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[SOLVED] Prove the following dilogarithmic value

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \text{Li}_{2}\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2} \log^2 (2) \)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I found the following functional equation :

\(\displaystyle \text{Li}_2(x)+\text{Li}_{2}(1-x) = \frac{\pi^2}{6}- \ln(x)\cdot \ln(1-x) \)

Substituting x = 1/2 gives us the result , does anybody know how to prove this functional equation ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I found the following functional equation :

\(\displaystyle \text{Li}_2(x)+\text{Li}_{2}(1-x) = \frac{\pi^2}{6}- \ln(x)\cdot \ln(1-x) \)

Substituting x = 1/2 gives us the result , does anybody know how to prove this functional equation ?
if we let \(\displaystyle \text { Li }_2{(x)}=- \int ^ { x}_0\frac {\log (1- u ) } {u} \, du\)

Then if we differentiated the functional equation we get the result . But that still unsatisfactory.(Tauri)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
ZaidAlyafey said:
But that still unsatisfactory
I don't see what's unsatisfactory to you. Galactus' derivation is perfectly logical and satisfactory as it occurs to me. If you want another proof, then you might be interested in a proof of Abel's identity which is a further generalization of the reflection formula.
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I don't see what's unsatisfactory to you. Galactus' derivation is perfectly logical and satisfactory as it occurs to me. If you want another proof, then you might be interested in a proof of Abel's identity which is a further generalization of the reflection formula.
Actually it is , but I posted this before seeing the derivation. Here is a link
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Actually it is , but I posted this before seeing the derivation...
The link given above seems to be to a post that was deleted by the OP.
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You can set your link to take you to to the post you mean...

Is this it?

clickety-click
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Ok , here is the full proof

\(\displaystyle \text{Li}_{2}(x) =-\int^{x}_0 \frac{\log (1-u)}{u}\,du\)

\(\displaystyle \frac{d}{dx} \left(\text{Li}_{2}(x)\right) =-\frac{\log (1-x)}{x}\)

Integrating by parts

\(\displaystyle \text{Li}_{2}(x) = -\log(1-x) \log(x) +\int^{1-x}_0 \frac{\log(1-u)}{u}du+ C\)

\(\displaystyle \text{Li}_{2}(x) = -\log(1-x) \log(x) - \text{Li}_2 (1-x) +C\)

Letting $x$ approaches 0 we get :

\(\displaystyle C=\text{Li}_2 (1)= \zeta(2) = \frac{\pi^2}{6}\)

\(\displaystyle \text{Li}_{2}(x) + \text{Li}_2 (1-x) = \frac{\pi^2}{6}-\log(1-x) \log(x) \)