高専の数学解説

第5章 5.37

(1)

∫kπ(k+1)πe−xsin⁡x dx\int_{k\pi}^{(k+1)\pi} e^{-x}\sin x\,dx 5.36から

=12{[−e−xsin⁡x]kπ(k+1)π+[−e−xcos⁡x]kπ(k+1)π}=12(e−kπ+e−(k+1)π)=12e−kπ(1+e−π)=\frac{1}{2}\left\{\left[-e^{-x}\sin x\right]_{k\pi}^{(k+1)\pi}+\left[-e^{-x}\cos x\right]_{k\pi}^{(k+1)\pi}\right\}=\frac{1}{2}\left(e^{-k\pi}+e^{-(k+1)\pi}\right)=\frac{1}{2}e^{-k\pi}(1+e^{-\pi})

(2)

∑k=0nSk=12(1+e−π)∑k=0ne−kπ=12(1+e−π){1+e−π+e−2π+⋯+e−nπ}=12(1+e−π)1−e−(n+1)π1−e−π\sum_{k=0}^{n}S_k=\frac{1}{2}(1+e^{-\pi})\sum_{k=0}^{n}e^{-k\pi}=\frac{1}{2}(1+e^{-\pi})\{1+e^{-\pi}+e^{-2\pi}+\cdots+e^{-n\pi}\}=\frac{1}{2}(1+e^{-\pi})\frac{1-e^{-(n+1)\pi}}{1-e^{-\pi}}

(3)

lim⁡n→∞∑k=0nSk=lim⁡n→∞12(1+e−π)1−e−(n+1)π1−e−π=12(1+e−π)11−e−π=1+e−π2(1−e−π)\lim_{n\to\infty}\sum_{k=0}^{n}S_k=\lim_{n\to\infty}\frac{1}{2}(1+e^{-\pi})\frac{1-e^{-(n+1)\pi}}{1-e^{-\pi}}=\frac{1}{2}(1+e^{-\pi})\frac{1}{1-e^{-\pi}}=\frac{1+e^{-\pi}}{2(1-e^{-\pi})}

解説: r31bn1z