多重ゼータ値とその類似　Part６

　
${\rm\bf{Proposition~6.1.}}$

$ \D[6.1]　　　\sum_{0< m\le n}\frac{C_n}{C_m m^2 n^2}=\pi^2\ln^2 2-\sum_{0< m\le n}\frac{1}{mn^3C_n} $

$\Pr$
$\BA \D　　　\sum_{0< m\le n}\frac{C_n}{C_m m^2 n^2} &=\sum_{0< n}\frac{C_n}{n^2}\L(-\frac{\pi^2}{2}+\frac{4}{C_n}\int_0^\frac{\pi}{2}t\sin^{2n}t\,dt \R)\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2}+\sum_{0< n}\frac{4}{n^2}\int_0^\frac{\pi}{2}t\sin^{2n}t\,dt\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2} +\sum_{0< n}\frac{4}{n^2}\L(\L[\frac{t^2}{2}\sin^{2n}t \R]_0^\frac{\pi}{2}-n\int_0^\frac{\pi}{2}t^2\sin^{2n-1}t\cos t\,dt \R)\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2}+\sum_{0< n}\frac{4}{n^2}\frac{\L(\frac{\pi}{2}\R)^2}{2} -\sum_{0< n}\frac{4}{n}\int_0^\frac{\pi}{2}t^2\sin^{2n-1}t\cos t\,dt\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2}+\frac{\pi^4}{12} -\sum_{0< n}\frac{4}{n}\int_0^1 t^{2n-1}(\sin^{-1}t)^2dt\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2}+\frac{\pi^4}{12} -\sum_{0< n}\frac{2}{n}\int_0^1 t^{2n-1}\sum_{0< m}\frac{t^{2m}}{m^2 C_m}\,dt\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2}+\frac{\pi^4}{12} -\sum_{0< m,n}\frac{2}{nm^2 C_m}\int_0^1 t^{2m+2n-1}\,dt\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_n}{n^2}+\frac{\pi^4}{12} -\sum_{0< m,n}\frac{1}{nm^2(m+n) C_m}\\ &=-\frac{\pi^2}{2}\L(\frac{\pi^2}{6}-2\ln^2 2 \R)+\frac{\pi^4}{12} -\sum_{0< m,n}\frac{1}{m^3 C_m}\L(\frac{1}{n}-\frac{1}{m+n} \R) \\ &=\pi^2\ln^2 2-\sum_{0< m\le n}\frac{1}{mn^3C_n} \EA$
$\qed$

　
$\bf Proposition~6.2.$

$ \D[6.2]　　　\sum_{0< m\le n}\frac{C_n x^{n}}{m^2 n^2 C_m} =\frac{\pi^2}{2}{\rm Li}_2(x)-\frac{\pi^2}{2}\sum_{0< n}\frac{C_nx^n}{n^2} -\sum_{0< m,n}\frac{x^n}{m^2 n(m+n)C_m} $

$\Pr$
$\BA \D　　　\sum_{0< m\le n}\frac{C_n x^{n}}{m^2 n^2 C_m} &=\sum_{0< n}\frac{C_n x^n}{n^2}\L(-\frac{\pi^2}{2}+\frac{4}{C_n}\int_0^\frac{\pi}{2}t\sin^{2n}t\,dt \R)\\ &=-\frac{\pi^2}{2}\sum_{0< n}\frac{C_nx^n}{n^2} +\sum_{0< n}\frac{4x^n}{n^2}\L(\L[\frac{t^2}{2}\sin^{2n}t \R]_0^\frac{\pi}{2}-n\int_0^\frac{\pi}{2}t^2\sin^{2n-1}t\cos t\,dt \R)\\ &=\frac{\pi^2}{2}{\rm Li}_2(x)-\frac{\pi^2}{2}\sum_{0< n}\frac{C_nx^n}{n^2} -\sum_{0< n}\frac{4x^n}{n}\int_0^1t^{2n-1}(\sin^{-1}t)^2dt\\ &=\frac{\pi^2}{2}{\rm Li}_2(x)-\frac{\pi^2}{2}\sum_{0< n}\frac{C_nx^n}{n^2} -\sum_{0< n}\frac{2x^n}{n}\int_0^1t^{2n-1}\sum_{0< m}\frac{t^{2m}}{m^2C_m} dt\\ &=\frac{\pi^2}{2}{\rm Li}_2(x)-\frac{\pi^2}{2}\sum_{0< n}\frac{C_nx^n}{n^2} -\sum_{0< n}\frac{2x^n}{m^2 nC_m}\int_0^1t^{2m+2n-1} dt\\ &=\frac{\pi^2}{2}{\rm Li}_2(x)-\frac{\pi^2}{2}\sum_{0< n}\frac{C_nx^n}{n^2} -\sum_{0< m,n}\frac{x^n}{m^2 n(m+n)C_m} \EA$
$\qed$

$\bf Proposition~6.3.$

$ \D[6.3]　　　\pi\sum_{0\le n} C_n^2 x^n =\sqrt{1-x}\sum_{0\le m< n}\frac{C_m^2}{nC_n}x^{n-m-1} $

$\Pr$
$\BA \D　　　\pi\sum_{0\le n}C_n^2x^n &=2\int_0^\frac{\pi}{2}\frac{d\theta}{\sqrt{1-x\sin^2 \theta}}\\ &=2\int_0^1 \frac{dt}{\sqrt{(1-xt^2)(1-t^2)}}\\ &=\int_0^1 \frac{dt}{\sqrt{(1-xt)(1-t)t}}\\ &=\frac{1}{\sqrt{x}}\int_0^x \frac{dt}{\sqrt{(1-t)\L(1-\frac{t}{x}\R)t}}\\ &=\frac{1}{\sqrt{x}}\sum_{0\le m}\frac{C_m}{x^m}\int_0^1 \frac{t^{m-\frac{1}{2}}}{\sqrt{1-t}}\,dt\\ &=\frac{1}{\sqrt{x}}\sum_{0\le m}\frac{C_m}{x^m}\,C_m\sum_{m< n}\frac{x^{n-\frac{1}{2}}\sqrt{1-x}}{nC_n}\\ &=\sqrt{1-x}\sum_{0\le m< n}\frac{C_m^2}{nC_n}x^{n-m-1} \EA$
$\qed$

$\bf Lemma~6.1.$

$ \D[6.4]　　　\int_0^\infty \frac{x}{\cosh^{2n}x}\,dx=\frac{\ln2}{2nC_n}+\frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1)^kH_k}{2k+1}\binom{n-1}{k} $

$\Pr$
$\BA \D　　　\int_0^z \frac{x}{\cosh^{2n}x}\,dx &=\L[x\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\binom{n-1}{k}\tanh^{2k+1}x \R]_0^z -\int_0^z\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\binom{n-1}{k}\tanh^{2k+1}x\,dx\\ &=z\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\binom{n-1}{k}\tanh^{2k+1}z -\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\binom{n-1}{k}\L[\ln\cosh x-\sum_{l=1}^{k}\frac{\tanh^{2l}x}{2l} \R]_0^z\\ &=\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\binom{n-1}{k}\L(z\tanh^{2k+1}z-\ln\cosh z+\sum_{l=1}^k\frac{\tanh^{2l}z}{2l} \R) \EA$
いま
$ \D　　　\lim_{z\to\infty}(z\tanh^{2k+1}z-\ln\cosh z)=\ln2 $
ので
$ \D　　　\int_0^\infty \frac{x}{\cosh^{2n}x}\,dx=\frac{\ln2}{2nC_n}+\frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1)^kH_k}{2k+1}\binom{n-1}{k} $
$\qed$

$\bf Proposition~6.4.$　　$\D -\frac{\pi}{2}\le x\le \frac{\pi}{2}$に対して

$ \D[6.5]　　　2x^2\ln2+\sum_{0< k\le m< n}\frac{(-1)^{m}\sin^{2n}x}{k(2m+1)n}\binom{n-1}{m} =\sum_{0< n}\frac{(-1)^{n-1}}{n^3}(1-\cos 2nx) $

$\Pr$
$\BA \D　　　\int_0^\infty t\ln\frac{1}{1-\frac{\sin^2 x}{\cosh^2 t}}\,dt &=\sum_{0< n}\frac{\sin^{2n}x}{n}\int_0^\infty \frac{t}{\cosh^{2n}t}\,dt\\ &=\sum_{0< n}\frac{\sin^{2n}x}{n}\L(\frac{\ln2}{2nC_n}+\frac{1}{2}\sum_{0< k\le m< n}\frac{(-1)^{m-1}}{k(2m+1)}\binom{n-1}{m} \R)\\ &=x^2\ln2+\frac{1}{2}\sum_{0< k\le m< n}\frac{(-1)^{m}\sin^{2n}x}{k(2m+1)n}\binom{n-1}{m} \EA$

$\BA \D　　　\int_0^\infty t\ln\frac{1}{1-\frac{\sin^2 x}{\cosh^2 t}}\,dt &=\int_0^\infty t\ln\frac{(1+e^{-2t})^2}{1+2e^{-2t}\sin2x+e^{-4t}}\,dt\\ &=\int_0^\infty t\cdot 2\sum_{0< n}\frac{(-1)^{n-1}e^{-2nt}}{n}(1-\cos2nx)\,dt\\ &=\sum_{0< n}\frac{(-1)^{n-1}}{2n^3}(1-\cos 2nx) \EA$
$\qed$