Closed-form of a special value dilogarithm identity
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Let $c$ be the following.
$$c = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)+operatornameLi_2left(frac 34 + fracisqrt 34right) + frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right),$$
where $operatornameLi_2$ is the dilogarithm function.
This $c$ appeared while combining answers to this question. Results from there we know that
$$Re(c) = -fracpi^254-frac12ln^2 2+fracpi,sqrt36 ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14psi_1left(frac13right)+frac16operatornameLi_2left(-frac13right),$$
where $psi_1$ is the trigamma function.
My questions.
$1^textst$ Question. Could we evaluate $Re(c)$ also via some dilogarithm identity?
$2^textnd$ Question. Could we specify a closed-form of $Im(c)$ too?
Edit. From the analysis by @David H we have more results. Let
$$beginalign
c_1 & = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)\
c_2 & = operatornameLi_2left(frac 34 + fracisqrt 34right)\
c_3 & = frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right).
endalign$$
Therefore of course $c=c_1+c_2+c_3$. David H has shown that
$$Re(c_2) = frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).$$
From here it's easy to see, that
$$Re(c_1+c_3) = -frac25,pi^2216-frac18ln^2 3 + fracpi,sqrt36ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14 psi_1left(frac13right) - frac112 operatornameLi_2 left( - frac13 right).$$
calculus special-functions closed-form polylogarithm
edited Apr 13 '17 at 12:20
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asked Oct 19 '14 at 0:01
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Let $c$ be the following.
$$c = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)+operatornameLi_2left(frac 34 + fracisqrt 34right) + frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right),$$
where $operatornameLi_2$ is the dilogarithm function.
This $c$ appeared while combining answers to this question. Results from there we know that
$$Re(c) = -fracpi^254-frac12ln^2 2+fracpi,sqrt36 ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14psi_1left(frac13right)+frac16operatornameLi_2left(-frac13right),$$
where $psi_1$ is the trigamma function.
My questions.
$1^textst$ Question. Could we evaluate $Re(c)$ also via some dilogarithm identity?
$2^textnd$ Question. Could we specify a closed-form of $Im(c)$ too?
Edit. From the analysis by @David H we have more results. Let
$$beginalign
c_1 & = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)\
c_2 & = operatornameLi_2left(frac 34 + fracisqrt 34right)\
c_3 & = frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right).
endalign$$
Therefore of course $c=c_1+c_2+c_3$. David H has shown that
$$Re(c_2) = frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).$$
From here it's easy to see, that
$$Re(c_1+c_3) = -frac25,pi^2216-frac18ln^2 3 + fracpi,sqrt36ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14 psi_1left(frac13right) - frac112 operatornameLi_2 left( - frac13 right).$$
calculus special-functions closed-form polylogarithm
edited Apr 13 '17 at 12:20
Community♦
1
asked Oct 19 '14 at 0:01
user153012
6,09522277
add a comment |Â
up vote
5
down vote
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up vote
5
down vote
favorite
Let $c$ be the following.
$$c = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)+operatornameLi_2left(frac 34 + fracisqrt 34right) + frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right),$$
where $operatornameLi_2$ is the dilogarithm function.
This $c$ appeared while combining answers to this question. Results from there we know that
$$Re(c) = -fracpi^254-frac12ln^2 2+fracpi,sqrt36 ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14psi_1left(frac13right)+frac16operatornameLi_2left(-frac13right),$$
where $psi_1$ is the trigamma function.
My questions.
$1^textst$ Question. Could we evaluate $Re(c)$ also via some dilogarithm identity?
$2^textnd$ Question. Could we specify a closed-form of $Im(c)$ too?
Edit. From the analysis by @David H we have more results. Let
$$beginalign
c_1 & = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)\
c_2 & = operatornameLi_2left(frac 34 + fracisqrt 34right)\
c_3 & = frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right).
endalign$$
Therefore of course $c=c_1+c_2+c_3$. David H has shown that
$$Re(c_2) = frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).$$
From here it's easy to see, that
$$Re(c_1+c_3) = -frac25,pi^2216-frac18ln^2 3 + fracpi,sqrt36ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14 psi_1left(frac13right) - frac112 operatornameLi_2 left( - frac13 right).$$
calculus special-functions closed-form polylogarithm
edited Apr 13 '17 at 12:20
Community♦
1
asked Oct 19 '14 at 0:01
user153012
6,09522277
Let $c$ be the following.
$$c = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)+operatornameLi_2left(frac 34 + fracisqrt 34right) + frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right),$$
where $operatornameLi_2$ is the dilogarithm function.
This $c$ appeared while combining answers to this question. Results from there we know that
$$Re(c) = -fracpi^254-frac12ln^2 2+fracpi,sqrt36 ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14psi_1left(frac13right)+frac16operatornameLi_2left(-frac13right),$$
where $psi_1$ is the trigamma function.
My questions.
$1^textst$ Question. Could we evaluate $Re(c)$ also via some dilogarithm identity?
$2^textnd$ Question. Could we specify a closed-form of $Im(c)$ too?
Edit. From the analysis by @David H we have more results. Let
$$beginalign
c_1 & = frac1+isqrt 33operatornameLi_2left(1-fracisqrt 33right)\
c_2 & = operatornameLi_2left(frac 34 + fracisqrt 34right)\
c_3 & = frac1-isqrt33operatornameLi_2left(frac12+fracisqrt 36right).
endalign$$
Therefore of course $c=c_1+c_2+c_3$. David H has shown that
$$Re(c_2) = frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).$$
From here it's easy to see, that
$$Re(c_1+c_3) = -frac25,pi^2216-frac18ln^2 3 + fracpi,sqrt36ln 2 - fracpi,sqrt39 ln 3 + frac16 ln 2 ln 3 + frac14 psi_1left(frac13right) - frac112 operatornameLi_2 left( - frac13 right).$$
calculus special-functions closed-form polylogarithm
calculus special-functions closed-form polylogarithm
edited Apr 13 '17 at 12:20
Community♦
1
asked Oct 19 '14 at 0:01
user153012
6,09522277
edited Apr 13 '17 at 12:20
Community♦
1
asked Oct 19 '14 at 0:01
user153012
6,09522277
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Oct 19 '14 at 0:01
user153012
6,09522277
asked Oct 19 '14 at 0:01
user153012
6,09522277
asked Oct 19 '14 at 0:01
user153012
6,09522277
6,09522277
add a comment |Â
add a comment |Â
1 Answer
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The following is a partial answer to the first question.
For any real number $ainmathbbR$ and any two complex numers $z_1,z_2inmathbbC$, the operations $Re(cdot)$ and $Im(cdot)$ obey the following algebraic properties:
$$begincases
Re(z_1):=fracz_1+barz_12\
Im(z_1):=fracz_1-barz_12i\
Re(z_1+z_2)=Re(z_1)+Re(z_2)\
Im(z_1+z_2)=Im(z_1)+Im(z_2)\
Re(az_1)=a,Re(z_1)\
Im(az_1)=a,Im(z_1)\
Re(iz_1)=-Im(z_1)\
Im(iz_1)=Re(z_1).\
endcases$$
Using the above properties, we may simplify $Re(c)$ as follows:
$$beginalign
Re(c)
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)+frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)+frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+Releft[frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +Releft[frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+frac13Releft[operatornameLi_2left(1-ifracsqrt33right)right]-frac1sqrt3Imleft[operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +frac13Releft[operatornameLi_2left(frac12+ifracsqrt36right)right]+frac1sqrt3Imleft[operatornameLi_2left(frac12+ifracsqrt36right)right].\
endalign$$
Now, Landen's dilogarithm identity states:
$$operatornameLi_2left(zright)=-operatornameLi_2left(fraczz-1right)-frac12ln^2left(1-zright);~znotin[1,infty).$$
Then, letting $z=frac34+ifracsqrt34$ we have $fraczz-1=-i,sqrt3$, and thus:
$$beginalign
operatornameLi_2left(frac34+ifracsqrt34right)
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac14-ifracsqrt34right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac12e^-fracipi3right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(lnleft(frac12right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-lnleft(2right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-fracpi^29+ln^2left(2right)+frac2ipiln(2)3right)\
&=-operatornameLi_2left(-i,sqrt3right)+fracpi^218-fracln^2left(2right)2-fracipiln(2)3\
&=fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right).\
endalign$$
Taking the real component of this dilogarithmic term yields:
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=Releft[fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-Releft[operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+overlineoperatornameLi_2left(-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(overline-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4,\
endalign$$
where in going from the third to the fourth line we've used the mirror symmetry of the dilogarithm, and to obtain the last line we've used the dilogarithmic identity,
$$operatornameLi_2left(zright)+operatornameLi_2left(-zright)=frac12operatornameLi_2left(z^2right).$$
One more dilogarithmic identity we shall need relates the dilogarithms of reciprocal arguments:
$$operatornameLi_2left(zright)=-operatornameLi_2left(frac1zright)-frac12ln^2(-z)-fracpi^26;~znotin[0,1].$$
Hence,
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4\
&=fracpi^218-fracln^2left(2right)2-frac-operatornameLi_2left(-frac13right)-frac12ln^2(3)-fracpi^264\
&=frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).\
endalign$$
answered Oct 19 '14 at 5:07
David H
21.2k24190
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
1
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
1
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
2
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
1
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
The following is a partial answer to the first question.
For any real number $ainmathbbR$ and any two complex numers $z_1,z_2inmathbbC$, the operations $Re(cdot)$ and $Im(cdot)$ obey the following algebraic properties:
$$begincases
Re(z_1):=fracz_1+barz_12\
Im(z_1):=fracz_1-barz_12i\
Re(z_1+z_2)=Re(z_1)+Re(z_2)\
Im(z_1+z_2)=Im(z_1)+Im(z_2)\
Re(az_1)=a,Re(z_1)\
Im(az_1)=a,Im(z_1)\
Re(iz_1)=-Im(z_1)\
Im(iz_1)=Re(z_1).\
endcases$$
Using the above properties, we may simplify $Re(c)$ as follows:
$$beginalign
Re(c)
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)+frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)+frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+Releft[frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +Releft[frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+frac13Releft[operatornameLi_2left(1-ifracsqrt33right)right]-frac1sqrt3Imleft[operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +frac13Releft[operatornameLi_2left(frac12+ifracsqrt36right)right]+frac1sqrt3Imleft[operatornameLi_2left(frac12+ifracsqrt36right)right].\
endalign$$
Now, Landen's dilogarithm identity states:
$$operatornameLi_2left(zright)=-operatornameLi_2left(fraczz-1right)-frac12ln^2left(1-zright);~znotin[1,infty).$$
Then, letting $z=frac34+ifracsqrt34$ we have $fraczz-1=-i,sqrt3$, and thus:
$$beginalign
operatornameLi_2left(frac34+ifracsqrt34right)
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac14-ifracsqrt34right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac12e^-fracipi3right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(lnleft(frac12right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-lnleft(2right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-fracpi^29+ln^2left(2right)+frac2ipiln(2)3right)\
&=-operatornameLi_2left(-i,sqrt3right)+fracpi^218-fracln^2left(2right)2-fracipiln(2)3\
&=fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right).\
endalign$$
Taking the real component of this dilogarithmic term yields:
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=Releft[fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-Releft[operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+overlineoperatornameLi_2left(-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(overline-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4,\
endalign$$
where in going from the third to the fourth line we've used the mirror symmetry of the dilogarithm, and to obtain the last line we've used the dilogarithmic identity,
$$operatornameLi_2left(zright)+operatornameLi_2left(-zright)=frac12operatornameLi_2left(z^2right).$$
One more dilogarithmic identity we shall need relates the dilogarithms of reciprocal arguments:
$$operatornameLi_2left(zright)=-operatornameLi_2left(frac1zright)-frac12ln^2(-z)-fracpi^26;~znotin[0,1].$$
Hence,
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4\
&=fracpi^218-fracln^2left(2right)2-frac-operatornameLi_2left(-frac13right)-frac12ln^2(3)-fracpi^264\
&=frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).\
endalign$$
answered Oct 19 '14 at 5:07
David H
21.2k24190
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
1
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
1
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
2
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
1
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
 |Â
show 7 more comments
up vote
7
down vote
The following is a partial answer to the first question.
For any real number $ainmathbbR$ and any two complex numers $z_1,z_2inmathbbC$, the operations $Re(cdot)$ and $Im(cdot)$ obey the following algebraic properties:
$$begincases
Re(z_1):=fracz_1+barz_12\
Im(z_1):=fracz_1-barz_12i\
Re(z_1+z_2)=Re(z_1)+Re(z_2)\
Im(z_1+z_2)=Im(z_1)+Im(z_2)\
Re(az_1)=a,Re(z_1)\
Im(az_1)=a,Im(z_1)\
Re(iz_1)=-Im(z_1)\
Im(iz_1)=Re(z_1).\
endcases$$
Using the above properties, we may simplify $Re(c)$ as follows:
$$beginalign
Re(c)
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)+frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)+frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+Releft[frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +Releft[frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+frac13Releft[operatornameLi_2left(1-ifracsqrt33right)right]-frac1sqrt3Imleft[operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +frac13Releft[operatornameLi_2left(frac12+ifracsqrt36right)right]+frac1sqrt3Imleft[operatornameLi_2left(frac12+ifracsqrt36right)right].\
endalign$$
Now, Landen's dilogarithm identity states:
$$operatornameLi_2left(zright)=-operatornameLi_2left(fraczz-1right)-frac12ln^2left(1-zright);~znotin[1,infty).$$
Then, letting $z=frac34+ifracsqrt34$ we have $fraczz-1=-i,sqrt3$, and thus:
$$beginalign
operatornameLi_2left(frac34+ifracsqrt34right)
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac14-ifracsqrt34right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac12e^-fracipi3right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(lnleft(frac12right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-lnleft(2right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-fracpi^29+ln^2left(2right)+frac2ipiln(2)3right)\
&=-operatornameLi_2left(-i,sqrt3right)+fracpi^218-fracln^2left(2right)2-fracipiln(2)3\
&=fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right).\
endalign$$
Taking the real component of this dilogarithmic term yields:
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=Releft[fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-Releft[operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+overlineoperatornameLi_2left(-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(overline-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4,\
endalign$$
where in going from the third to the fourth line we've used the mirror symmetry of the dilogarithm, and to obtain the last line we've used the dilogarithmic identity,
$$operatornameLi_2left(zright)+operatornameLi_2left(-zright)=frac12operatornameLi_2left(z^2right).$$
One more dilogarithmic identity we shall need relates the dilogarithms of reciprocal arguments:
$$operatornameLi_2left(zright)=-operatornameLi_2left(frac1zright)-frac12ln^2(-z)-fracpi^26;~znotin[0,1].$$
Hence,
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4\
&=fracpi^218-fracln^2left(2right)2-frac-operatornameLi_2left(-frac13right)-frac12ln^2(3)-fracpi^264\
&=frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).\
endalign$$
answered Oct 19 '14 at 5:07
David H
21.2k24190
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
1
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
1
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
2
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
1
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
 |Â
show 7 more comments
up vote
7
down vote
up vote
7
down vote
The following is a partial answer to the first question.
For any real number $ainmathbbR$ and any two complex numers $z_1,z_2inmathbbC$, the operations $Re(cdot)$ and $Im(cdot)$ obey the following algebraic properties:
$$begincases
Re(z_1):=fracz_1+barz_12\
Im(z_1):=fracz_1-barz_12i\
Re(z_1+z_2)=Re(z_1)+Re(z_2)\
Im(z_1+z_2)=Im(z_1)+Im(z_2)\
Re(az_1)=a,Re(z_1)\
Im(az_1)=a,Im(z_1)\
Re(iz_1)=-Im(z_1)\
Im(iz_1)=Re(z_1).\
endcases$$
Using the above properties, we may simplify $Re(c)$ as follows:
$$beginalign
Re(c)
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)+frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)+frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+Releft[frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +Releft[frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+frac13Releft[operatornameLi_2left(1-ifracsqrt33right)right]-frac1sqrt3Imleft[operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +frac13Releft[operatornameLi_2left(frac12+ifracsqrt36right)right]+frac1sqrt3Imleft[operatornameLi_2left(frac12+ifracsqrt36right)right].\
endalign$$
Now, Landen's dilogarithm identity states:
$$operatornameLi_2left(zright)=-operatornameLi_2left(fraczz-1right)-frac12ln^2left(1-zright);~znotin[1,infty).$$
Then, letting $z=frac34+ifracsqrt34$ we have $fraczz-1=-i,sqrt3$, and thus:
$$beginalign
operatornameLi_2left(frac34+ifracsqrt34right)
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac14-ifracsqrt34right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac12e^-fracipi3right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(lnleft(frac12right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-lnleft(2right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-fracpi^29+ln^2left(2right)+frac2ipiln(2)3right)\
&=-operatornameLi_2left(-i,sqrt3right)+fracpi^218-fracln^2left(2right)2-fracipiln(2)3\
&=fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right).\
endalign$$
Taking the real component of this dilogarithmic term yields:
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=Releft[fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-Releft[operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+overlineoperatornameLi_2left(-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(overline-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4,\
endalign$$
where in going from the third to the fourth line we've used the mirror symmetry of the dilogarithm, and to obtain the last line we've used the dilogarithmic identity,
$$operatornameLi_2left(zright)+operatornameLi_2left(-zright)=frac12operatornameLi_2left(z^2right).$$
One more dilogarithmic identity we shall need relates the dilogarithms of reciprocal arguments:
$$operatornameLi_2left(zright)=-operatornameLi_2left(frac1zright)-frac12ln^2(-z)-fracpi^26;~znotin[0,1].$$
Hence,
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4\
&=fracpi^218-fracln^2left(2right)2-frac-operatornameLi_2left(-frac13right)-frac12ln^2(3)-fracpi^264\
&=frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).\
endalign$$
answered Oct 19 '14 at 5:07
David H
21.2k24190
The following is a partial answer to the first question.
For any real number $ainmathbbR$ and any two complex numers $z_1,z_2inmathbbC$, the operations $Re(cdot)$ and $Im(cdot)$ obey the following algebraic properties:
$$begincases
Re(z_1):=fracz_1+barz_12\
Im(z_1):=fracz_1-barz_12i\
Re(z_1+z_2)=Re(z_1)+Re(z_2)\
Im(z_1+z_2)=Im(z_1)+Im(z_2)\
Re(az_1)=a,Re(z_1)\
Im(az_1)=a,Im(z_1)\
Re(iz_1)=-Im(z_1)\
Im(iz_1)=Re(z_1).\
endcases$$
Using the above properties, we may simplify $Re(c)$ as follows:
$$beginalign
Re(c)
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)+frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)+frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+Releft[frac1+isqrt33operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +Releft[frac1-isqrt33operatornameLi_2left(frac12+ifracsqrt36right)right]\
&=Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]+frac13Releft[operatornameLi_2left(1-ifracsqrt33right)right]-frac1sqrt3Imleft[operatornameLi_2left(1-ifracsqrt33right)right]\
&~~~~~ +frac13Releft[operatornameLi_2left(frac12+ifracsqrt36right)right]+frac1sqrt3Imleft[operatornameLi_2left(frac12+ifracsqrt36right)right].\
endalign$$
Now, Landen's dilogarithm identity states:
$$operatornameLi_2left(zright)=-operatornameLi_2left(fraczz-1right)-frac12ln^2left(1-zright);~znotin[1,infty).$$
Then, letting $z=frac34+ifracsqrt34$ we have $fraczz-1=-i,sqrt3$, and thus:
$$beginalign
operatornameLi_2left(frac34+ifracsqrt34right)
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac14-ifracsqrt34right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12ln^2left(frac12e^-fracipi3right)\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(lnleft(frac12right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-lnleft(2right)-fracipi3right)^2\
&=-operatornameLi_2left(-i,sqrt3right)-frac12left(-fracpi^29+ln^2left(2right)+frac2ipiln(2)3right)\
&=-operatornameLi_2left(-i,sqrt3right)+fracpi^218-fracln^2left(2right)2-fracipiln(2)3\
&=fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right).\
endalign$$
Taking the real component of this dilogarithmic term yields:
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=Releft[fracpi^218-fracln^2left(2right)2-fracipiln(2)3-operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-Releft[operatornameLi_2left(-i,sqrt3right)right]\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+overlineoperatornameLi_2left(-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(overline-i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-i,sqrt3right)+operatornameLi_2left(i,sqrt3right)2\
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4,\
endalign$$
where in going from the third to the fourth line we've used the mirror symmetry of the dilogarithm, and to obtain the last line we've used the dilogarithmic identity,
$$operatornameLi_2left(zright)+operatornameLi_2left(-zright)=frac12operatornameLi_2left(z^2right).$$
One more dilogarithmic identity we shall need relates the dilogarithms of reciprocal arguments:
$$operatornameLi_2left(zright)=-operatornameLi_2left(frac1zright)-frac12ln^2(-z)-fracpi^26;~znotin[0,1].$$
Hence,
$$beginalign
Releft[operatornameLi_2left(frac34+ifracsqrt34right)right]
&=fracpi^218-fracln^2left(2right)2-fracoperatornameLi_2left(-3right)4\
&=fracpi^218-fracln^2left(2right)2-frac-operatornameLi_2left(-frac13right)-frac12ln^2(3)-fracpi^264\
&=frac7pi^272+fracln^2(3)8-fracln^2left(2right)2+frac14operatornameLi_2left(-frac13right).\
endalign$$
answered Oct 19 '14 at 5:07
David H
21.2k24190
answered Oct 19 '14 at 5:07
David H
21.2k24190
answered Oct 19 '14 at 5:07
David H
21.2k24190
answered Oct 19 '14 at 5:07
David H
21.2k24190
21.2k24190
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
1
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
1
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
2
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
1
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
 |Â
show 7 more comments
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
1
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
1
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
2
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
1
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
Nice answer David H again! Thank you! +1.
– user153012
Oct 19 '14 at 8:49
1
1
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
Impressive! +1 for the answer and +1 for the OP @user153012. I am really too kind for upvoting post so easily (ã£＾▿＾)💨
– Anastasiya-Romanova 秀
Oct 19 '14 at 9:24
1
1
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
Thank you @Anastasiya-Romanova and with this, yes, you really have a really nice expression to your hypergeom series there. Finally. ;>
– user153012
Oct 19 '14 at 9:26
2
2
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
@Anastasiya-Romanova $$_3F_2left(frac13,frac13,frac13;frac43,frac43; frac12 right)$$ equals to $$fracsqrt[3]23left( -frac5,pi^272 - fracpi,sqrt39ln 2 + fracpi,sqrt36ln 3+frac18ln^2 3 - frac16 ln^2 2 + frac14 psi_1left(frac13right) + frac13operatornameLi_2left(-frac13right) right)$$
– user153012
Oct 19 '14 at 10:00
1
1
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
@Anastasiya-Romanova The proof comes from the answers by M.N.C.E. and David H here, and the analysis by David H of this question.
– user153012
Oct 19 '14 at 10:07
 |Â
show 7 more comments
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