{n=0}^{\infty} \frac{\cos(4n+2)\theta}{2n+1} $ to yield

    \begin{align*} \int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta &= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) - 2 \sum_{n=0}^{\infty} \frac{\sin\left( \frac{\pi}{2} (2n+1) \right) - \sin\left( \frac{\pi}{6} (2n+1) \right)}{(2n+1)^2} \\ &= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) - \frac{2}{3}G. \end{align*}

Combining two result, we obtain the desired answer.

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    $\begingroup$ This is beauty ! $\endgroup$ – Claude Leibovici Jan 22 at 16:10
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On the path of Kemono Chen...

\begin{align}J&=\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx\end{align}

Perform the change of variable $y=\operatorname{arcsinh}(2\tan x)$,

\begin{align}J&=\int_0^{+\infty}\frac{2x\cosh x}{4+\sinh^2 x}\,dx\\ &=\int_0^{+\infty}\frac{4x\left(\text{e}^{x}+\text{e}^{-x}\right)}{14+\text{e}^{2x}+\text{e}^{-2x}}\,dx\\ &=\int_0^{+\infty}\frac{4x\text{e}^{-x}\left(\text{e}^{2x}+1\right)}{14+\text{e}^{2x}+\text{e}^{-2x}}\,dx\\ \end{align}

Perform the change of variable $y=\text{e}^{-x}$,

\begin{align}J&=-\int_0^1 \frac{4\ln x\left(1+\frac{1}{x^2}\right)}{14+x^2+\frac{1}{x^2}}\\ &=-\int_0^1 \frac{4\ln x\left(1+x^2\right)}{x^4+14x^2+1}\\ &=\left[-\arctan\left(\frac{4x}{1-x^2}\right)\ln x\right]_0^1+\int_0^1 \frac{\arctan\left(\frac{4x}{1-x^2}\right)}{x}\,dx\\ &=\int_0^1 \frac{\arctan\left(\frac{4x}{1-x^2}\right)}{x}\,dx\\ &=\int_0^1 \frac{\arctan\left(\left(2+\sqrt{3}\right)x\right)}{x}\,dx+\int_0^1 \frac{\arctan\left(\left(2-\sqrt{3}\right)x\right)}{x}\,dx\\ \end{align}

In the first integral perform the change of variable $y=\left(2+\sqrt{3}\right)x$,

In the second integral perform the change of variable $y=\left(2-\sqrt{3}\right)x$,

\begin{align}J&=\int_0^{2+\sqrt{3}}\frac{\arctan x}{x}\,dx+\int_0^{2-\sqrt{3}}\frac{\arctan x}{x}\,dx\\ &=\Big[\arctan x\ln x\Big]_0^{2+\sqrt{3}}-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx+\Big[\arctan x\ln x\Big]_0^{2-\sqrt{3}}-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\ &=\frac{5\pi}{12}\ln\left(2+\sqrt{3}\right)-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx+\frac{\pi}{12}\ln\left(2-\sqrt{3}\right)-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\ &=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx \end{align}

In the first integral perform the change of variable $y=\dfrac{1}{x}$,

\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\int_{2-\sqrt{3}}^{+\infty}\frac{\ln x}{1+x^2}\,dx-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\ &=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\int_0^{+\infty}\frac{\ln x}{1+x^2}\,dx-2\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\ &=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\ \end{align}

Perform the change of variable $y=\tan x$,

\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\int_0^{\frac{\pi}{12}}\ln\left(\tan x\right)\,dx\\ \end{align}

It is well known that,

\begin{align} \int_0^{\frac{\pi}{12}}\ln\left(\tan x\right)\,dx=-\frac{2}{3}\text{G} \end{align}

(see: Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )

Thus,

\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\times -\frac{2}{3}\text{G}\\ &=\boxed{\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\frac{4}{3}\text{G}} \end{align}

NB:

Observe that,

\begin{align}2-\sqrt{3}&=\frac{1}{2+\sqrt{3}}\\ \ln\left(2-\sqrt{3}\right)&=-\ln\left(2+\sqrt{3}\right)\\ \int_0^\infty \frac{\ln x}{1+x^2}\,dx&=0 \end{align} (perform the change of variable $y=\dfrac{1}{x}$ )

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  • $\begingroup$ This is beauty ! $\endgroup$ – Claude Leibovici Jan 22 at 16:10
  • $\begingroup$ How did you do $$\int\arctan\frac{4x}{1-x^2}\frac{\mathrm dx}{x}=\int\arctan(2+\sqrt3)x\frac{\mathrm dx}{x}+\int\arctan(2-\sqrt3)x\frac{\mathrm dx}{x}$$ $\endgroup$ – clathratus Jan 22 at 16:10
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    $\begingroup$ @Clathratus: if $ab<1$then \begin{align}\arctan a+\arctan b=\arctan\left(\frac{a+b}{1-ab}\right)\end{align} $\endgroup$ – FDP Jan 22 at 16:23
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Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.

Let $$I(a) = \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (a \tan x) \, dx, \qquad a > 1.$$ We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.

For $a = 1$ we have \begin{align} I(1) &= \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (\tan x) \, dx\\ &= \int_0^{\frac{\pi}{2}} \ln \left (\frac{1 + \sin x}{\cos x} \right ) \, dx\\ &= \int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx - \int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx. \end{align} Now the first of these integrals can be found, for example, by rewriting it as \begin{align} \int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx &= \int_0^{\frac{\pi}{2}} \ln (1 + \cos x) \, dx\\ &= \int_0^{\frac{\pi}{2}} \ln \left (\frac{1}{2} \cos^2 \frac{x}{2} \right ) \, dx\\ &= \frac{\pi}{2} \ln 2 + 4 \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx \end{align} and then employing the Fourier series representation for $\ln (\cos x)$ found here. The final result is $$\int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx = 2 \mathbf{G} - \frac{\pi}{2} \ln 2.$$ Here $\mathbf{G}$ is Catalan's constant.

The second of the integrals is very well known. Here $$\int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx = - \frac{\pi}{2} \ln 2.$$ Thus $I(1) = 2 \mathbf{G}$.

Moving to the main event, by differentiating under the integral sign with respect to $a$ we have \begin{align} I'(a) &= \int_0^{\frac{\pi}{2}} \frac{\tan x}{\sqrt{a^2 \tan^2 x + 1}} \, dx\\ &= \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{a^2 - (a^2 - 1) \cos^2 x}} \, dx. \end{align} Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = \cos x$ one has $$I'(a) = \int_0^1 \frac{du}{\sqrt{a^2 - (a^2 - 1) u^2}} = \frac{1}{\sqrt{a^2 - 1}} \sin^{-1} \left (\frac{\sqrt{a^2 - 1}}{a} \right ).$$

As we require $I(2)$ observe that $$I(2) - I(1) = \int_1^2 I'(a) \, da.$$ Thus $$I(2) = 2 \mathbf{G} + \int_1^2 \frac{1}{\sqrt{a^2 - 1}} \sin^{-1} \left (\frac{\sqrt{a^2 - 1}}{a} \right ) \, da.$$ Integrating by parts leads to $$I(2) = 2 \mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - \int_1^2 \frac{\cosh^{-1} a}{a \sqrt{a^2 - 1}} \, da.$$ Now let $a = \cosh t$. This gives $$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - \int_0^{\ln(2 + \sqrt{3})} \frac{t}{\cosh t} \, dt.$$ Then let $t = \ln y$. This gives $$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\int_1^{2 + \sqrt{3}} \frac{\ln y}{1 + y^2} \, dy.$$ Now let $y = \tan \theta$. Then we have $$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\int_{\frac{\pi}{4}}^{\frac{5\pi}{12}} \ln (\tan \theta) \, d\theta.$$ Finally, enforcing a substitution of $\theta \mapsto \dfrac{\pi}{2} - \theta$ leads to \begin{align} I(2) &= 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) + 2 \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln (\tan \theta) \, d\theta\\ &= 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) + 2\int_0^{\frac{\pi}{4}} \ln (\tan \theta) \, d\theta - 2 \int_0^{\frac{\pi}{12}} \ln (\tan \theta) \, d\theta. \end{align} For the first of the integrals we have $$\int_0^{\frac{\pi}{4}} \ln (\tan x) \, dx = -\mathbf{G}.$$ While for the second of the integrals we have $$\int_0^{\frac{\pi}{12}} \ln (\tan x) \, dx = -\frac{2}{3} \mathbf{G}.$$ So finally $$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\mathbf{G} + \frac{4}{3} \mathbf{G},$$ or $$\int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (2 \tan x) \, dx = \frac{\pi}{3} \ln (2 + \sqrt{3}) + \frac{4}{3} \mathbf{G},$$ as announced.

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    $\begingroup$ \begin{align}A&=\int_0^{\frac{\pi}{4}}\ln(\cos x)\,dx\\ B&=\int_0^{\frac{\pi}{4}}\ln(\sin x)\,dx\\ B-A&=\int_0^{\frac{\pi}{4}}\ln(\tan x)\,dx\\ &=-\text{G}\\ A+B&=\int_0^{\frac{\pi}{4}}\ln(\sin x\cos x)\,dx\\ &=\int_0^{\frac{\pi}{4}}\ln\left(\frac{\sin(2x)}{2}\right)\,dx\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin x}{2}\right)\,dx\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\left(\sin x\right)\,dx-\frac{\pi}{4}\ln 2\\ &=-\frac{1}{2}\pi\ln 2\\ A&=\frac{1}{2}\left(A+B-\left(B-A\right)\right)\\ &=\frac{1}{2}\text{G}-\frac{1}{4}\pi\ln 2 \end{align} $\endgroup$ – FDP Jan 23 at 13:44
  • $\begingroup$ Yes, very nice. $\endgroup$ – omegadot Jan 24 at 1:29
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Sorry in a rush, so only a PARTIAL SOLUTION:

Here I will employ Feynman's Trick: \begin{equation} I = \int_0^{\infty}\frac{\operatorname{arcsinh(2x)}}{1 + x^2}\:dx \end{equation}

Let \begin{equation} J(t) = \int_0^{\infty}\frac{\operatorname{arcsinh(tx)}}{1 + x^2}\:dx \end{equation}

We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':

\begin{equation} J'(t) = \int_0^{\infty}\frac{x}{\sqrt{1 + t^2x^2}}\frac{1}{1 + x^2}\:dx = \left[\frac{1}{\sqrt{t^2 - 1}} \cdot \arctan\left(\sqrt{\frac{1 + t^2x^2}{t^2 - 1}} \right) \right]_0^{\infty} = \frac{1}{\sqrt{t^2 - 1}}\left[\frac{\pi}{2} - \arctan\left(\frac{1}{\sqrt{t^2 -1}} \right) \right] \end{equation}

Thus,

\begin{align} J(t) &= \int \frac{1}{\sqrt{t^2 - 1}}\left[\frac{\pi}{2} - \arctan\left(\frac{x}{\sqrt{t^2 -1}} \right) \right]\:dt \\ &= \int \frac{\pi}{2}\cdot \frac{1}{\sqrt{t^2 - 1}}\:dt - \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt = I_1 - I_2 \end{align}

For $I_1$:

\begin{equation} I_1 = \int \frac{\pi}{2}\cdot \frac{1}{\sqrt{t^2 - 1}}\:dt = \frac{\pi}{2}\ln\left| \sqrt{t^2 - 1} + t\right| + C_1 \end{equation}

Where $C_1$ is the constant of integration. Note that $I_1(2) = \ln\left|2 + \sqrt{3} \right|$

For $I_2$:

\begin{equation} I_2 = \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt \end{equation}

Unfortunately this is not so easy to evaluate. I will first make the substitution $u = \frac{1}{\sqrt{t^2 - 1}}$:

\begin{align} I_2 &= \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt = \int u \cdot \arctan(u) \cdot \frac{-u^4}{\sqrt{1 +u^2}}\:du\\ & = - \int \frac{-u^5}{\sqrt{1 +u^2}} \cdot \arctan(u) \:du \end{align}

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  • $\begingroup$ Will post up full solution later. $\endgroup$ – user150203 Jan 22 at 10:08
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$$ \begin{align}\newcommand{\arcsinh}{\operatorname{arcsinh}} \int_0^\infty\frac{\arcsinh(x)}{1+x^2}\,\mathrm{d}x &=\int_0^\infty\frac{x\,\mathrm{d}x}{\cosh(x)}\tag{1a}\\ &=\int_0^\infty\frac{2x\,\mathrm{d}x}{e^x+e^{-x}}\tag{1b}\\ &=2\sum_{k=0}^\infty(-1)^k\int_0^\infty x\,e^{-(2k+1)x}\,\mathrm{d}x\tag{1c}\\ &=2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\tag{1d}\\[6pt] &=2\mathrm{G}\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: substitute $x\mapsto\sinh(x)$
$\text{(1b)}$: write $\cosh(x)$ as exponentials
$\text{(1c)}$: expand $\left(e^x+e^{-x}\right)^{-1}$ as a power series
$\text{(1d)}$: evaluate the integral
$\text{(1e)}$: use the definition of Catalan's Constant $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{\arcsinh(ax)}{1+x^2}\,\mathrm{d}x &=\int_0^\infty\frac{x}{1+x^2}\frac{\mathrm{d}x}{\sqrt{1+a^2x^2}}\tag{2a}\\ &=\int_0^\infty\frac{x}{a^2+x^2}\frac{\mathrm{d}x}{\sqrt{1+x^2}}\tag{2b}\\ &=\int_0^{\pi/2}\frac{\mathrm{d}\sec(x)}{a^2+\tan^2(x)}\tag{2c}\\ &=\int_1^\infty\frac{\mathrm{d}x}{x^2+\left(a^2-1\right)}\tag{2d}\\ &=\frac{\arctan\left(\sqrt{a^2-1}\right)}{\sqrt{a^2-1}}\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: $\frac{\mathrm{d}}{\mathrm{d}a}\arcsinh(ax)=\frac{x}{\sqrt{1+a^2x^2}}$
$\text{(2b)}$: substitute $x\mapsto x/a$
$\text{(2c)}$: substitute $x\mapsto\tan(x)$
$\text{(2d)}$: substitute $\sec(x)\mapsto x$
$\text{(2e)}$: arctan integral

Therefore, $$ \begin{align}\newcommand{\Li}{\operatorname{Li}} \int_0^\infty\frac{\arcsinh(2x)}{1+x^2}\,\mathrm{d}x &=2\mathrm{G}+\int_1^2\frac{\arctan\left(\sqrt{a^2-1}\right)}{\sqrt{a^2-1}}\,\mathrm{d}a\tag{3a}\\ &=2\mathrm{G}+\int_0^{\pi/3}\frac{a}{\cos(a)}\,\mathrm{d}a\tag{3b}\\ &=2\mathrm{G}+2\int_0^{\pi/3}\frac{a}{e^{ia}+e^{-ia}}\,\mathrm{d}a\tag{3c}\\ &=2\mathrm{G}+2\sum_{k=0}^\infty(-1)^k\int_0^{\pi/3}ae^{-i(2k+1)a}\,\mathrm{d}a\tag{3d}\\ &=2\mathrm{G}+2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\int_0^{(2k+1)\pi/3}ae^{-ia}\,\mathrm{d}a\tag{3e}\\ &=2\mathrm{G}+2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\left[(1+i(2k+1)\pi/3)e^{-i(2k+1)\pi/3}-1\right]\tag{3f}\\ &=2\mathrm{G}+\frac\pi3\log\left(\frac{2+\sqrt3+i}{2-\sqrt3-i}\right)+2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\left[e^{-i(2k+1)\pi/3}-1\right]\tag{3g}\\ &=2\mathrm{G}+\frac\pi3\log\left(\left(2+\sqrt3\right)i\right) +\sum_{k=0}^\infty(-1)^k\scriptsize\left[\frac{-1-i\sqrt3}{(6k+1)^2}-\frac{-4}{(6k+3)^2}+\frac{-1+i\sqrt3}{(6k+5)^2}\right]\tag{3h}\\ &=\frac43\mathrm{G}+\frac\pi3\log\left(2+\sqrt3\right)+i\left[\frac{\pi^2}6-\sqrt3\sum_{k=0}^\infty(-1)^k\scriptsize\left(\frac1{(6k+1)^2}-\frac1{(6k+5)^2}\right)\right]\tag{3i}\\ &=\frac43\mathrm{G}+\frac\pi3\log\left(2+\sqrt3\right)+i\left[\frac{\pi^2}6-\frac{\sqrt3}{36}\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{\left(k+\frac16\right)^2}\right]\tag{3j}\\ &=\bbox[5px,border:2px solid #C0A000]{\frac43\mathrm{G}+\frac\pi3\log\left(2+\sqrt3\right)}\tag{3k} \end{align} $$ Explanation:
$\text{(3a)}$: apply $(1)$ and $(2)$
$\text{(3b)}$: substitute $a\mapsto\sec(a)$
$\text{(3c)}$: write $cos(x)$ as exponentials
$\text{(3d)}$: expand $\left(e^{ix}+e^{-ix}\right)^{-1}$ as a power series
$\text{(3e)}$: substitute $a\mapsto a/(2k+1)$
$\text{(3f)}$: integrate
$\text{(3g)}$: recognize the series for $\arctan(x)=\frac1{2i}\log\left(\frac{1+ix}{1-ix}\right)$
$\text{(3h)}$: evaluate the exponentials
$\text{(3i)}$: separate the real and imaginary parts
$\text{(3j)}$: rewrite the sum
$\text{(3k)}$: $\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(k+x)^2}=\pi^2\cot(\pi x)\csc(\pi x)$

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  • $\begingroup$ Seems like there should be a ^2 in the denominator of the sum of the explanation (3k) $\endgroup$ – Kemono Chen Jan 24 at 9:03
  • $\begingroup$ @KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$. $\endgroup$ – robjohn Jan 24 at 9:25
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This is not an answer.

Your premonition seems to be good. Using another CAS, $$2\int_0^\infty\frac{x\cosh( x)}{3+\cosh^2(x)}\,dx=-\frac{\pi}{4} \log \left(7-4 \sqrt{3}\right)-i \left(\text{Li}_2\left(-i \left(-2+\sqrt{3}\right)\right)-\text{Li}_2\left(i \left(-2+\sqrt{3}\right)\right)\right)$$ Now, ???

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