MATH

ヒント：難しい積分は解き方を覚えてないとなかなか解けない．今回は周期性を考えた解き方になります．

問題

\(\displaystyle \lim_{n\rightarrow\infty}n\int_{0}^{1}e^{-nx}|\sin n\pi x|dx\)の値を求めよ．（オリジナル）

これは一回やっておけば似ているような問題が出ても大丈夫です．

解説

まず，\(y=|\sin n\pi x|\)のグラフは\(\displaystyle \frac{1}{n}\)の周期です．これを考慮すると，\(\displaystyle \frac{k}{n}\leqq x\leqq\frac{k+1}{n}\)の範囲で積分を考えればよい．

\(\displaystyle \int e^{-nx}\sin n\pi xdx=\int e^{-nx}\frac{-1}{n\pi}(\cos n\pi x)^{\prime}dx\)\(=-\displaystyle \frac{1}{n\pi}e^{-nx}\cos n\pi x-\frac{1}{\pi}\int e^{-nx}\cos n\pi xdx\)\(=-\displaystyle \frac{1}{n\pi}e^{-nx}\cos n\pi x-\frac{1}{\pi}\left(\frac{1}{n\pi}e^{-nx}\sin n\pi x+\frac{1}{\pi}\int e^{-nx}\sin n\pi xdx\right)\)

よって，\(\displaystyle \int e^{-nx}\sin n\pi xdx=\frac{\pi^{2}}{\pi^{2}+1}\left(-\frac{1}{n\pi}e^{-nx}\cos n\pi x-\frac{1}{n\pi^{2}}e^{-nx}\sin n\pi x\right)+C\)

ゆえに，\(\displaystyle \int_{\frac{k}{n}}^{\frac{k+1}{n}}e^{-nx}|\sin n\pi x|dx\)\(=\displaystyle \left|\int_{\frac{k}{n}}^{\frac{k+1}{n}}e^{-nx}\sin n\pi xdx\right|\)\(=\displaystyle \left|\frac{\pi^{2}}{\pi^{2}+1}\left[-\frac{1}{n\pi}e^{-nx}\cos n\pi x-\frac{1}{n\pi^{2}}e^{-nx}\sin n\pi x\right]_{\frac{k}{n}}^{\frac{k+1}{n}}\right|\)\(=\left|\displaystyle \frac{\pi^{2}}{\pi^{2}+1}\left[-\frac{1}{n\pi}e^{-(k+1)}\cos(k+1)\pi-\frac{1}{n\pi^{2}}e^{-(k+1)}\sin(k+1)\pi+\frac{1}{n\pi}e^{-k}\cos k\pi+\frac{1}{n\pi^{2}}e^{-k}\sin k\pi\right]\right|\)\(=\displaystyle \frac{\pi^{2}}{\pi^{2}+1}\frac{e^{-k}}{n\pi}(e^{-1}+1)\)

ゆえに\(n\displaystyle \int_{0}^{1}e^{-nx}|\sin n\pi x|dx\)\(=\displaystyle n\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}e^{-nx}|\sin n\pi x|dx\)\(=\displaystyle n\sum_{k=0}^{n-1}\frac{\pi^{2}}{\pi^{2}+1}\frac{e^{-k}}{n\pi}(e^{-1}+1)\)\(=\displaystyle \frac{\pi}{\pi^{2}+1}\frac{1-e^{-n}}{1-e^{-1}}(e^{-1}+1)\)

したがって\(\displaystyle \lim_{n\rightarrow\infty}n\int_{0}^{1}e^{-nx}|\sin n\pi x|dx\)\(=\displaystyle \frac{\pi}{\pi^{2}+1}\frac{(e^{-1}+1)}{1-e^{-1}}\)\(=\displaystyle \frac{\pi}{\pi^{2}+1}\frac{e+1}{e-1}\)

参考入試問題

自然数\(n\)に対して

\(I_{n}=\displaystyle \int_{0}^{1}x^{2}|\sin n\pi x|dx\)とおく．極限値\(\displaystyle \lim_{n\rightarrow\infty}I_{n}\)を求めよ．（08　東工大後期）

解答

前述の解答通り区間に分けて考えます．

\(y=|\sin n\pi x|\)のグラフは\(\displaystyle \frac{1}{n}\)の周期です．これを考慮すると，\(\displaystyle \frac{k}{n}\leqq x\leqq\frac{k+1}{n}\)の範囲で積分を考えればよい．

\(\displaystyle \int x^{2}\sin n\pi xdx\)\(=\displaystyle -\frac{1}{n\pi}\int x^{2}(\cos n\pi x)^{\prime}dx\)\(=-\displaystyle \frac{1}{n\pi}\left(x^{2}\cos n\pi x-\int 2x\cos n\pi xdx\right)\)\(=-\displaystyle \frac{1}{n\pi}\left(x^{2}\cos n\pi x-\frac{2}{n\pi}\left(x\sin n\pi x-\int \sin n\pi xdx\right)\right)\)\(=-\displaystyle \frac{1}{n\pi}\left(x^{2}\cos n\pi x-\frac{2}{n\pi}\left(x\sin n\pi x+\frac{1}{n\pi}\cos n\pi x\right)\right)\)

\(\left|\displaystyle \int_{\frac{k}{n}}^{\frac{k+1}{n}}x^{2}\sin n\pi xdx\right|\)\(=\displaystyle \left|-\frac{1}{n\pi}\left(\left(\frac{k+1}{n}\right)^{2}(-1)^{k+1}-\frac{2}{n\pi}\left(\frac{k+1}{n}\times 0+\frac{1}{n\pi}(-1)^{k+1}\right)\right)+\frac{1}{n\pi}\left(\left(\frac{k}{n}\right)^{2}(-1)^{k}-\frac{2}{n\pi}\left(\frac{k}{n}\times 0+\frac{1}{n\pi}(-1)^{k}\right)\right)\right|\) \(=\displaystyle \frac{|(-1)^{k}|}{n^{3}\pi}\left|\left((k+1)^{2}+k^{2}-\frac{4}{\pi^{2}}\right)\right|\)\(=\displaystyle \frac{1}{n^{3}\pi}\left(2k^{2}+2k+1-\frac{4}{\pi^{2}}\right)\)

∴\(I_{n}=\displaystyle \sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}x^{2}|\sin n\pi x|dx\)\(=\displaystyle \sum_{k=0}^{n-1}\left|\int_{\frac{k}{n}}^{\frac{k+1}{n}}x^{2}\sin n\pi xdx\right|\)\(=\displaystyle \frac{1}{n^{3}\pi}\sum_{k=0}^{n-1}\left(2k^{2}+2k+1-\frac{4}{\pi^{2}}\right)\)\(=\displaystyle \frac{2}{n^{3}\pi}\sum_{k=0}^{n-1}k^{2}+\frac{1}{n^{3}\pi}\sum_{k=0}^{n-1}\left(2k+1-\frac{4}{\pi^{2}}\right)\)

ここで，\(\displaystyle \frac{1}{n^{3}\pi}\sum_{k=0}^{n-1}\left(2k+1-\frac{4}{\pi^{2}}\right)\)は\(n\rightarrow \infty\)で0に収束する．一方\(\displaystyle \frac{2}{n^{3}\pi}\sum_{k=0}^{n-1}k^{2}=\frac{2}{3\pi}\frac{(n-1)(n-\frac{1}{2})}{n^{2}}\)→\(\displaystyle \frac{2}{3\pi}\)

以上から\(\displaystyle \lim_{n\rightarrow\infty}I_{n}=\frac{2}{3\pi}\)