第9章 9.14

(1)

∫−ππdx=[x]−ππ=π−(−π)=2π\int_{-\pi}^{\pi} dx = [x]_{-\pi}^{\pi} = \pi - (-\pi) = 2\pi

(2)

∫−ππsin⁡(nx) dx=1n[−cos⁡(nx)]−ππ=1n{−cos⁡(nπ)+cos⁡(−nπ)}=1n{−(−1)n+(−1)n}=0\int_{-\pi}^{\pi} \sin(nx)\,dx = \frac{1}{n}[-\cos(nx)]_{-\pi}^{\pi} = \frac{1}{n}\{-\cos(n\pi) + \cos(-n\pi)\} = \frac{1}{n}\{-(-1)^n + (-1)^n\} = 0

(3)

∫−ππcos⁡(nx) dx=1n[sin⁡(nx)]−ππ=1n{sin⁡(nπ)−sin⁡(−nπ)}=1n(0−0)=0\int_{-\pi}^{\pi} \cos(nx)\,dx = \frac{1}{n}[\sin(nx)]_{-\pi}^{\pi} = \frac{1}{n}\{\sin(n\pi) - \sin(-n\pi)\} = \frac{1}{n}(0-0) = 0

(4)

m = n のとき

∫−ππsin⁡(mx)⋅sin⁡(mx) dx=∫−ππ1−cos⁡2mx2 dx=12[x−12msin⁡(2mx)]−ππ=12{π−(−π)}=π\int_{-\pi}^{\pi} \sin(mx)\cdot\sin(mx)\,dx = \int_{-\pi}^{\pi} \frac{1-\cos 2mx}{2}\,dx = \frac{1}{2}\Big[x - \frac{1}{2m}\sin(2mx)\Big]_{-\pi}^{\pi} = \frac{1}{2}\{\pi-(-\pi)\} = \pi

m ≠ n のとき

∫−ππsin⁡(mx)⋅sin⁡(nx) dx=12∫−ππ{cos⁡(m−n)−cos⁡(m+n)}dx=12{∫−ππcos⁡(m−n) dx−∫−ππcos⁡(m+n) dx}=0\int_{-\pi}^{\pi} \sin(mx)\cdot\sin(nx)\,dx = \frac{1}{2}\int_{-\pi}^{\pi}\{\cos(m-n)-\cos(m+n)\}dx = \frac{1}{2}\left\{\int_{-\pi}^{\pi}\cos(m-n)\,dx - \int_{-\pi}^{\pi}\cos(m+n)\,dx\right\} = 0

(5)

m = n のとき

∫−ππcos⁡(mx)⋅cos⁡(mx) dx=∫−ππcos⁡2(mx) dx=∫−ππ1+cos⁡(2mx)2 dx=12{∫−ππdx−∫−ππcos⁡(2mx) dx}\int_{-\pi}^{\pi} \cos(mx)\cdot\cos(mx)\,dx = \int_{-\pi}^{\pi}\cos^2(mx)\,dx = \int_{-\pi}^{\pi}\frac{1+\cos(2mx)}{2}\,dx = \frac{1}{2}\left\{\int_{-\pi}^{\pi}dx - \int_{-\pi}^{\pi}\cos(2mx)\,dx\right\}

=12(2π−0)=π= \frac{1}{2}(2\pi - 0) = \pi

m ≠ n のとき

∫−ππcos⁡(mx)⋅cos⁡(nx) dx=12∫−ππ{cos⁡(m−n)+cos⁡(m+n)}dx=12(0−0)=0\int_{-\pi}^{\pi}\cos(mx)\cdot\cos(nx)\,dx = \frac{1}{2}\int_{-\pi}^{\pi}\{\cos(m-n)+\cos(m+n)\}dx = \frac{1}{2}(0-0) = 0

(6)

∫−ππsin⁡(mx)⋅cos⁡(nx) dx=12∫−ππ{sin⁡(m−n)+sin⁡(m+n)}dx=12(0−0)=0\int_{-\pi}^{\pi}\sin(mx)\cdot\cos(nx)\,dx = \frac{1}{2}\int_{-\pi}^{\pi}\{\sin(m-n)+\sin(m+n)\}dx = \frac{1}{2}(0-0) = 0

解説: r31bn1z