Oncelikle PoliLogaritma fonksiyonunu integral formunda tanimlayalim (genel hali kompleks sayilar uzerinde tanimli ama bizim icin reel sayilarda tanimli olmasi yeterli).
$$\text{Li}_{1}(x)=-\ln(1-x)$$ olmak uzere
$$\text{Li}_{s+1}(x)=\int_{0}^{x}\frac{\text{Li}_s(t)}{t}dt,\qquad x\in\mathbb{R},\quad s=1,2,...$$
Ozel hali olan DiLogaritma fonksiyonu $s=1$ icin
$$\text{Li}_{2}(x)=\int_{0}^{x}\frac{-\ln(1-t)}{t}dt\qquad (1)$$
$u=1-t$ donusumu yaparsak,
$$\text{Li}_{2}(1-x)=\int_{1}^{x}\frac{\ln(u)}{1-u}du=\int_{0}^{x}\frac{\ln(u)}{1-u}du-\int_{0}^{1}\frac{\ln(u)}{1-u}du\qquad (2)$$
$u=t$ donusumu yaparsak $$\text{Li}_{2}(1-x)=\int_{0}^{x}\frac{\ln(t)}{1-t}dt-\int_{0}^{1}\frac{\ln(t)}{1-t}dt\qquad (3)$$
(1) ve (3) esitliklerini taraf trafa toplarsak
$$\text{Li}_{2}(x)+\text{Li}_{2}(1-x)=\int_{0}^{x}\frac{-\ln(1-t)}{t}+\frac{\ln(t)}{1-t}dt-\int_{0}^{1}\frac{\ln(t)}{1-t}dt$$
$$\text{Li}_{2}(x)+\text{Li}_{2}(1-x)=-\ln(x)\ln(1-x)-\int_{0}^{1}\frac{\ln(t)}{1-t}dt\qquad (4)$$
(4) esitliginde $x=1/2$ koyalim. (sondaki integral onceki soruda gosterilmisti).
$$\text{Li}_{2}(1/2)+\text{Li}_{2}(1/2)=-\ln(1/2)\ln(1/2)+\frac{\pi^2}{6}$$
$$2\text{Li}_{2}(1/2)=-\ln(2)^2+\frac{\pi^2}{6}$$
$$\text{Li}_{2}(1/2)=\int_{0}^{1/2}\frac{-\ln(1-t)}{t}dt=\frac{\pi^2}{12}-\frac12\ln(2)^2$$
Kaynakca:
1) https://en.wikipedia.org/wiki/Polylogarithm
2) Abel'in makalesi
http://www.abelprisen.no/nedlastning/verker/oeuvres_1881_del2/oeuvres_completes_de_abel_nouv_ed_2_kap14_opt.pdf
