Legendre's Chi- Function

Question

I want to get the numerical value(twenty at thirty decimals) of $$\operatorname\chi_{2}(\frac{1}{\sqrt{2}})$$ Thanks you very much.

Answer 1

The Legendre chi function is defined as the following

$$\chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^\nu}.$$

By using the Dirichlet series of the polylogarithm we can deduce the following identity

$$\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right].$$

Because you need a numerical value of the chi function at order $2$, the identity above turn into a dilogarithm identity.

$$\chi_2(z) = \frac{1}{2}\left[\operatorname{Li}_2(z) - \operatorname{Li}_2(-z)\right].$$

At $z=\sqrt{2}/2$ we could reduce it into the form

$$\chi_2\left(\frac{\sqrt{2}}{2}\right) = \chi_2\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2}\left[\operatorname{Li}_2\left(\frac{\sqrt{2}}{2}\right) - \operatorname{Li}_2\left(-\frac{\sqrt{2}}{2}\right)\right] = \operatorname{Li}_2\left(\frac{\sqrt{2}}{2}\right)-\frac{\pi^2}{48}+\frac{\ln^2 2}{8}.$$

The first $500$ digits of $\chi_2\left(\frac{\sqrt{2}}{2}\right)$ are the following.

$0.75609964689557998357406116157876985057285511798172214985738130157864573099\\ 66741725489537316251012365900537093263317726910769815441504106893716207216441\\ 09451098075219623435380238129498706729232480051698906278774001597288338615606\\ 09852481717932468084405924871137380393707224388991221330477236231438246612477\\ 77147546089451586432184926119404048907999162303125077730144258390374720834726\\ 52427894740049923850286746477964962527589567518516877950301895647704933931095\\ 84164619205084063565822266178875549517875$

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