解説8
@integralsbot さんがツイートした こちらの定理 の解説です．
以下の等式が成り立ちます．ただし$s\in\mathbb{C}$とし，$\mathfrak{R}s>-1$を満たすとします．
$\displaystyle\int_0^1\left\lfloor\frac{1}{x}\right\rfloor^{-1}\left(\left(1+s\right)x^s-sx^{-1+s}\right)\mathrm{d}x=\zeta\left(2+s\right)-1$
ツイート内容では$\mathfrak{R}s>0$となっていますが，より広い範囲$\mathfrak{R}s>-1$で成り立ちます．恐らく
$\lim_{\varepsilon\rightarrow+0}\frac{\varepsilon^s}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor} =\lim_{\varepsilon\rightarrow+0}\frac{\frac{1}{\varepsilon}}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\varepsilon^{1+s}=1\cdot0=0$
と出来るところを
$\lim_{\varepsilon\rightarrow+0}\frac{\varepsilon^s}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}=\lim_{\varepsilon\rightarrow+0}\frac{1}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\varepsilon^s=0\cdot0=0$
としたことに起因すると思われます．
解説
\begin{align*} &\int_0^1\left\lfloor\frac{1}{x}\right\rfloor^{-1}\left(\left(1+s\right)x^s-sx^{-1+s}\right)\mathrm{d}x\\ =&\lim_{\varepsilon\rightarrow+0}\int_\varepsilon^1\left\lfloor\frac{1}{x}\right\rfloor^{-1}\left(\left(1+s\right)x^s-sx^{-1+s}\right)\mathrm{d}x\\ =&\lim_{\varepsilon\rightarrow+0}\int_\frac{1}{\varepsilon}^1\left\lfloor t\right\rfloor^{-1}\left(\left(1+s\right)\left(\frac{1}{t}\right)^s-s\left(\frac{1}{t}\right)^{-1+s}\right)\left(-\frac{\mathrm{d}t}{t^2}\right)\\ =&\lim_{\varepsilon\rightarrow+0}\left(\sum_{k=1}^{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\int_k^{k+1}\frac{1}{k}\left(\left(1+s\right)t^{-2-s}-st^{-1-s}\right)\mathrm{d}t-\int_\frac{1}{\varepsilon}^{\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1}\frac{1}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\left(\left(1+s\right)t^{-2-s}-st^{-1-s}\right)\mathrm{d}t\right)\\ =&\lim_{\varepsilon\rightarrow+0}\left(\sum_{k=1}^{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\frac{1}{k}\left[-t^{-1-s}+t^{-s}\right]_{t=k}^{k+1}-\frac{1}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\left[-t^{-1-s}+t^{-s}\right]_{t=\frac{1}{\varepsilon}}^{\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1}\right)\\ =&\lim_{\varepsilon\rightarrow+0}\left(\sum_{k=1}^{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\left(\frac{1}{k^{2+s}}+\frac{1}{\left(k+1\right)^{1+s}}-\frac{1}{k^{1+s}}\right)-\frac{1}{\left(\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1\right)^{1+s}}-\frac{\varepsilon^{1+s}}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}+\frac{\varepsilon^s}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\right)\\ =&\lim_{\varepsilon\rightarrow+0}\left(H_{\left\lfloor\frac{1}{\varepsilon}\right\rfloor,2+s}+\frac{1}{\left(\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1\right)^{1+s}}-1-\frac{1}{\left(\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1\right)^{1+s}}-\frac{\varepsilon^{1+s}}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}+\frac{\varepsilon^s}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\right)\\ =&\lim_{\varepsilon\rightarrow+0}\left(H_{\left\lfloor\frac{1}{\varepsilon}\right\rfloor,2+s}-1-\frac{\varepsilon^{1+s}}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}+\frac{\varepsilon^s}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\right)\\ =&\lim_{\varepsilon\rightarrow+0}\left(H_{\left\lfloor\frac{1}{\varepsilon}\right\rfloor,2+s}-1-\frac{1}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\varepsilon^{1+s}+\frac{\frac{1}{\varepsilon}}{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}\varepsilon^{1+s}\right)\\ =&\zeta\left(2+s\right)-1-0\cdot0+1\cdot0\\ =&\zeta\left(2+s\right)-1 \end{align*}
なので，$\displaystyle\int_0^1\left\lfloor\frac{1}{x}\right\rfloor^{-1}\left(\left(1+s\right)x^s-sx^{-1+s}\right)\mathrm{d}x=\zeta\left(2+s\right)-1$です．$\blacksquare$