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The inverse tangent integral is defined as

$$\operatorname{Ti}_2(x)=\Im\operatorname{Li}_2\left(ix\right)$$

Because this we have some special value identitiy.

Let $c_1 = \operatorname{Li}_2(i)$, then $\Im c_1 = G$, where $G$ is Catalan's constant.

Let $c_2 = 4\operatorname{Li}_2\left(\frac{i}{2}\right) + 2\operatorname{Li}_2\left(\frac{i}{3}\right)+\operatorname{Li}_2\left(\frac{3i}{4} \right)$, then $\Im c_2=6G - \pi \ln 2$.

Let $c_3 = \operatorname{Li}_2\left(2i-\sqrt 3\,i\right)$, then $\Im c_3 = \frac{2G}{3}-\frac{\pi}{12}\ln(2+\sqrt 3)$.

Let $c_4 = \operatorname{Li}_2\left(2i+\sqrt 3\,i\right)$, then $\Im c_4 = \frac{2G}{3}+\frac{5\pi}{12}\ln(2+\sqrt 3)$.

Let $c_5 = \operatorname{Li}_2\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt 2 +1 }i\right)-\operatorname{Li}_2\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt 2 -1 }i\right)+\frac{2}{3}\operatorname{Li}_2\left((\sqrt 2 - 1)i\right)$, then $\Im c_5 = \frac{\pi}{6}\ln\left(\frac{\sqrt{2} - 1}{(\sqrt3 - \sqrt 2)(\sqrt 2 + 1)}\right)$.

Furthermore we know that $\Re c_1 = -\frac{\pi^2}{48}$.

Question. Is there a closed-form of $\Re c_i$, $i=2\dots5$ ? If we couldn't specify closed-forms of them, then of any of their combinations?

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$$\Re\,\operatorname{Li}_2\left(\tfrac i2\right)=-\frac14\operatorname{Li}_2\left(\tfrac15\right)-\frac12\ln^22-\frac18\ln^25+\frac12\ln2\cdot\ln5$$

$$\Re\,\operatorname{Li}_2\left(\tfrac i3\right)=\operatorname{Li}_2\left(\tfrac13\right)-\frac14\operatorname{Li}_2\left(\tfrac15\right)-\frac12\operatorname{Li}_2\left(\tfrac25\right)-\frac{\pi^2}{48}\\ -\frac38\ln^25-\frac12\ln2\cdot\ln3+\frac12\ln2\cdot\ln5+\frac12\ln3\cdot\ln5$$

$$\Re\,\operatorname{Li}_2\left(\tfrac{3i}4\right)=\operatorname{Li}_2\left(\tfrac13\right)-\frac12\operatorname{Li}_2\left(\tfrac15\right)+\operatorname{Li}_2\left(\tfrac25\right)-\frac{\pi^2}{8}\\ -2\ln^22+\frac12\ln^23+\frac14\ln^25+\ln2\cdot\ln3+\ln2\cdot\ln5-\ln3\cdot\ln5$$

$$\Re\,\operatorname{Li}_2\left(i\left(2+\sqrt3\right)\right)=-2\operatorname{Li}_2\left(2-\sqrt3\right)+\frac{\pi^2}{16}\\ -\frac34\ln^22-3\ln^2\left(1+\sqrt3\right)+3\ln2\cdot\ln\left(1+\sqrt3\right)$$

$$\Re\,\operatorname{Li}_2\left(i\left(2-\sqrt3\right)\right)=2\operatorname{Li}_2\left(2-\sqrt3\right)-\frac{5\pi^2}{48}+\frac14\ln^22+\ln^2\left(1+\sqrt3\right)-\ln2\cdot\ln\left(1+\sqrt3\right)$$

$$\Re\,\operatorname{Li}_2\left(\frac{i\left(\sqrt3-\sqrt2\right)}{\sqrt2+1}\right)=2\operatorname{Li}_2\left(\sqrt2-1\right)-\operatorname{Li}_2\left(-\sqrt2-\sqrt3\right)\\ +\frac14\operatorname{Li}_2\left(\frac{\sqrt3-\sqrt2}{\sqrt{2}-1}\right) +\frac14\operatorname{Li}_2\left(\frac{\sqrt3-\sqrt2}{\sqrt2+1}\right)-\frac{29\pi^2}{96}\\+\frac34\ln^2\left(1+\sqrt2\right) -\frac14 \ln^2\left(\sqrt2+\sqrt3\right)-\frac12\ln\left(1+\sqrt{2}\right)\cdot\ln\left(\sqrt2+\sqrt3\right)$$

$$\Re\,\operatorname{Li}_2\left(\frac{i\left(\sqrt3-\sqrt2\right)}{\sqrt2-1}\right)=-2\operatorname{Li}_2\left(\sqrt2-1\right)+\operatorname{Li}_2\left(\sqrt3-\sqrt2\right)\\ +\frac14\operatorname{Li}_2\left(\frac{\sqrt3-\sqrt2}{\sqrt2-1}\right)+\frac14\operatorname{Li}_2\left(\frac{\sqrt3-\sqrt2}{\sqrt2+1}\right)+\frac{\pi^2}{32}\\ -\frac54\ln^2\left(1+\sqrt2\right)+\frac14\ln^2\left(\sqrt2+\sqrt3\right)+\frac12 \ln\left(1+\sqrt2\right)\cdot\ln\left(\sqrt2+\sqrt3\right)$$

$$\Re\,\operatorname{Li}_2\left(i\left(\sqrt2-1\right)\right)=\frac32\operatorname{Li}_2\left(\sqrt2-1\right)-\frac{11\pi^2}{96}+\frac12\ln^2\left(1+\sqrt2\right)$$

Proofs are left to the reader.

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  • 4
    $\begingroup$ Proofs are left to the reader - Ah, that good Russian humor... :-$)$ $\endgroup$ – Lucian Jan 3 '16 at 3:02
  • 6
    $\begingroup$ @Lucian Somewhere out there, there is a Russian mathematician who's made the joke, "Proofs are left to the vodka." =) $\endgroup$ – David H Jan 3 '16 at 3:38

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