【微分積分】5-3-2 項別微分・項別積分｜問題集
1.自然数\(n\)に対して\(f_n(x)=e^{-nx}-2e^{-2nx}\)とおくとき、次の値を求めなさい。
(1)\(\displaystyle \sum_{n=1}^{\infty}\int_0^{\infty}f_n(x)dx\)
\(\displaystyle =\sum_{n=1}^{\infty}\int_0^{\infty}(e^{-nx}-2e^{-2nx})dx\)
\(\displaystyle =\sum_{n=1}^{\infty}\left[-\frac{1}{n}e^{-nx}+\frac{1}{n}e^{-2nx}\right]_0^{\infty}\)
\(\displaystyle =\sum_{n=1}^{\infty}\lim_{t\to\infty}\left(-\frac{1}{n}e^{-nt}+\frac{1}{n}e^{-2nt}\right)\)
\(\displaystyle =\sum_{n=1}^{\infty}0\)
\(\displaystyle =0\)
\(\displaystyle =\sum_{n=1}^{\infty}\left[-\frac{1}{n}e^{-nx}+\frac{1}{n}e^{-2nx}\right]_0^{\infty}\)
\(\displaystyle =\sum_{n=1}^{\infty}\lim_{t\to\infty}\left(-\frac{1}{n}e^{-nt}+\frac{1}{n}e^{-2nt}\right)\)
\(\displaystyle =\sum_{n=1}^{\infty}0\)
\(\displaystyle =0\)
(2)\(\displaystyle \int_0^{\infty}\sum_{n=1}^{\infty}f_n(x)dx\)
\(\displaystyle =\int_0^{\infty}\sum_{n=1}^{\infty}\{(e^{-x})^n-2(e^{-2x})^n\}dx\)
\(\displaystyle =\int_0^{\infty}\left(\frac{e^{-x}}{1-e^{-x}}-\frac{2e^{-2x}}{1-e^{-2x}}\right)dx\)
\(\displaystyle =\int_0^{\infty}\left(\frac{1}{e^x-1}-\frac{2}{e^{2x}-1}\right)dx\)
\(\displaystyle =\int_0^{\infty}\frac{1}{e^x+1}dx\)
\(\displaystyle =\int_1^{\infty}\frac{1}{t(t+1)}dt\)
\(\displaystyle =\int_1^{\infty}\left(\frac{1}{t}-\frac{1}{t+1}\right)dt\)
\(\displaystyle =\left[\log\left|\frac{t}{t+1}\right|\right]_1^{\infty}\)
\(\displaystyle =\lim_{s\to\infty}\left(\log\frac{1}{1+\frac{1}{s}}+\log2\right)\)
\(\displaystyle =\log2\)
\(\displaystyle =\int_0^{\infty}\left(\frac{e^{-x}}{1-e^{-x}}-\frac{2e^{-2x}}{1-e^{-2x}}\right)dx\)
\(\displaystyle =\int_0^{\infty}\left(\frac{1}{e^x-1}-\frac{2}{e^{2x}-1}\right)dx\)
\(\displaystyle =\int_0^{\infty}\frac{1}{e^x+1}dx\)
\(\displaystyle =\int_1^{\infty}\frac{1}{t(t+1)}dt\)
\(\displaystyle =\int_1^{\infty}\left(\frac{1}{t}-\frac{1}{t+1}\right)dt\)
\(\displaystyle =\left[\log\left|\frac{t}{t+1}\right|\right]_1^{\infty}\)
\(\displaystyle =\lim_{s\to\infty}\left(\log\frac{1}{1+\frac{1}{s}}+\log2\right)\)
\(\displaystyle =\log2\)
次の学習に進もう！