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大学数学基礎解説
解説4
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@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$\alpha\in\left]0,1\right[$とします．
$\displaystyle\int_0^1\left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm{d}x=\alpha\ln\alpha$
まず補題を証明します．
以下の等式が成り立ちます．ただし$r\in\R_{>0}$とします．
$\displaystyle\int_1^r\frac{\left\lfloor t\right\rfloor}{t^2}\mathrm{d}t=H_\left\lfloor r\right\rfloor-\frac{\left\lfloor r\right\rfloor}{r}$
解説
\begin{align*} &\int_1^r\frac{\left\lfloor t\right\rfloor}{t^2}\mathrm{d}t\\ =&\sum_{k=2}^{\left\lfloor r\right\rfloor+2}\int_{k-1}^k\frac{\left\lfloor t\right\rfloor}{t^2}dt-\int_r^{\left\lfloor r\right\rfloor+1}\frac{\left\lfloor t\right\rfloor}{t^2}dt-\int_{\left\lfloor r\right\rfloor+1}^{\left\lfloor r\right\rfloor+2}\frac{\left\lfloor t\right\rfloor}{t^2}dt\\ =&\sum_{k=2}^{\left\lfloor r\right\rfloor+2}\int_{k-1}^k\frac{k-1}{t^2}dt-\int_r^{\left\lfloor r\right\rfloor+1}\frac{\left\lfloor r\right\rfloor}{t^2}dt-\int_{\left\lfloor r\right\rfloor+1}^{\left\lfloor r\right\rfloor+2}\frac{\left\lfloor r\right\rfloor+1}{t^2}dt\\ =&\sum_{k=2}^{\left\lfloor r\right\rfloor+2}\left[-\frac{k-1}{t}\right]_{t=k-1}^k-\left[-\frac{\left\lfloor r\right\rfloor}{t}\right]_{t=r}^{\left\lfloor r\right\rfloor+1}-\left[-\frac{\left\lfloor r\right\rfloor+1}{t}\right]_{t=\left\lfloor r\right\rfloor+1}^{\left\lfloor r\right\rfloor+2}\\ =&\sum_{k=2}^{\left\lfloor r\right\rfloor+2}\left(-\frac{k-1}{k-1}+\frac{k-1}{k}\right)-\left(-\frac{\left\lfloor r\right\rfloor}{\left\lfloor r\right\rfloor+1}+\frac{\left\lfloor r\right\rfloor}{r}\right)-\left(-\frac{\left\lfloor r\right\rfloor+1}{\left\lfloor r\right\rfloor+2}+\frac{\left\lfloor r\right\rfloor+1}{\left\lfloor r\right\rfloor+1}\right)\\ =&\sum_{k=2}^{\left\lfloor r\right\rfloor+2}\frac{1}{k}+1-\frac{1}{\left\lfloor r\right\rfloor+1}-\frac{\left\lfloor r\right\rfloor}{r}-\frac{1}{\left\lfloor r\right\rfloor+2}\\ =&H_{\left\lfloor r\right\rfloor+2}-\frac{1}{\left\lfloor r\right\rfloor+1}-\frac{1}{\left\lfloor r\right\rfloor+2}-\frac{\left\lfloor r\right\rfloor}{r}\\ =&H_\left\lfloor r\right\rfloor-\frac{\left\lfloor r\right\rfloor}{r} \end{align*}
なので，$\displaystyle\int_1^r\frac{\left\lfloor t\right\rfloor}{t^2}\mathrm{d}t=H_\left\lfloor r\right\rfloor-\frac{\left\lfloor r\right\rfloor}{r}$です．$\blacksquare$
では，定理の証明に移ります．
解説
\begin{align*} &\int_0^1\left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm{d}x\\ =&\lim_{\varepsilon\rightarrow+0}\int_\varepsilon^1\left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm{d}x\\ =&\lim_{\varepsilon\rightarrow+0}\left(\int_\frac{\alpha}{\varepsilon}^\alpha\left\lfloor t\right\rfloor\left(-\frac{\alpha\mathrm{d}t}{t^2}\right)-\alpha\int_\frac{1}{\varepsilon}^1\left\lfloor t\right\rfloor\left(-\frac{\mathrm{d}t}{t^2}\right)\right)\\ =&\lim_{\varepsilon\rightarrow+0}\alpha\left(-\int_1^\alpha\frac{\left\lfloor t\right\rfloor}{t^2}\mathrm{d}t+\int_1^\frac{\alpha}{\varepsilon}\frac{\left\lfloor t\right\rfloor}{t^2}\mathrm{d}t-\int_1^\frac{1}{\varepsilon}\frac{\left\lfloor t\right\rfloor}{t^2}\mathrm{d}t\right)\\ =&\lim_{\varepsilon\rightarrow+0}\alpha\left(-\left(H_\left\lfloor\alpha\right\rfloor-\frac{\left\lfloor\alpha\right\rfloor}{\alpha}\right)+\left(H_\left\lfloor\frac{\alpha}{\varepsilon}\right\rfloor-\frac{\left\lfloor\frac{\alpha}{\varepsilon}\right\rfloor}{\frac{\alpha}{\varepsilon}}\right)-\left(H_\left\lfloor\frac{1}{\varepsilon}\right\rfloor-\frac{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}{\frac{1}{\varepsilon}}\right)\right)\\ =&\lim_{\varepsilon\rightarrow+0}\alpha\left(-H_\left\lfloor\alpha\right\rfloor+\frac{\left\lfloor\alpha\right\rfloor}{\alpha}+\left(H_\left\lfloor\frac{\alpha}{\varepsilon}\right\rfloor-\ln\left\lfloor\frac{\alpha}{\varepsilon}\right\rfloor\right)+\ln\frac{\left\lfloor\frac{\alpha}{\varepsilon}\right\rfloor}{\frac{\alpha}{\varepsilon}}-\frac{\left\lfloor\frac{\alpha}{\varepsilon}\right\rfloor}{\frac{\alpha}{\varepsilon}}-\left(H_\left\lfloor\frac{1}{\varepsilon}\right\rfloor-\ln\left\lfloor\frac{1}{\varepsilon}\right\rfloor\right)-\ln\frac{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}{\frac{1}{\varepsilon}}+\frac{\left\lfloor\frac{1}{\varepsilon}\right\rfloor}{\frac{1}{\varepsilon}}+\ln\alpha\right)\\ =&\alpha\left(-H_\left\lfloor\alpha\right\rfloor+\frac{\left\lfloor\alpha\right\rfloor}{\alpha}+\gamma+\ln1-1-\gamma-\ln1+1+\ln\alpha\right)\\ =&-\alpha H_\left\lfloor\alpha\right\rfloor+\left\lfloor\alpha\right\rfloor+\alpha\ln\alpha\\ =&-\alpha H_0+0+\alpha\ln\alpha\\ =&\alpha\ln\alpha \end{align*}
なので，$\displaystyle\int_0^1\left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm{d}x=\alpha\ln\alpha$です．$\blacksquare$
投稿日：2021年2月6日
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