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$\qquad\qquad$ 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 !


Apparently, $~\Im~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)~+~\Im~\text{Li}_3(1+i)~=~\dfrac7{128}\cdot\pi^3~+~\dfrac3{32}\cdot\pi\cdot\ln^22,~$ so the question

is equivalent to asking for the closed form of the imaginary part of $~\Im~\text{Li}_3(1+i).$

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    $\begingroup$ Any particular reason you chose to italicize most of this question? $\endgroup$
    – abiessu
    Sep 4, 2014 at 0:27
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    $\begingroup$ No. Just a simple matter of aesthetics. $\endgroup$
    – Lucian
    Sep 4, 2014 at 2:10

7 Answers 7

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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 $$\Im\Big[\operatorname{Li}_3\left(1+i\right)\Big]=\frac{3\pi^3}{64}+\frac\pi{16}\ln^22-\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$

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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

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

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.

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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

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

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  • $\begingroup$ 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, 2014 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, 2014 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, 2014 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, 2014 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, 2014 at 23:05
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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, $$\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\color{red}{\tfrac14}\right) = \frac{32}{27}\,\Im\left[\operatorname{Li}_3\left(\frac{1+i}{\sqrt2}\right)\right]=\frac{7}{216}\pi^3$$ $\endgroup$ Jun 24, 2019 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, 2019 at 23:01
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    $\begingroup$ 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, 2019 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. $\endgroup$ Jul 13, 2019 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. $\endgroup$ Jul 13, 2019 at 13:35
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Using the relation between 3 trilogarithms given here: \begin{equation} \operatorname{Li}_3(z)=-\operatorname{Li}_3(\frac{z}{z-1})-\operatorname{Li}_3(1-z)+\frac{1}{6}\log^3(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{\pi^2}{6}\log(1-z)+\zeta(3) \end{equation} with $z=(1+i)/2$, we have $z/(z-1)=-i$ and $1-z=(1-i)/2$. With the known specific value (see here) \begin{align} \operatorname{Li}_3(-i)&=-\frac{3}{32}\zeta(3)-i\frac{\pi^3}{32} \end{align}and remarking that $\operatorname{Li}_n(z)=\overline{\operatorname{Li}_n(\overline{z})}$, \begin{equation} \operatorname{Li}_3(z)+\operatorname{Li}_3(1-z)=2\Re\operatorname{Li}_3(\frac{1+i}{2}) \end{equation} and \begin{align} \log\left( z \right)&=-\frac{\log 2}{2}+i\frac{\pi}{4}\\ \log\left(1- z \right)&=-\frac{\log 2}{2}-i\frac{\pi}{4} \end{align} we deduce \begin{align} \Re\operatorname{Li}_3(\frac{1+i}{2})&=\frac{1}{2}\left[ \frac{3}{32}\zeta(3)+i\frac{\pi^3}{32} + \frac{1}{6}\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)^3-\right.\\ &\left.\frac{1}{2}\left( -\frac{\log 2}{2}+i\frac{\pi}{4} \right))\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)^2+\frac{\pi^2}{6}\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)+\zeta(3) \right]\\ &=\frac{35}{64}\zeta(3)+\frac{\log^32}{48}-\frac{5}{192}\pi^2\ln 2 \end{align} as expected.

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  • $\begingroup$ (+1) It looks fine, thanks! $\endgroup$ Dec 7, 2019 at 4:33
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    $\begingroup$ Another way to see that $\displaystyle \Re\biggr \{\operatorname{Li}_3\left(\frac{1-i}{2}\right)\biggr\}=\Re\biggr \{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}$ is by using that $\displaystyle -\frac{1}{2}\sum_{n=1}^{\infty}\frac{x^n}{n}(H_n^2+H_n^{(2)})=\operatorname{Li}_3\left(\frac{x}{x-1}\right)$, and plugging $x=\pm i$. $\endgroup$ Dec 10, 2019 at 18:18
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    $\begingroup$ Indeed, this is neat. $\endgroup$
    – Paul Enta
    Dec 10, 2019 at 21:03
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I'll sketch a simple solution to the fact that $$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3),$$ as the OP stated.

First step: We easily prove this (possibly new?) result

$$\int_0^1 \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x$$ $$=\zeta (2)\frac{ \log (1-a)}{a}+\frac{\log ^3(1-a)}{6 a}+\frac{\text{Li}_3(a)}{a}-\frac{\text{Li}_3\left(\frac{a}{a-1}\right)}{a}. \tag1$$ The result above is easy to prove with the algebraic identities and the result $\displaystyle \int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$ from (Almost) Impossible Integrals, Sums, and Series, page $4$.

Second step: Plug in $(1)$ $a=-i$.

Third step: We need the integral

$$\int_0^1 \frac{x\log(1- x)\log(x)}{1+x^2}\textrm{d}x=\frac{1}{16}\left(\frac{41}{4}\zeta(3)-9\log(2)\zeta(2)\right),$$ and this is calculated by real methods in the book (Almost) Impossible Integrals, Sums, and Series

Fourth step: We combine the results above and extract the desired value

$$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).$$ In a similar style we derive other values mentioned by user153012 in his/her post.

More details may be found in the preprint, A special way of extracting the real part of the Trilogarithm, Li_3((1±i)/2))


For those interested in various forms of $(1)$, with slightly different numerators, see these new entries in the mathematical literature,

Let $a\le1$ be a real number. The following equalities hold: \begin{equation*} i) \ \int_0^1 \frac{\log (x)\operatorname{Li}_2(x) }{1-a x} \textrm{d}x=\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}; \end{equation*} \begin{equation*} ii) \ \int_0^1 \frac{\log^2(x)\operatorname{Li}_3(x) }{1-a x} \textrm{d}x=20\frac{\operatorname{Li}_6(a)}{a}-12 \zeta(2)\frac{\operatorname{Li}_4(a)}{ a}+\frac{(\operatorname{Li}_3(a))^2}{a}, \end{equation*}

which appears in the preprint A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by Cornel I. Valean.

Just observe the result from $i)$ is the result in $(1)$ with $\log(1-x)$ replaced by $\operatorname{Li}_2(x)$.

Further Research:

The new formulae might open the possibility of extracting the real parts of the more advanced versions, $\Re\bigg[\text{Li}_4\bigg(\dfrac{1\pm i}2\bigg)\bigg]$, which are known in the mathematical literature, or even approach the general case $\Re\bigg[\text{Li}_n\bigg(\dfrac{1\pm i}2\bigg)\bigg]$.


Related and conjectured cases you may find here On the simplification of $\Re\operatorname{Li}_5(1+i)$ and its generalization, like $\Re\{\operatorname{Li}_4(1+i)\}$ and $\Re\{\operatorname{Li}_5(1+i)\}$.

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    $\begingroup$ A note: this question of mine, which relates $\Re\operatorname{Li}_n\left(\frac{1+i}2\right)$ to $\operatorname{Li}_n\left(\frac12\right)$, is related to your last paragraph. $\endgroup$ Dec 10, 2019 at 6:58
  • $\begingroup$ @KemonoChen Thank you for letting me know! Great question! $\endgroup$ Dec 10, 2019 at 18:04
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Start with the trilog identity

$$\text{Li}_3(x)+\text{Li}_3(1-x)+\text{Li}_3\left(\frac{x-1}{x}\right)$$ $$=\zeta(3)+\frac16\ln^3(x)+\zeta(2)\ln(x)-\frac12\ln^2(x)\ln(1-x)\tag1$$

set $x=i$ and consider the real parts of the two sides we have

$$\Re\left\{\text{Li}_3(i)+\text{Li}_3(1-i)+\text{Li}_3\left(\frac{i-1}{i}\right)\right\}$$ $$=\zeta(3)+\Re\left\{\frac16\ln^3(i)+\zeta(2)\ln(i)-\frac12\ln^2(i)\ln(1-i)\right\}$$

note that $\frac{i-1}{i}=1+i$ and that $\Re \text{Li}_3(1+i)=\Re \text{Li}_3(1-i)$. This gives

$$\Re\left\{\text{Li}_3(i)+2\text{Li}_3(1+i)\right\}=\zeta(3)+\Re\left\{\frac16\ln^3(i)+\zeta(2)\ln(i)-\frac12\ln^2(i)\ln(1-i)\right\}$$

Substitute $\Re\text{Li}_3(i)=-\frac{3}{32}\zeta(3)$, $\ln(i)=\frac{\pi}{2}i$ and $\ln(1-i)=\frac12\ln(2)-\frac{\pi}{4}i$ we get

$$\Re\text{Li}_3(1+i)=\frac{3}{16}\ln(2)\zeta(2)+\frac{35}{64}\zeta(3)$$

now setting $x=1+i$ in $(1)$ and considering the real parts of both sides yields

$$\Re\left\{\text{Li}_3(1+i)+\text{Li}_3(-i)+\text{Li}_3\left(\frac{i}{1+i}\right)\right\}$$ $$=\zeta(3)+\Re\left\{\frac16\ln^3(1+i)+\zeta(2)\ln(1+i)-\frac12\ln^2(1+i)\ln(-i)\right\}$$

Note $\frac{i}{1+i}=\frac{1+i}{2}$ and substitute $\Re\text{Li}_3(1+i)=\frac{3}{16}\ln(2)\zeta(2)+\frac{35}{64}\zeta(3)$, $\Re\text{Li}_3(-i)=-\frac{3}{32}\zeta(3)$, $\ln(-i)=-\frac{\pi}{2}i$ and $\ln(1+i)=\frac12\ln(2)+\frac{\pi}{4}i$ we obtain

$$\Re\text{Li}_3\bigg(\dfrac{1+i}2\bigg)=\dfrac{\ln^3(2)}{48}-\dfrac{\ln(2)}{32}\zeta(2)+\dfrac{35}{64}\zeta(3)=\Re\text{Li}_3\bigg(\dfrac{1-i}2\bigg)$$

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    $\begingroup$ Imaginary part? $\endgroup$
    – FDP
    May 3, 2020 at 11:41
  • $\begingroup$ @FDP I dont think a closed form exists for the imaginary part of the trilog in the question. $\endgroup$ May 3, 2020 at 11:46

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