(1)

\begin{align*} \eta\left(s\right) & =-\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k^{s}}\\ & =-\frac{1}{2}-\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k}}{k^{s}}+\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^{s}}\right) \end{align*}
\[ \eta'\left(s\right)=\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k}}{k^{s}}\Log k+\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^{s}}\Log\left(k+1\right)\right) \]
より、
\begin{align*} \eta'\left(0\right) & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\left(-1\right)^{k}\Log k+\left(-1\right)^{k+1}\Log\left(k+1\right)\right)\\ & =-\frac{1}{2}\sum_{k=1}^{\infty}\left(\left(-1\right)^{k}\Log\frac{k+1}{k}\right)\\ & =-\frac{1}{2}\sum_{k=1}^{\infty}\left(\left(-1\right)^{2k-1}\Log\frac{\left(2k-1\right)+1}{\left(2k-1\right)}+\left(-1\right)^{2k}\Log\frac{2k+1}{2k}\right)\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\Log\frac{\left(2k\right)\left(2k\right)}{\left(2k-1\right)\left(2k+1\right)}\right)\\ & =\frac{1}{2}\Log\prod_{k=1}^{\infty}\frac{\left(2k\right)\left(2k\right)}{\left(2k-1\right)\left(2k+1\right)}\\ & =\frac{1}{2}\Log\frac{\pi}{2}\cmt{\text{ウォリス積}}\\ & =\Log\sqrt{\frac{\pi}{2}} \end{align*}

(2)

\begin{align*} \zeta'\left(0\right) & =\left[\frac{d}{ds}\zeta\left(s\right)\right]_{s=0}\\ & =\left[\frac{d}{ds}\frac{\eta\left(s\right)}{1-2^{1-s}}\right]_{s=0}\\ & =\left[\frac{\eta'\left(s\right)\left(1-2^{1-s}\right)-\eta\left(s\right)\left(2^{1-s}\Log2\right)}{\left(1-2^{1-s}\right)^{2}}\right]_{s=0}\\ & =-\left(\eta'\left(0\right)+2\eta\left(0\right)\Log2\right)\\ & =-\left(\Log\sqrt{\frac{\pi}{2}}+2\left(\frac{1}{2}\right)\Log2\right)\\ & =-\left(\Log\sqrt{\frac{\pi}{2}}+\Log2\right)\\ & =-\Log\sqrt{2\pi} \end{align*}

(3)

リーマン・ゼータ関数\(\zeta\left(s\right)\)を\(s=1\)周りでローラン展開すると、
\[ \zeta\left(s\right)=\frac{1}{s-1}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{k!}\gamma_{k}\left(s-1\right)^{k} \]
となるので、これより、
\begin{align*} \eta'\left(1\right) & =\left[\frac{d}{ds}\eta\left(s\right)\right]_{s\rightarrow1}\\ & =\left[\frac{d}{ds}\left(1-2^{1-s}\right)\zeta\left(s\right)\right]_{s\rightarrow1}\\ & =\left[2^{1-s}\log2\cdot\zeta\left(s\right)+\left(1-2^{1-s}\right)\zeta'\left(s\right)\right]_{s\rightarrow1}\\ & =\left[2^{1-s}\log2\left(\frac{1}{s-1}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{k!}\gamma_{k}\left(s-1\right)^{k}\right)+\left(1-2^{1-s}\right)\left(\frac{-1}{\left(s-1\right)^{2}}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{\left(k-1\right)!}\gamma_{k}\left(s-1\right)^{k-1}\right)\right]_{s\rightarrow1}\\ & =\left[2^{1-s}\log2\left(\frac{1}{s-1}+\gamma_{0}\right)+\left(1-2^{1-s}\right)\left(\frac{-1}{\left(s-1\right)^{2}}-\gamma_{1}\right)\right]_{s\rightarrow1}\\ & =\left[2^{1-s}\log2\left(\frac{1}{s-1}\right)-\frac{1-2^{1-s}}{\left(s-1\right)^{2}}\right]_{s\rightarrow1}+\gamma_{0}\log2\\ & =\left[\frac{\left(s-1\right)2^{1-s}\log2-\left(1-2^{1-s}\right)}{\left(s-1\right)^{2}}\right]_{s\rightarrow1}+\gamma_{0}\log2\\ & =\left[\frac{\left(2^{1-s}-\left(s-1\right)2^{1-s}\log2\right)\log2-2^{1-s}\log2}{2\left(s-1\right)}\right]_{s\rightarrow1}+\gamma_{0}\log2\\ & =\left[\frac{-\left(s-1\right)2^{1-s}\log2\cdot\log2}{2\left(s-1\right)}\right]_{s\rightarrow1}+\gamma_{0}\log2\\ & =\left[-2^{-s}\log^{2}2\right]_{s\rightarrow1}+\gamma_{0}\log2\\ & =-\frac{1}{2}\log^{2}2+\gamma_{0}\log2\\ & =-\frac{1}{2}\log^{2}2+\gamma\log2 \end{align*}
となる。
従って与式は成り立つ。