In mathematics, the Legendre chi function (named after Adrien-Marie Legendre) is a special function whose Taylor series is also a Dirichlet series, given by
<math display="block">\chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1{(2k+1)^\nu}.</math>
thumb|Legendre chi function
As such, it resembles the Dirichlet series for the polylogarithm, and, indeed, is trivially expressible as the odd part of the polylogarithm
<math display="block">\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right].</math>
The Legendre chi function appears as the discrete Fourier transform, with respect to the order ν, of the Hurwitz zeta function, and also of the Euler polynomials, with the explicit relationships given in those articles.
The Legendre chi function is a special case of the Lerch transcendent, and is given by
<math display="block">\chi_\nu(z)=2^{-\nu}z\,\Phi (z^2,\nu,1/2).</math>
Identities
<math display="block">\chi_{-1}\left(x\right)=\frac{x\left(1+x^{2}\right)}{\left(1-x^{2}\right)^{2 </math>
<math display="block">\chi_{0}\left(x\right)=\frac{x}{1-x^{2 </math>
<math display="block">\chi_1(x) =\operatorname{arctanh}(x) =\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) </math>
<math display="block">\chi_2(x) = \int_{0}^{x}\frac{\operatorname{arctanh}(t)}t \mathrm dt</math>
<math display="block">\chi_\infty(x) = x</math>
<math display="block">\frac{\mathrm d}{\mathrm dx}\chi_\nu(x) = \frac{\chi_{\nu-1}(x)}{x}</math>
<math display="block">\frac{\mathrm d}{\mathrm dx}\chi_2(x) = \frac{\operatorname{arctanh}(x)}{x}</math>
<math display="block">\chi_2(x) + \chi_2(1/x)= \frac{\pi^2}{4}-\frac{i \pi}{2}\ln |x| </math>
<math display="block">\chi_2(x)+\chi_2\left(\frac{1-x}{1+x}\right)=\frac{\pi^2}{8}+\ln(x)\operatorname{arctanh}(x), \quad x\in(0,1)</math>
Special Values
It takes the special values:
<math display="block">\chi_2(1) = \frac{\pi^2}{8}</math>
<math display="block">\chi_2(-1) = -\frac{\pi^2}{8}</math>
<math display="block">\chi_2(\sqrt{2}-1) = \frac{\pi^2}{16}-\frac14\ln^2(\sqrt{2}+1)</math>
<math display="block">\chi_2\left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2\left(\frac{\sqrt{5}+1}{2}\right)</math>
<math display="block">\chi_2(\sqrt{5}-2) = \frac{\pi^2}{24}-\frac{3}{4}\ln^2\left(\frac{\sqrt{5}+1}{2}\right)</math>
<math display="block">\chi_2(i) = iK,</math>
where <math display>i</math> is the imaginary unit and K is Catalan's constant.
Integral relations
<math display="block">\int_0^{\pi/2} \arcsin (r \sin \theta)\, \mathrm d\theta
= \chi_2(r),
\qquad\int_0^{\pi/2} \arccos (r \cos \theta)\, \mathrm d\theta
= \left(\frac{\pi}{2}\right)^2-\chi_2(r)\qquad {\rm if}~~|r|\leq 1</math>
<math display="block">\int_0^{\pi/2} \arctan (r \sin \theta) \,\mathrm d\theta
= -\frac{1}{2}\int_0^{\pi} \frac{ r \theta \cos \theta}{1+ r^2 \sin^2 \theta} \,\mathrm d\theta
= 2 \chi_2\!\!\left(\frac{\sqrt{1+r^2}- 1}{r}\right)</math>
<math display="block">\int_0^{\pi/2} \arctan (p \sin \theta)\arctan (q \sin \theta) \,\mathrm d\theta = \pi \chi_2\!\!\left(\frac{\sqrt{1+p^2}- 1}{p}\cdot\frac{\sqrt{1+q^2}- 1}{q}\right)</math>
<math display="block">\int_0^{\alpha}\int_0^{\beta} \frac{\mathrm dx\, \mathrm dy}{1-x^2 y^2} = \chi_2(\alpha\beta)\qquad {\rm if}~~|\alpha\beta|\leq 1</math>