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- Jan 17, 2013

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

- Thread starter ZaidAlyafey
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- Jan 17, 2013

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

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

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

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if we let \(\displaystyle \text { Li }_2{(x)}=- \int ^ { x}_0\frac {\log (1- u ) } {u} \, du\)

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

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

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

- Mar 22, 2013

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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.ZaidAlyafey said:But that still unsatisfactory

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Actually it is , but I posted this before seeing the derivation. Here is a linkI 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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The link given above seems to be to a post that was deleted by the OP.Actually it is , but I posted this before seeing the derivation...

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scroll down it is not the first post .The link given above seem to be to a post that was deleted by the OP.

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

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

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 :

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

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