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The following integral appears in More (Almost) Impossible Integrals, Sums, and Series (2023) (evaluation details on pages $301$-$306$), the sequel of (Almost) Impossible Integrals, Sums, and Series (2019): $$\int_0^1 \arctan \left(\sqrt{\frac{1+x^2}{x(1-x) }}\right) \frac{\log (1+x) }{x \sqrt{1+x^2}}\textrm{d}x$$ $$=\frac{1}{2}\log ^2(\sqrt{2}-1)\pi-\frac{\pi^3}{8}-3\pi\operatorname{Li}_2(1-\sqrt{2}),\tag1$$ where $\operatorname{Li}_2$ represents the Dilogarithm function.

Now, in the book, there are two preliminary steps needed before evaluating the preceding integral, which are as follows: $$\int_0^{\pi/2}\frac{\arctan(\sin(x))\log(1+\sin^2(x))}{\sin(x)} \textrm{d}x$$ $$=\frac{5}{24}\pi ^3-\frac{1}{2}\log ^2(\sqrt{2}-1)\pi+4\pi \operatorname{Li}_2(1-\sqrt{2}) \tag2$$ and $$\small \int_0^{\pi/2}\frac{\arctan(\sin(x))\log(1+\sin^2(x))}{\sin(x)} \textrm{d}x+2 \int_0^1 \arctan \left(\sqrt{\frac{1+x^2}{x(1-x) }}\right) \frac{\log (1+x) }{x \sqrt{1+x^2}} \textrm{d}x$$ $$=\frac{1}{2}\log^2(\sqrt{2}-1)\pi-\frac{\pi^3}{24}-2\pi\operatorname{Li}_2(1-\sqrt{2}). \tag3$$

Question_1: How would we like to go differently in $(1)$ without involving the mentioned preliminary steps in $(2)$ and $(3)$ (maybe in a more direct way)?

Question_2: I also find interesting the version with the squared arctan. Ideas for its evaluation? $$\int_0^1 \arctan^2 \left(\sqrt{\frac{1+x^2}{x(1-x) }}\right) \frac{\log (1+x) }{x \sqrt{1+x^2}}\textrm{d}x,$$

or more generally,

$$\mathcal{I}=\int_0^1 \arctan^n \left(\sqrt{\frac{1+x^2}{x(1-x) }}\right) \frac{\log (1+x) }{x \sqrt{1+x^2}}\textrm{d}x, \ n\ge2,\ n \in \mathbb{N}.$$

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  • $\begingroup$ Maybe it can help to rewrite the integral as: $$\int_{-\infty}^\infty \frac{\ln(1+\frac{1}{1+x^2})\arctan\left(\sqrt{(x-\sqrt 2/x)^2+2(1+\sqrt 2)}\right)}{\sqrt{(x-\sqrt 2/x)^2+2(1+\sqrt 2)}}dx$$ $$=\int_0^1\int_0^1\int_{-\infty}^\infty \frac{1}{1+a^2(t^2+2(1+\sqrt 2))}\frac{1}{1+\sqrt 2 +y +\frac{1+y}{1+\sqrt 2+y}t^2}dtdady$$ Evaluating the first integral w.r.t. $t$ provides the $\pi$ term (which is found in all 3 terms of the closed form), however the resulting double integral is quite difficult. $\endgroup$
    – Zacky
    Nov 12, 2023 at 23:35
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    $\begingroup$ @Zacky I'll check the resulting double integrals and see what could be done further. Thanks! (the appearance of $\pi$ factor might be a kind of good news) $\endgroup$ Nov 13, 2023 at 7:23
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    $\begingroup$ Here's another form that looks promising: $$I=4\pi\int_{s}^{\large {\sqrt{\frac{s}{2}}}}\frac{\operatorname{arctanh}\left(\frac{x}{\sqrt{1+x^2}+1/\sqrt s}\right)}{\sqrt{x^2-s^2}}dx, s= \sqrt 2-1$$ I would expect even different values of $s$ to have a closed form.. that resembles $$a\pi^2+b\log ^2(s)+c\operatorname{Li}_2(\pm s)$$ $\endgroup$
    – Zacky
    Nov 18, 2023 at 13:02
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    $\begingroup$ @Zacky This is a nice alternative way of rewriting the integral and possibly promising in the sense of making use of it for a full derivation. I don't even mention extracting the generalization, but that would be the best. $\endgroup$ Nov 18, 2023 at 16:53
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    $\begingroup$ Another possible idea would be to consider $I=\int_0^1 \arctan \left(\sqrt{\frac{1+x^2}{x(1-x) }}\right) \frac{\log (1+x) }{x \sqrt{1+x^2}}\textrm{d}x$ and $J=\int_0^1 \arctan \left(\sqrt{\frac{1+x^2}{x(1-x) }}\right) \frac{\log (1-x) }{x \sqrt{1+x^2}}\textrm{d}x$, and then try to calculate $I+J$ and $I-J$. It must be well investigated before telling how good such an option is. $\endgroup$ Nov 18, 2023 at 16:59

1 Answer 1

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$$\boxed{\int_0^1 \arctan\left(\sqrt{\frac{1+x^2}{x(1-x)}}\right)\frac{\ln(1+x)}{x\sqrt{1+x^2}}dx=\frac{\pi^3}{48}-\frac{\pi}{2}\operatorname{Li}_2 \left(-\frac{1}{(\delta_S)^2}\right)}$$ Where $\delta_S=1+\sqrt 2\, $ is the silver ratio and $\operatorname{Li}_2(x)$ is the dilogarithm.

I felt like using only $2$ terms for the result is more aesthetic, however it can be expanded into the original closed form by using: $$\operatorname{Li}_2 \left(-\frac{1}{(\delta_S)^2}\right)=\frac{7\pi^2}{24}-\ln^2\left(\delta_S\right) + 6\operatorname{Li}_2 \left(-\frac{1}{\delta_S}\right)$$


$$\int_0^1 \color{blue}{\operatorname{arccot}\left(\sqrt{\frac{1+x^2}{x(1-x)}}\right)}\frac{\ln(1+x)}{\color{blue}{x\sqrt{1+x^2}}}dx =\int_0^1 \color{blue}{\int_0^{\large \sqrt{\frac{1-x}{x}}}} \frac{\ln(1+x)}{\color{blue}{1+x^2(1+y^2)}}\color{blue}{dy}dx$$

$$=\int_0^\infty \int_{0}^{\large \frac{1}{1+y^2}} \frac{\ln(1+x)}{1+(1+y^2)x^2}dxdy\overset{\large (*)}=\int_0^\infty \int_0^\infty \frac{\ln x}{1+(1+y^2)(1+x)^2}dxdy$$

$$=\frac{\pi}{2}\int_0^\infty \frac{\ln x}{(1+x)\sqrt{1+(1+x)^2}}dx\overset{\large x\to \frac{1-x}{x}}=\frac{\pi}{2}\int_0^1 \frac{\ln\left(\frac{1-x}{x}\right)}{\sqrt{1+x^2}}dx$$

Above was essential to start with the $\operatorname{arccot }$ version of the integral as it allows to use the identity:

$$\int_0^{\large \frac{1}{a}} \frac{\ln(1+x)}{1+ax^2}dx=\int_0^\infty \frac{\ln x}{1+a(1+x)^2}dx \tag{*}$$


$$\Rightarrow \int_0^1 \arctan\left(\sqrt{\frac{1+x^2}{x(1-x)}}\right)\frac{\ln(1+x)}{x\sqrt{1+x^2}}dx=\frac{\pi}{2}\underbrace{\int_0^1 \frac{\ln(1+x)}{x\sqrt{1+x^2}}dx}_{\large x\to \frac{2x}{1-x^2}}-\frac{\pi}{2}\underbrace{\int_0^1 \frac{\ln\left(\frac{1-x}{x}\right)}{\sqrt{1+x^2}}dx}_{\large x\to \frac{1-x^2}{2x}}$$

$$\require{cancel} =\frac{\pi}{2}\int_0^{1/\delta_S} \cancel{\ln\left(\frac{(1+\delta_S x)(1-x/\delta_S)}{1-x^2}\right)}\frac{dx}{x}-\frac{\pi}{2}\int_{1/\delta_S}^1 \ln\left(\frac{(\delta_Sx-1)(1+x/\delta_S)}{\cancel{1-x^2}}\right)\frac{dx}{x}$$

$$=\frac{\pi}{2}\left(\operatorname{Li}_2\left(-\frac{1}{\delta_S}\right)-\operatorname{Li}_2 \left(1-\delta_S\right)-\ln \left(\delta_S\right) \ln \left(\delta_S-1 \right) -\operatorname{Li}_2 \left(-\frac{1}{(\delta_S)^2}\right)\right)$$ $$=\frac{\pi}{2}\left(\frac{\pi^2}{24}-\operatorname{Li}_2 \left(-\frac{1}{(\delta_S)^2}\right)\right)$$

The simplification from above follows by plugging $x=\delta_S-1\, $ in $(6)$ from here.


It might be also worth to mention that the $(*)$ identity can be further employed to generate other interesting integrals, by using: $$\int_a^b g(y)\int_0^{\large \frac{1}{f(y)}} \frac{\ln(1+x)}{1+f(y) x^2}dxdy=\int_0^\infty\ln x\int_a^b \frac{g(y)}{1+f(y)(1+x)^2}dydx$$

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    $\begingroup$ (+1) In essence, you turn the integral into a double integral (the same as in More (Almost) Impossible Integrals, Sums, and Series (2023), page $305$, slightly rearranged with variable changes since, $$\color{blue}{\int_0^1 \frac{\log(1+y)}{1+y^2} \left(\int_{\arcsin{(\sqrt{y})}}^{\pi/2} \frac{(\tan(a))'}{(y/\sqrt{1+y^2})^2+\tan^2(a)}\textrm{d}a\right)\textrm{d}y}$$ $$=\int _0^1\int _{\sqrt{\frac{x}{1-x}}}^{\infty }\frac{\log(1+x)}{x^2+(1+x^2) y^2}dydx= \int _0^1\int _0^{\sqrt{\frac{1-x}{x}}}\frac{\log(1+x)}{1+ x^2(1+y^2)}dydx,$$ and the novelty element is to use later the identity (*). $\endgroup$ Feb 8 at 7:25
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    $\begingroup$ While the focus might not necessarily be on getting that specific closed form, the identity $\operatorname{Li}_2 \left(-\frac{1}{(\delta_S)^2}\right)=\frac{7\pi^2}{24}-\ln^2\left(\delta_S\right) + 6\operatorname{Li}_2 \left(-\frac{1}{\delta_S}\right)$ can be challenging without the right tools in hand. $\endgroup$ Feb 8 at 7:31
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    $\begingroup$ The way I found it was by substituting $x\to \frac{1}{x}$ followed by $x-1\to x$ to arrive at: $$\int_0^\infty \frac{\arctan \left(\frac{1}{\sqrt x}\sqrt{1+(1+x)^2}\right)\ln\left(1+\frac{1}{1+x}\right)}{\sqrt{1+(1+x)^2}}dx$$ And now it's easy to notice that we're having an integral of the form $$\frac{\arctan\left(g(x) f(x)\right)}{f(x)}= \int_0^{g(x)} \frac{1}{1+f^2(x) y} dy$$ Then I switched to the arccot version of the integral, but yes, both are the same in disguise. $\endgroup$
    – Zacky
    Feb 8 at 10:40
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    $\begingroup$ Regarding the identity $(*)$ we should also note that more integrals can be found by replacing $f(y)$ with something more exotic: $$\int_0^\infty g(y)\int_0^{\large \frac{1}{f(y)}} \frac{\ln(1+x)}{1+f(y) x^2}dxdy=\int_0^\infty g(y)\int_0^\infty \frac{\ln x}{1+f(y)(1+x)^2}dxdy$$ Also, everything about this integral is extremely challenging without the right tools in hand - not only challenging, but also stubborn. I had so many attempts that although felt right to use, all hit a dead end. $\endgroup$
    – Zacky
    Feb 8 at 10:50
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    $\begingroup$ It would be useful to include the last integral relation in your answer - also in other situations, it could be proved useful. $\endgroup$ Feb 8 at 11:04

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