Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$

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.

Solution 1:

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 $y=\dfrac{1}{x}$ )

Solution 2:

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.