Difference between revisions of "Dilogarithm"

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(Created page with "The dilogarithm is the function $$L_2(z)=\displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{k^2},$$ which is a special case of the polylogarithm. =Properties= <div class="tocco...")
 
(Properties)
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<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$L_2(z)=-L_2 \left( \dfrac{1}{z} \right) - \dfrac{1}{2} \left( \log(z) \right)^2 + \pi i \log(z) + \dfrac{\pi^2}{3}.$$
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$$\mathrm{Li}_2(z)=-\mathrm{Li}_2 \left( \dfrac{1}{z} \right) - \dfrac{1}{2} \left( \log(z) \right)^2 + \pi i \log(z) + \dfrac{\pi^2}{3}.$$
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> The following formula holds:
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$$\mathrm{Li}_2(x)+\mathrm{Li}_2(-x)=\dfrac{1}{2}\mathrm{Li}_2(x^2).$$
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> The following formula holds:
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$$\mathrm{Li}_2(1-x)+\mathrm{Li}_2 \left(1-\dfrac{1}{x} \right)=-\dfrac{1}{2}\left( \log x \right)^2.$$
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> The following formula holds:
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$$\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\dfrac{\pi^2}{6} - (\log x) \log(1-x).$$
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> The following formula holds:
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$$\mathrm{Li}_2(-x)-\mathrm{Li}_2(1-x)+\dfrac{1}{2}\mathrm{Li}_2(1-x^2)=-\dfrac{\pi^2}{12} - (\log x) \log(x+1).$$
 
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<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  

Revision as of 00:16, 20 May 2015

The dilogarithm is the function $$L_2(z)=\displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{k^2},$$ which is a special case of the polylogarithm.

Properties

Theorem: The following formula holds: $$\mathrm{Li}_2(z)=-\mathrm{Li}_2 \left( \dfrac{1}{z} \right) - \dfrac{1}{2} \left( \log(z) \right)^2 + \pi i \log(z) + \dfrac{\pi^2}{3}.$$

Proof:

Theorem: The following formula holds: $$\mathrm{Li}_2(x)+\mathrm{Li}_2(-x)=\dfrac{1}{2}\mathrm{Li}_2(x^2).$$

Proof:

Theorem: The following formula holds: $$\mathrm{Li}_2(1-x)+\mathrm{Li}_2 \left(1-\dfrac{1}{x} \right)=-\dfrac{1}{2}\left( \log x \right)^2.$$

Proof:

Theorem: The following formula holds: $$\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\dfrac{\pi^2}{6} - (\log x) \log(1-x).$$

Proof:

Theorem: The following formula holds: $$\mathrm{Li}_2(-x)-\mathrm{Li}_2(1-x)+\dfrac{1}{2}\mathrm{Li}_2(1-x^2)=-\dfrac{\pi^2}{12} - (\log x) \log(x+1).$$

Proof:

References

(page 31)