A possible new representation of $\text{Ti}_2(x)$

Question

Do we have the following inverse tangent integral representation in the literature or remember some particular cases of it?

$$\small \frac{4}{\pi}\int_0^1 \frac{\displaystyle \arctan\left(\frac{4 x y}{(1+x^2)(1-y^2)}\right)\operatorname{arctanh}(x)}{1+x^2}\textrm{d}x=\int_0^y \frac{\arctan(t)}{t}\textrm{d}t=\operatorname{Ti_2}(y), \ |y|<1.$$

I tried to use https://approach0.xyz/, but it seems to be down at the moment. It's proposed by Cornel I. V., and his solution is immediately obtained by combining the odd case below

\begin{equation*} \ \color{blue}{\small\int_0^{\pi/4} \sin(2 n x)\operatorname{arctanh}(\tan(x))\textrm{d}x}= \begin{cases} \displaystyle \quad \dfrac{\pi}{8} (-1)^{m-1}\dfrac{1}{2m-1}, \qquad \qquad \quad n=2m-1, \ m \in \mathbb{N}; \\[10pt] \displaystyle \quad \dfrac{1}{4}(-1)^{m-1}\dfrac{ H_{2m}}{m}-\frac{1}{8}(-1)^{m-1}\dfrac{ H_m}{m}\\[10pt] \displaystyle =\dfrac{1}{4} (-1)^{m-1}\dfrac{\overline{H}_{2m}}{m}+\frac{1}{8}(-1)^{m-1}\dfrac{ H_m}{m}, \ n=2m, \ m \in \mathbb{N}, \end{cases} \end{equation*} where $H_n=\sum_{k=1}^n \frac{1}{k}$ is the $n$th harmonic number and $\overline{H}_n=\sum_{k=1}^n(-1)^{k-1}\frac{1}{k}$ represents the $n$th skew-harmonic number, found with an elementary proof here, and $\displaystyle \sum_{n=1}^{\infty}y^{2n-1} \frac{\sin((2n-1)x)}{2n-1}= \frac{1}{2}\arctan\left(\frac{2 y\sin(x)}{1-y^2}\right), \ 0<x<2\pi, |y|<1$.

I would like to see solutions that utilize alternative methods, with no use of series if possible.

(no hurry, whenever it seems interesting enough to deserve some efforts)

A first note: Exploiting the main result, we can get some curious results involving the golden ratio, $$\int_0^1 \frac{\displaystyle \text{Ti}_2\left(\frac{4 x}{1+x^2}\right) \text{arctanh}(x)}{1+x^2} \textrm{d}x= \frac{\pi }{4} \int_0^{1/\phi } \frac{\displaystyle \arctan(x) \log \left(\frac{1-x^2}{x}\right)}{x} \textrm{d}x, \tag1$$ which can be developed further and brought to a closed form, but more interestingly, I believe it can be exploited to get sums of integrals with nice closed forms.

Another (optional) request (if I'm allowed)

I think I would also love to see how to go with evaluating $$\int_0^1 \frac{\displaystyle \text{Ti}_2\left(\frac{4 x}{1+x^2}\right) \text{arctanh}(x)}{1+x^2} \textrm{d}x,$$ on the left-hand side of $(1)$ from a very different perspective, without exploiting the main result (there is no need for full solutions, but ideas that clearly work).

A second note: If we are used to some well-known Cauchy products involving harmonic numbers, then it is not hard to related the closed form of the integral in blue, the even case, to $\displaystyle \arctan^2(x)=\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}x^{2n} \frac{2H_{2n}-H_n}{n}, \ |x|\le1$, by also using the following familiar result, $\displaystyle \sum_{n=1}^{\infty}y^n \sin(nx)=\frac{y\sin(x)}{1-2y\cos(x)+y^2}, \ |y|<1$, and then we get to the so-beautiful result,

$$\int_0^{\pi/4} \frac{\sin (4 x) \text{arctanh}(\tan (x))}{1-2 y^2 \cos (4 x)+y^4} \textrm{d}x=\frac{1}{4}\frac{\arctan^2(y)}{y^2}, \ |y|<1.$$

A third note: we probably might also enjoy a related result that involves a product of two inverse hyperbolic tangents, like $$\int_0^1 \frac{\displaystyle \text{arctanh}\left(\frac{\sqrt{2} x}{1+x^2}\right)\text{arctanh}(x)}{1+x^2} \textrm{d}x=\frac{\pi ^3}{64}-\frac{\pi}{16}\log ^2\left(\sqrt{2}-1\right).$$

Yes, it looks great!

Answer 1

Reference to a more general result

The result is already known in a more general form, as shown here by Sangchul Lee:

$$\int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) d\theta = \pi \chi_{2}(rs)$$

Where $\chi_2(x)$ is Legendre's Chi function. Shifting $s\to is$ and using $\operatorname{Ti}_2(x)=-i\chi_2(ix)$ yields:

$$\int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \operatorname{arctanh}\left( \frac{2s\sin\theta}{1+s^{2}} \right) d\theta = \pi \operatorname{Ti}_{2}(rs)$$

$$\overset{\large \tan \frac{\theta}{2} = x} \Rightarrow \int_{0}^{1} \frac{\arctan\left(\frac{2r}{1-r^{2}} \frac{2x}{1+x^2}\right) \operatorname{arctanh}\left(\frac{2s}{1+s^{2}} \frac{2x}{1+x^2}\right)}{1+x^2} dx = \frac{\pi}{2} \operatorname{Ti}_{2}(rs)$$

Finally, by setting $s=1$ and using $2\operatorname{arctanh} x = \operatorname{arctanh} \frac{2x}{1+x^2}$, we arrive at the particular case seen in the question. Admittedly, I haven't seen it stated in terms of $\operatorname{Ti}_2(x)$.

Perhaps it's worth to mention that it can also be written in the following form:

$$\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}\left( \frac{2r\sin\theta}{1+r^{2}} \right)\operatorname{arctanh}\left( \frac{2s\sin\theta}{1+s^{2}} \right) d\theta = \pi \chi_{2}(rs)$$

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A direct approach

In order to also have it here self-contained, we can quickly adapt the approach for the $\operatorname{arctanh}$ function, without using any kind of series.

$$\int_{0}^{\frac{\pi}{2}} \color{blue}{\arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right)} \color{red}{\operatorname{arctanh}\left( \frac{2s\sin\theta}{1+s^{2}} \right)} d\theta$$

$$=\color{red}{\int_0^{\large\frac{2s}{1+s^2}}}\color{blue}{\int_0^{\large \frac{2r}{1-r^2}}}\int_0^\frac{\pi}{2} \frac{\color{blue}{\sin \theta}\color{red}{\sin \theta}}{\color{blue}{(1+x^2 \sin^2 \theta)}\color{red}{(1-y^2\sin^2 \theta)}}d\theta \color{blue}{dx}\color{red}{dy}$$

$$=\frac{\pi}{2}\int_0^{\large \frac{2s}{1+s^2}}\int_0^{\large \frac{2r}{1-r^2}} \frac{\sqrt{1+x^2}-\sqrt{1-y^2}}{x^2+y^2}\frac{1}{\sqrt{1+x^2}\sqrt{1-y^2}}dxdy$$

$$\underset{\large y\to \frac{2y}{1+y^2}}{\overset{\large x\to \frac{2x}{1-x^2}}=}\ \pi \int_0^s \int_0^r \frac{dxdy}{1+x^2 y^2}=\pi\int_0^s\frac{\arctan(ry)}{y}dy\overset{ry\to y}=\boxed{\pi\operatorname{Ti}_2(rs)}$$