床関数を含む積分です
床関数を含む積分です
次の積分を求めよ。
\[ \int_{0}^{\frac{\pi}{2}}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx=? \]
次の積分を求めよ。
\[ \int_{0}^{\frac{\pi}{2}}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx=? \]
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\(\left\lfloor x\right\rfloor \)は床関数\begin{align*}
\int_{0}^{\frac{\pi}{2}}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx & =\int_{0}^{\tan^{\bullet}\infty}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx\\
& =\sum_{k=0}^{\infty}\int_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx\\
& =\sum_{k=0}^{\infty}\int_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\frac{k}{\tan x}dx\\
& =\sum_{k=1}^{\infty}k\int_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\frac{1}{\tan x}dx\\
& =\sum_{k=1}^{\infty}k\left[\log\left(\sin x\right)\right]_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\\
& =\sum_{k=1}^{\infty}k\left\{ \log\left(\sin\tan^{\bullet}\left(k+1\right)\right)-\log\left(\sin\tan^{\bullet}\left(k\right)\right)\right\} \\
& =\sum_{k=1}^{\infty}k\left\{ \log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-\log\frac{k}{\sqrt{1+k^{2}}}\right\} \cmt{\because\sin\tan^{\bullet}z=\frac{z}{\sqrt{1+z^{2}}}}\\
& =\sum_{k=1}^{\infty}\left\{ \left(k+1\right)\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-k\log\frac{k}{\sqrt{1+k^{2}}}\right\} \\
& =\sum_{k=1}^{\infty}\left\{ \left(k+1\right)\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-k\log\frac{k}{\sqrt{1+k^{2}}}\right\} -\sum_{k=1}^{\infty}\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}\\
& =-\log\frac{1}{\sqrt{2}}-\sum_{k=1}^{\infty}\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}\\
& =\frac{1}{2}\log2-\frac{1}{2}\sum_{k=1}^{\infty}\log\frac{\left(k+1\right)^{2}}{1+\left(k+1\right)^{2}}\\
& =\frac{1}{2}\log2+\frac{1}{2}\sum_{k=1}^{\infty}\log\frac{\left(k+1\right)^{2}+1}{\left(k+1\right)^{2}}\\
& =\frac{1}{2}\log2+\frac{1}{2}\sum_{k=1}^{\infty}\log\left(1+\frac{1}{\left(k+1\right)^{2}}\right)\\
& =\frac{1}{2}\log2+\frac{1}{2}\log\prod_{k=1}^{\infty}\left(1+\frac{1}{\left(k+1\right)^{2}}\right)\\
& =\frac{1}{2}\log2+\frac{1}{2}\log\prod_{k=2}^{\infty}\left(1+\frac{1}{k^{2}}\right)\\
\\
& =\frac{1}{2}\log2+\frac{1}{2}\log\left(\frac{1}{2}\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)\right)\\
& =\frac{1}{2}\log2-\frac{1}{2}\log2+\frac{1}{2}\log\left(\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)\right)\\
& =\frac{1}{2}\log\left(\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)\right)\\
& =\frac{1}{2}\log\frac{\sinh\pi}{\pi}\cmt{\because\pi z\prod_{k=1}^{\infty}\left(1+\frac{z}{k^{2}}\right)=\sinh\left(\pi z\right)}
\end{align*}
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