$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?

Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3)$, so I

was wondering whether a similar closed form expression might also exist for its imaginary part as well. Thank you !

  • $\begingroup$ Any particular reason you chose to italicize most of this question? $\endgroup$ – abiessu Sep 4 '14 at 0:27
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    $\begingroup$ No. Just a simple matter of aesthetics. $\endgroup$ – Lucian Sep 4 '14 at 2:10

If you consider a hypergeometric function to be a closed form, you can have the following result: $$\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right]=\frac{\pi^3}{128}+\frac\pi{32}\ln^22+\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$ And for the polylogarithm value that appears in another answer, you can have $$\Im\Big[\operatorname{Li}_3\left(1+i\right)\Big]=\frac{3\pi^3}{64}+\frac\pi{16}\ln^22-\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$

  • $\begingroup$ Very nice indeed! Unfortunately, I was unable to find a closed form for $\mathrm{Im} \; \mathrm{Li}_3(1+i)$, though I was able to discern that it's value was very much related to that of $\mathrm{Im} \; \mathrm{Li}_3(1/2+i/2)$, as confirmed by your calculations. Regardless, I am going to continue my efforts and see if I can simplify these results any further. $\endgroup$ – Gahawar Sep 6 '14 at 1:51
  • $\begingroup$ I really have the feeling that this result is related. $\endgroup$ – user153012 Sep 17 '14 at 12:21
  • $\begingroup$ @user153012 You're feeling may be correct since, $$\Im\left[\operatorname{Li}_3\big(1+i\big)\right] =\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac1{192}\pi^3$$ $\endgroup$ – Tito Piezas III Jul 1 at 5:04

Form here we know that $$\operatorname{Li}_3(z)-\operatorname{Li}_3\left(\frac{1}{z}\right)=-\frac{1}{6} \ln^3(z)-\frac{\pi\sqrt{-(z-1)^2}}{2(z-1)}\ln^2(z)+\frac{\pi^2}{3}\ln(z)\tag{1}.$$

If we put $z:=(1+i)/2$ into $(1)$ and get the imaginary part of it, we get

$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] - \Im \left[\operatorname{Li}_3(1-i)\right] = \frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2 2.$$

@Tunk-Fey said that $\displaystyle\Im[\operatorname{Li}_3(1-i)]=-\Im[\operatorname{Li}_3(1+i)]$, so it is also true, that

$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] + \Im \left[\operatorname{Li}_3(1+i)\right] = \frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2 2.$$

Sadly the only two exact complex value of $\operatorname{Li}_3$ function what I found is $\operatorname{Li}_3(\pm i).$ We could use it, but with known identities we coudn't break out of the prison of $\operatorname{Li}_3(1+i)$ or $\operatorname{Li}_3(1-i)$ with this approach.

I found nothing about it, but I think @Tunk-Fey's result is in general true, and I think that is true, that $\Im \operatorname{Li}_3(z)+\Im \operatorname{Li}_3(\overline{z}) = 0$ for all $z$ which have complex part. It isn't a good news, because using identities we get an unknown complex value of $\operatorname{Li}_3$ or a complex conjugate pair of the variable what we are looking for.

Of course there is relationship between polylogarithm function and generalized hypergeometric function. For $\operatorname{Li}_3$ we have $$\operatorname{Li}_3(z) = z \;_{4}F_{3} (1,1,\dots,1; \,2,2,\dots,2; \,z).$$

So we could write the problem also into the form

$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] = \Im\left[\frac{1+i}{2}{_4F_3}\left(\begin{array}c\ 1,1,1,1\\2,2,2\end{array}\middle|\,\frac{1+i}{2}\right) \right].$$

I don't know how @Cleo transformed it into the form that is given above, but it is really nice, how the imaginary part is eliminated.

If somebody could give some more complex valued exact solutions of $\operatorname{Li}_3$ function, then maybe I could get something more. But for now, I'm also waiting for a solution.

By the way, if your result about the real part is correct, then we can get a really beautiful closed formula for $\Re[\operatorname{Li}_3(1 \pm i)]$ using the partial results of your problem.

If we put again $z:=(1+i)/2$ into $(1)$ and now we get the real part of it, we have

$$\Re\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] - \Re \left[\operatorname{Li}_3(1-i)\right] = \frac{\ln^3 2}{48}-\frac{11\pi^2}{192}\ln 2.$$

Now if you're right, and


then we get for $\Re \left[\operatorname{Li}_3(1 \pm i)\right]$ the following.

$$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$$

This seems to me numerically correct, and via your problem I could solve a really interesting related diamond.


This is equivalent to finding a closed form for the following series

$$I :=\mathrm{Im} \left[\mathrm{Li}_3\left(\frac{1+i}{2}\right)\right] = \sum_{k=1}^{\infty} \frac{\sin \pi k/4}{2^{k/2}k^3}.$$

I was able to find the following expression, which seems to be numerically correct

$$I=\frac{7\pi^3}{256}+\frac{3\pi}{64}\log^22+\frac{1}{2} \mathrm{Im}\left[\mathrm{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)-\mathrm{Li}_3(1+i)\right].$$

Update: Taking the advice of V.Rosssetto into account, we are reduced to

$$I=\frac{7\pi^3}{128}+\frac{3\pi}{32}\log^22-\mathrm{Im} \; \mathrm{Li}_3(1+i).$$

  • $\begingroup$ If your result is correct, you can remark that the $\mathrm{Li}_3(\frac12+\frac{\mathrm i}2)$ is actually equal to $I$ and you get $I=\frac{7\pi^3}{128}-\frac{7\pi}{32}\ln^22-\mathrm{Im}\left(\mathrm{Li}_3(1 +{\mathrm i})\right)$ ! $\endgroup$ – Tom-Tom Sep 5 '14 at 15:26
  • $\begingroup$ @V.Rossetto I just corrected a small error in the second term. Indeed, I am making use of such a fact right now and seeing where it leads me. $\endgroup$ – Gahawar Sep 5 '14 at 15:28
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    $\begingroup$ @Gahawar I just stumbled upon the value $\mathrm{Im} \left[\mathrm{Li}_3\left(\frac{1+i}{\sqrt{2}}\right)\right]=\frac{7\pi^3}{256}$. I surprised how simple this "in-between" value turned out to be, given the difficulty of finding the two values you descried. $\endgroup$ – David H Sep 12 '14 at 8:59
  • $\begingroup$ @DavidH That is quite interesting indeed. I am often amused at how a small change within an expression can lead to wildly different results. I find it curious that $\frac{7\pi^3}{256}$ appeared in my unsuccessful attempt. $\endgroup$ – Gahawar Sep 12 '14 at 12:21
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    $\begingroup$ @DavidH: Don't be! $\sqrt2$ is the absolute value of $1+i$, making the entire expression a root of unity. $\dfrac{1+i}2$, however, enjoys no such “special” privileges. :-) $\endgroup$ – Lucian Sep 12 '14 at 23:05

I'm seriously late for this party, but to complement Cleo's answer, if we consider a hypergeometric as a closed-form, then using a different hypergeometric, the family is,

$$\Im\left[\operatorname{Li}_\color{red}2\big(1+i\big)\right] =\sqrt2 \;{_3F_2}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)$$

$$\Im\left[\operatorname{Li}_\color{red}2\left(\frac{1+i}2\right)\right] =\frac{-1}{\sqrt2} \;{_3F_2}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)+\frac{3K}2$$

with Catalan's constant $K=\Im\left[\operatorname{Li}_\color{blue}2(i)\right]$ and,

$$\Im\left[\operatorname{Li}_\color{red}3\big(1+i\big)\right] =\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac{\pi^3}{192}$$

$$\Im\left[\operatorname{Li}_\color{red}3\left(\frac{1+i}2\right)\right] =-\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)+\frac{23\pi^3}{384}+\frac{3\pi\ln^2 2}{32}$$

where $\frac{\pi^3}{32}=\Im\left[\operatorname{Li}_\color{blue}3(i)\right]$. It is tempting to speculate what comes next.

Edit: July 2019. With insights from this answer, it turns out,

$$\Im\left[\operatorname{Li}_\color{red}4\big(1+i\big)\right] =\sqrt2 \;{_5F_4}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac{\pi^3\ln 2}{382}-\frac{\pi\zeta(3)}8+\frac{\Im\left[\operatorname{Li}_\color{blue}4(i)\right]}4$$

P.S. The first two hypergeometrics actually have a closed-form, but I didn't want to clutter my answer.

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    $\begingroup$ Note that the similar hypergeometric below has a nice closed-form, $$\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\color{red}{\tfrac14}\right) = \frac{32}{27}\,\Im\left[\operatorname{Li}_3\left(\frac{1+i}{\sqrt2}\right)\right]=\frac{7}{216}\pi^3$$ $\endgroup$ – Tito Piezas III Jun 24 at 15:36
  • $\begingroup$ $2^{-1/2}\cdot\beta(2)+2^{-5/2}\cdot\Gamma^2(1/2)\cdot\ln(2)~$ has a nice ring to it. $\endgroup$ – Lucian Jun 24 at 23:01
  • $\begingroup$ Does $~_5F_4\Big(\{1/2\}_5~;~\{3/2\}_4~;~{\color{blue}{1/8}}\Big)~$ also possess a closed form expression ? $\endgroup$ – Lucian Jul 13 at 9:40
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    $\begingroup$ @Lucian: A general closed-form with the most number of terms seems to be $$\,_3F_2\left(\frac12,\frac12,\frac12;\frac32,\frac32;x\right) = -\,\frac{{\rm Li}_2(1-\beta^2)}{2\sqrt{-x}}-\frac{\ln^2 \beta}{2\sqrt{-x}}$$ where, $$\beta = \sqrt{1-x}+\sqrt{-x}$$ Beyond that is harder. $\endgroup$ – Tito Piezas III Jul 13 at 10:17
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    $\begingroup$ Another representation $$\Im\operatorname{Li}_4(1+i)=\frac{1}{8} \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)+\frac{\beta(4)}{2}+\frac{1}{96} \pi \log ^3(2)+\frac{3}{128} \pi ^3 \log (2)$$ Don't know this should be the comment of the answer of Cleo or this answer. $\endgroup$ – Kemono Chen Jul 13 at 13:35

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