sin(mx)sin(nx), cos(mx)cos(nx), sin(mx)cos(nx) の0から2πまでの積分

準備

準備の準備

$$\begin{array}{rcl}\cos (\alpha +\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\\ \cos (\alpha -\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)\\ \sin (\alpha +\beta )& =& \sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)\\ \sin (\alpha -\beta )& =& \sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)\end{array}$$

準備1

$$\begin{array}{r}\\ \cos (\alpha -\beta )-\cos (\alpha +\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)-\{\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\}\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& \cancel{\cos \left(\alpha \right)\cos \left(\beta \right)}+\sin \left(\alpha \right)\sin \left(\beta \right)\cancel{-\cos \left(\alpha \right)\cos \left(\beta \right)}+\sin \left(\alpha \right)\sin \left(\beta \right)\\ & =& 2\sin \left(\alpha \right)\sin \left(\beta \right)\\ \sin \left(\alpha \right)\sin \left(\beta \right)& =& \frac{\cos (\alpha -\beta )-\cos (\alpha +\beta )}{2}\end{array}$$

準備2

$$\begin{array}{r}\\ \cos (\alpha -\beta )+\cos (\alpha +\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)+\{\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\}\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& \cos \left(\alpha \right)\cos \left(\beta \right)+\cancel{\sin \left(\alpha \right)\sin \left(\beta \right)}+\cos \left(\alpha \right)\cos \left(\beta \right)-\cancel{\sin \left(\alpha \right)\sin \left(\beta \right)}\\ & =& 2\cos \left(\alpha \right)\cos \left(\beta \right)\\ \cos \left(\alpha \right)\cos \left(\beta \right)& =& \frac{\cos (\alpha -\beta )+\cos (\alpha +\beta )}{2}\end{array}$$

準備3

$$\begin{array}{rcl}\cos \left(2x\right)& =& \cos (x+x)\\ & =& \cos \left(x\right)\cos \left(x\right)-\sin \left(x\right)\sin \left(x\right)\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& {\cos }^2\left(x\right)-{\sin }^2(x)\\ & =& (1-{\sin }^2(x))-{\sin }^2(x)\\ & =& 1-2{\sin }^2(x)\\ \cos \left(2x\right)-1& =& -2{\sin }^2(x)\\ 1-\cos \left(2x\right)& =& 2{\sin }^2(x)\\ {\sin }^2(x)& =& \frac{1-\cos \left(2x\right)}{2}=\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\end{array}$$

準備4

$$\begin{array}{rcl}\cos \left(2x\right)& =& \cos (x+x)\\ & =& \cos \left(x\right)\cos \left(x\right)-\sin \left(x\right)\sin \left(x\right)\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& {\cos }^2\left(x\right)-{\sin }^2(x)\\ & =& {\cos }^2\left(x\right)-(1-{\cos }^2\left(x\right))\\ & =& 2{\cos }^2(x)-1\\ \cos \left(2x\right)+1& =& 2{\cos }^2(x)\\ {\cos }^2(x)& =& \frac{1+\cos \left(2x\right)}{2}=\frac{1}{2}+\frac{\cos \left(2x\right)}{2}\end{array}$$

準備5

$$\begin{array}{r}\\ \sin (\alpha -\beta )+\sin (\alpha +\beta )& =& \sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)+\{\sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)\}\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& \sin \left(\alpha \right)\cos \left(\beta \right)\cancel{-\cos \left(\alpha \right)\sin \left(\beta \right)}+\sin \left(\alpha \right)\cos \left(\beta \right)\cancel{+\cos \left(\alpha \right)\sin \left(\beta \right)}\\ & =& 2\sin \left(\alpha \right)\cos \left(\beta \right)\\ \sin \left(\alpha \right)\cos \left(\beta \right)& =& \frac{\sin (\alpha -\beta )+\sin (\alpha +\beta )}{2}\end{array}$$

準備6

$$\begin{array}{rcl}\sin \left(2x\right)& =& \sin (x+x)\\ & =& \sin \left(x\right)\cos \left(x\right)+\cos \left(x\right)\sin \left(x\right)\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& 2\sin \left(x\right)\cos \left(x\right)\\ \sin \left(x\right)\cos \left(x\right)& =& \frac{\sin \left(2x\right)}{2}\end{array}$$

$\sin \left(mx\right)\sin \left(nx\right)$

$m,n\in \mathbb{N},m\ne n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m\ne n\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos (mx-nx)-\cos (mx+nx)]\mathrm{d}x\cdots \mathrm{準}\mathrm{備}1\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos \left((m-n)x\right)-\cos \left((m+n)x\right)]\mathrm{d}x\\ & =& \frac{1}{2}[{\int }_0^{2\pi }\cos \left((m-n)x\right)\mathrm{d}x-{\int }_0^{2\pi }\cos \left((m+n)x\right)\mathrm{d}x]\\ & =& \frac{1}{2}[{[\frac{1}{m-n}\sin \left((m-n)x\right)]}_0^{2\pi }-{[\frac{1}{m+n}\sin \left((m+n)x\right)]}_0^{2\pi }]\\ & =& \frac{1}{2}[\frac{1}{m-n}[\sin \left((m-n)2\pi \right)-\sin \left((m-n)0\right)]-\frac{1}{m+n}[\sin \left((m+n)2\pi \right)-\sin \left((m+n)0\right)]]\\ & =& \frac{1}{2}[\frac{1}{m-n}[0-0]-\frac{1}{m+n}[0-0]]\\ & =& \frac{1}{2}[\frac{0}{m-n}-\frac{0}{m+n}]\\ & =& \frac{1}{2}0\\ & =& 0\end{array}$$

$m,n\in \mathbb{N},m=n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m=n\\ & =& {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(mx\right)\mathrm{d}x\\ & =& {\int }_0^{2\pi }{\sin }^2\left(mx\right)\mathrm{d}x\\ & =& \frac{1}{m}{\int }_0^{2m\pi }{\sin }^2\left(u\right)\mathrm{d}u\cdots u=mx,\frac{\mathrm{d}u}{\mathrm{d}x}=m,\mathrm{d}x=\frac{1}{m}\mathrm{d}u\\ & =& \frac{1}{m}{\int }_0^{2m\pi }[\frac{1}{2}-\frac{1}{2}\cos \left(2u\right)]\mathrm{d}u\cdots \mathrm{準}\mathrm{備}3\\ & =& \frac{1}{2m}[{\int }_0^{2m\pi }\mathrm{d}u-{\int }_0^{2m\pi }\cos \left(2u\right)\mathrm{d}u]\\ & =& \frac{1}{2m}[{\left[u\right]}_0^{2m\pi }-{[\frac{1}{2}\sin \left(2u\right)]}_0^{2m\pi }]\\ & =& \frac{1}{2m}[[2m\pi -0]-[\frac{1}{2}\sin (2\cdot 2m\pi )-\frac{1}{2}\sin (2\cdot 0)]]\\ & =& \frac{1}{2m}[2m\pi -[0-0]]\\ & =& \frac{1}{\cancel{2m}}\cancel{2m}\pi \\ & =& \pi \end{array}$$

$\cos \left(mx\right)\cos \left(nx\right)$

$m,n\in \mathbb{N},m\ne n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m\ne n\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos (mx-nx)+\cos (mx+nx)]\mathrm{d}x\cdots \mathrm{準}\mathrm{備}2\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos \left((m-n)x\right)+\cos \left((m+n)x\right)]\mathrm{d}x\\ & =& \frac{1}{2}[{\int }_0^{2\pi }\cos \left((m-n)x\right)\mathrm{d}x+{\int }_0^{2\pi }\cos \left((m+n)x\right)\mathrm{d}x]\\ & =& \frac{1}{2}[{[\frac{1}{m-n}\sin \left((m-n)x\right)]}_0^{2\pi }+{[\frac{1}{m+n}\sin \left((m+n)x\right)]}_0^{2\pi }]\\ & =& \frac{1}{2}[\frac{1}{m-n}[\sin \left((m-n)2\pi \right)-\sin \left((m-n)0\right)]+\frac{1}{m+n}[\sin \left((m+n)2\pi \right)-\sin \left((m+n)0\right)]]\\ & =& \frac{1}{2}[\frac{1}{m-n}[0-0]+\frac{1}{m+n}[0-0]]\\ & =& \frac{1}{2}[\frac{0}{m-n}+\frac{0}{m+n}]\\ & =& \frac{1}{2}0\\ & =& 0\end{array}$$

$m,n\in \mathbb{N},m=n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m=n\\ & =& {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(mx\right)\mathrm{d}x\\ & =& {\int }_0^{2\pi }{\cos }^2\left(mx\right)\mathrm{d}x\\ & =& \frac{1}{m}{\int }_0^{2m\pi }{\cos }^2\left(u\right)\mathrm{d}u\cdots u=mx,\frac{\mathrm{d}u}{\mathrm{d}x}=m,\mathrm{d}x=\frac{1}{m}\mathrm{d}u\\ & =& \frac{1}{m}{\int }_0^{2m\pi }[\frac{1}{2}+\frac{1}{2}\cos \left(2u\right)]\mathrm{d}u\cdots \mathrm{準}\mathrm{備}4\\ & =& \frac{1}{2m}[{\int }_0^{2m\pi }\mathrm{d}u+{\int }_0^{2m\pi }\cos \left(2u\right)\mathrm{d}u]\\ & =& \frac{1}{2m}[{\left[u\right]}_0^{2m\pi }+{[\frac{1}{2}\sin \left(2u\right)]}_0^{2m\pi }]\\ & =& \frac{1}{2m}[[2m\pi -0]+[\frac{1}{2}\sin (2\cdot 2m\pi )-\frac{1}{2}\sin (2\cdot 0)]]\\ & =& \frac{1}{2m}[2m\pi +[0-0]]\\ & =& \frac{1}{\cancel{2m}}\cancel{2m}\pi \\ & =& \pi \end{array}$$

$\sin \left(mx\right)\cos \left(nx\right)$

$m,n\in \mathbb{N},m\ne n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m\ne n\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\sin (mx-nx)+\sin (mx+nx)]\mathrm{d}x\cdots \mathrm{準}\mathrm{備}5\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\sin ((m-n)x)+\sin ((m+n)x)]\mathrm{d}x\\ & =& \frac{1}{2}[{\int }_0^{2\pi }\sin ((m-n)x)\mathrm{d}x+{\int }_0^{2\pi }\sin ((m+n)x)\mathrm{d}x]\\ & =& \frac{1}{2}[{[\frac{-1}{m-n}\cos ((m-n)x)]}_0^{2\pi }+{[\frac{-1}{m+n}\cos ((m+n)x)]}_0^{2\pi }]\\ & =& \frac{1}{2}[\frac{-1}{m-n}[\cos ((m-n)2\pi )-\cos ((m-n)0)]+\frac{-1}{m+n}[\cos ((m-n)2\pi )-\cos ((m-n)0)]]\\ & =& \frac{1}{2}[\frac{-1}{m-n}[1-1]+\frac{-1}{m+n}[1-1]]\\ & =& \frac{1}{2}[\frac{-1}{m-n}0+\frac{-1}{m+n}0]\\ & =& \frac{1}{2}[0+0]\\ & =& \frac{1}{2}0\\ & =& 0\end{array}$$

$m,n\in \mathbb{N},m=n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m=n\\ & =& {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(mx\right)\mathrm{d}x\\ & =& {\int }_0^{2\pi }\frac{1}{2}\sin \left(2mx\right)\mathrm{d}x\cdots \mathrm{準}\mathrm{備}6\\ & =& \frac{1}{2}{\int }_0^{4m\pi }\frac{1}{2m}\sin \left(u\right)\mathrm{d}u\cdots u=2mx,\frac{\mathrm{d}u}{\mathrm{d}x}=2m,\mathrm{d}x=\frac{1}{2m}\mathrm{d}u\\ & =& \frac{1}{4m}{[-\cos \left(u\right)]}_0^{4m\pi }\\ & =& \frac{1}{4m}[-\cos \left(4m\pi \right)-(-\cos \left(0\right))]\\ & =& \frac{1}{4m}[-1-(-1)]\\ & =& \frac{1}{4m}[-1+1]\\ & =& \frac{1}{4m}0\\ & =& 0\end{array}$$

まとめ

$$\begin{array}{r}m,n\in \mathbb{N}\\ {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x& =& \{\begin{array}{l}\pi \cdots m=n\\ 0\cdots m\ne n\end{array}\\ {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(nx\right)\mathrm{d}x& =& \{\begin{array}{l}\pi \cdots m=n\\ 0\cdots m\ne n\end{array}\\ {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(nx\right)\mathrm{d}x& =& 0\end{array}$$