気に入っている積分1
気に入っている積分
$$\int_0^1\frac{{\rm{Li}}_2(\frac{1+x}{2})-\frac{\pi^2}{12}+\frac{1}{2}(\ln2)^2}{x} dx=\frac{13}{8}\zeta(3)-\frac{\pi^2}{6}\ln2$$
昔、何かの級数の値を求めようとしたら出てきた積分です。
今回はこの等式を示します。
$$\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\zeta(2)-\ln(z)\ln(1-z)$$
\begin{align} \mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)&=-\int_{0}^{z}\frac{\ln(1-t)}{t}dt-\int_{0}^{1-z}\frac{\ln(1-t)}{t}dt\\ &=-\int_{0}^{1}\frac{\ln(1-t)}{t}dt-\int_{1}^{z}\frac{\ln(1-t)}{t}dt-\int_{0}^{1-z}\frac{\ln(1-t)}{t}dt\\ &=\zeta(2)-\int_{1}^{z}\frac{\ln(1-t)}{t}dt+\int_{1}^{z}\frac{\ln(u)}{1-u}du \quad (t=1-u)\\ &=\zeta(2)-\int_{1}^{z}\frac{\ln(1-t)}{t}dt- \Big\lbrack \ln(u)\ln(1-u) \Big\rbrack_{1}^{z}+\int_{1}^{z}\frac{\ln(1-u)}{u}du\\ &=\zeta(2)-\ln(z)\ln(1-z)\blacksquare \end{align}
$$\mathrm{Li}_2\Big(\frac{1}{2}\Big)=\frac{\pi^2}{12}-\frac{1}{2}(\ln2)^2$$
命題1で$z=\frac{1}{2}$を代入すればよい$\blacksquare$
$$\zeta(\bar{2})=-\frac{\pi^2}{12}$$
$$\zeta(\bar{1},\bar{2})=\frac{\pi^2}{4}\ln2-\frac{13}{8}\zeta(3)$$
証明はこちらの記事を参照してください weight 3以下の交代多重ゼータ値を全て求める
証明
$$\int_0^1\frac{{\rm{Li}}_2(\frac{1+x}{2})-\frac{\pi^2}{12}+\frac{1}{2}(\ln2)^2}{x} dx=\frac{13}{8}\zeta(3)-\frac{\pi^2}{6}\ln2$$
\begin{align} \int_0^1\frac{{\rm{Li}}_2(\frac{1+x}{2})-\frac{\pi^2}{12}+\frac{1}{2}(\ln2)^2}{x} dx&=\int_0^1\frac{{\rm{Li}}_2(\frac{1+x}{2})-\mathrm{Li}_2(\frac{1}{2})}{x} dx \quad (\because \text{補題2})\\ &=-\int_0^1\frac{dx}{x}\int_{\frac{1}{2}}^{\frac{1+x}{2}}\frac{\ln(1-t)}{t}dt\\ &=-\int_0^1\frac{dx}{x}\int_{0}^{x}\frac{\ln(\frac{1-u}{2})}{1+u}du \quad \Big(t=\frac{1+u}{2}\Big)\\ &=-\int_0^1\frac{dx}{x}\int_{0}^{x}\frac{\ln({1-u})}{1+u}du+\ln2\int_0^1\frac{dx}{x}\int_{0}^{x}\frac{du}{1+u}\\ &=\int_0^1\frac{dx}{x}\int_{0}^{x}\frac{du}{1+u}\int_{0}^{u}\frac{ds}{1-s}+\ln2\int_0^1\frac{dx}{x}\int_{0}^{x}\frac{du}{1+u}\\ &=\int_{0\lt s \lt u \lt x\lt1}\frac{dx}{x}\frac{du}{1+u}\frac{ds}{1-s}+\ln2\int_{0 \lt u \lt x\lt1}\frac{dx}{x}\frac{du}{1+u}\\ &=-\zeta(\bar{1},\bar{2})-\zeta(\bar{2})\ln2\\ &=\frac{13}{8}\zeta(3)-\frac{\pi^2}{6}\ln2\quad (\because \text{補題3})\blacksquare \end{align}