Difference between revisions of "Legendre chi"
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=Properties= | =Properties= | ||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Proposition:</strong> The following formula holds for real $x>0$: | ||
| + | $$\chi_2(x)+\chi_2 \left( \dfrac{1}{x} \right) = \dfrac{\pi^2}{4} - \dfrac{i\pi}{2}|\log x|.$$ | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> █ | ||
| + | </div> | ||
| + | </div> | ||
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| + | {{:Derivative of Legendre chi}} | ||
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{{:Legendre chi in terms of polylogarithm}} | {{:Legendre chi in terms of polylogarithm}} | ||
{{:Catalan's constant using Legendre chi}} | {{:Catalan's constant using Legendre chi}} | ||
Revision as of 01:32, 21 March 2015
The Legendre chi function is defined by $$\chi_{\nu}(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^{2k+1}}{(2k+1)^{\nu}}.$$
Contents
Properties
Proposition: The following formula holds for real $x>0$: $$\chi_2(x)+\chi_2 \left( \dfrac{1}{x} \right) = \dfrac{\pi^2}{4} - \dfrac{i\pi}{2}|\log x|.$$
Proof: █
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \chi_2(z) = \dfrac{\mathrm{arctanh}(z)}{z},$$ where $\chi$ denotes the Legendre chi function and $\mathrm{arctanh}$ denotes the inverse hyperbolic tangent function.
Proof
References
Theorem
The following formula holds: $$\chi_{\nu}(z)=\dfrac{1}{2}[\mathrm{Li}_{\nu}(z)-\mathrm{Li}_{\nu}(-z)] = \mathrm{Li}_{\nu}(z)-2^{-\nu}\mathrm{Li}_{\nu}(z^2),$$ where $\chi$ denotes the Legendre chi function and $\mathrm{Li}_{\nu}$ denotes the polylogarithm.
Proof
References
Theorem
The following formula holds: $$K=-i\chi_2(i),$$ where $K$ is Catalan's constant and $\chi$ denotes the Legendre chi function.
Proof
Recall, by definition, $$K=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^2},$$ and $$\chi_2(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^{2k+1}}{(2k+1)^{\nu}}.$$ Since $i^{2k+1}=i^{2k}i=(-1)^ki$, plugging in $i$ into $\chi_2$ yields $$\chi_2(i) = \displaystyle\sum_{k=0}^{\infty} \dfrac{i^{2k+1}}{(2k+1)^{\nu}}=i \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^{\nu}}=iK.$$ Hence, $$-i\chi_2(i)=-i(iK)=-i^2K=-(-1)K=K,$$ completing the proof. $\blacksquare$
