Prove the following dilogarithmic value

[alyafey22]

\(\displaystyle \text{Li}_{2}\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2} \log^2 (2) \)

 

[alyafey22]

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 ?

 

[alyafey22]

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]

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.

 

[alyafey22]

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]

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.

 

[alyafey22]

The link given above seem to be to a post that was deleted by the OP.

scroll down it is not the first post .

 

[MarkFL]

You can set your link to take you to to the post you mean...

Is this it?

clickety-click

 

[alyafey22]

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 :

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

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