8.1 解答

8.1

1.

(a)

$${\cal F}_{s}[f^{\prime}(x)] = \int_{0}^{\infty}f^{\prime}(x)\sin{\omega x} dx \ \left(\begin{ar... ...dv = f^{\prime}(x) \\ du = \omega \cos{\omega x} & v = f(x) \end{array}\right)$$

$$= f(x) \sin{\omega x}\mid_{0}^{\infty -} - \omega \int_{0}^{\infty}f(x)\cos{\omega x} dx$$

$$= \lim_{x \rightarrow \infty}f(x)\sin{\omega x} - \omega {\cal F}_{c}[f(x)]$$

$$= - \omega {\cal F}_{c}[f(x)] \ \ \framebox{終}$$

(b)

$${\cal F}_{c}[f^{\prime}(x)] = \int_{0}^{\infty}f^{\prime}(x)\cos{\omega x} dx \ \left(\begin{ar... ...v = f^{\prime}(x) \\ du = -\omega \sin{\omega x} & v = f(x) \end{array}\right)$$

$$= f(x) \cos{\omega x}\mid_{0}^{\infty -} + \omega \int_{0}^{\infty}f(x)\sin{\omega x} dx$$

$$= \lim_{x \rightarrow \infty}f(x)\cos{\omega x} - \lim_{x \rightarrow 0+}f(x)\cos{\omega x} + \omega {\cal F}_{s}[f(x)]$$

$$= -f(0+) + \omega {\cal F}_{c}[f(x)] \ \ \framebox{終}$$

2.

(a)

$$F[\omega] = {\cal F}\{f\} = \int_{-\infty}^{\infty}f(x) e^{-i\omega x} dx$$

$$= \int_{-a}^{a} e^{-i \omega x} dx$$

$$= \frac{1}{-i\omega} e^{-i \omega x}\mid_{-a}^{a}$$

$$= \frac{1}{-i\omega}(e^{-i \omega a} - e^{i\omega a})$$

$$= \frac{1}{i\omega}(e^{i \omega a} - e^{-i \omega a})$$

$$= \frac{2}{\omega}(\frac{e^{i \omega a} - e^{-i \omega a}}{2i})$$

$$= \frac{2}{\omega}\sin {\omega a} \ \ \framebox{終}$$

(b)

$$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F[\omega] e^{i\omega x} d\omega$$

$$= \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{2\sin{\omega a}}{\omega} e^{i\omega x} d\omega$$

$$= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin{\omega a}(\cos{\omega x} + i\sin{\omega x})}{\omega} d\omega$$

$$= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin{\omega a}\cos{\omega x}}{\omega} d\omega \ \ (\frac{\sin{\omega a} \sin{\omega x}}{\omega} \mbox{は$\omega$について奇関数} )$$

よって

$$\int_{-\infty}^{\infty} \frac{\sin{\omega a}\cos{\omega x}}{\omeg... ... \vert x\vert < a\\ 0 , & \vert x\vert > a \end{array}\right. \ \ \framebox{終}$$

(c) $\int_{-\infty}^{\infty} \frac{\sin{\omega a}\cos{\omega x}}{\omega} d\omega$において $a = 1, x = 0$とおくと

$$\int_{-\infty}^{\infty}\frac{\sin{\omega}}{\omega} d\omega = \pi$$

よって

$$\int_{0}^{\infty}\frac{\sin{u}}{u} du = \frac{\pi}{2} \ \framebox{終}$$

3.

(a) $F_{c}(\omega) = \int_{0}^{\infty}f(x)\cos{\omega x}d \omega$よりフーリエ積分公式を用いると

$$f(x) = \frac{2}{\pi}\int_{0}^{\infty}F_{c}(\omega)\cos{\omega x}d\omega$$

$$= \frac{2}{\pi}\int_{0}^{1}(1 - \omega)\cos{\omega x}d\omega \ \lef... ... x} d\omega\\ du = - d\omega & v = \frac{\sin{\omega x}}{x} \end{array}\right)$$

$$= \frac{2}{\pi}[(1-\omega)\frac{\sin{\omega x}}{x} \mid_{0}^{1} + \frac{1}{x}\int_{0}^{1}\sin{\omega x }d\omega ]$$

$$= \frac{2}{\pi}[\frac{1}{x}(-\frac{1}{x})\cos{\omega x} \mid_{0}^{1}]$$

$$= \frac{2}{\pi}[-\frac{1}{x^2}(\cos{x} - 1) ]$$

$$= \frac{2}{\pi x^2}(1 - \cos{x}) \ \ \framebox{終}$$

(b) (a)の結果より

$$\int_{0}^{\infty} \frac{2}{\pi x^2}(1 - \cos{x})\cos{\omega x}dx ... ...ll} 1 - \omega & 0 \leq \omega \leq 1\\ 0 & \omega > 1 \end{array} \right.$$

ここで $\omega \rightarrow 0$ととると

$$\frac{2}{\pi}\int_{0}^{\infty}\frac{1 - \cos{x}}{x^2}dx = 1$$

よって

$$\int_{0}^{\infty}\frac{1 - \cos{x}}{x^2}dx = \frac{\pi}{2}$$

なお $1 - \cos{x} = 2\sin^{2}{x/2}$に注意すると

$$\int_{0}^{\infty}\frac{1 - \cos{x}}{x^2}dx = \int_{0}^{\infty}\frac{2\sin^{2}{\frac{x}{2}}}{x^2}dx \ \ \left(\begin{array}{l} u = \frac{x}{2}\\ du = \frac{1}{2}dx \end{array}\right)$$

$$= \int_{0}^{\infty}\frac{\sin^{2}{u}}{u^2}du \ \ \framebox{終}$$

4.

(a) $f(x) = \left\{\begin{array}{ll} \frac{\pi}{2}, & 0 < x < a\\ \frac{\pi}{4}, & x = a\\ 0, & x > a \end{array} \right.$とおき，奇関数拡張する．つまり

$$F_{s}(\omega) = \int_{0}^{\infty}f(x)\sin{\omega x}dx = \int_{0}^{a}\frac{\pi}{2}\sin{\omega x}dx$$

$$= \frac{-\pi}{2 \omega}\cos{\omega x}\mid_{0}^{a} = \frac{\pi }{2\omega}(1 - \cos{\omega a})$$

よって

$$f(x) = \frac{2}{\pi}\int_{0}^{\infty} \frac{\pi}{2 \omega}(1 - \cos{\omega a})\sin{\omega x}d \omega$$

$$= \int_{0}^{\infty}\frac{1 - \cos{\omega a}}{\omega} \sin{\omega x} d\omega \ \ \ \framebox{終}$$

(b) $f(x) = \left\{\begin{array}{ll} \frac{\pi \sin{x}}{2}, & \vert x\vert < \pi\\ 0, & \vert x\vert > \pi \end{array} \right.$とおくと，$f(x)$は奇関数．よってフーリエ積分公式より

$$A(\omega) = \int_{-\infty}^{\infty}f(x)\cos{\omega x} dx = 0$$

$$B(\omega) = \int_{-\infty}^{\infty}f(x)\sin{\omega x}dx = \int_{-\pi}^{\pi}\frac{\pi \sin{x}}{2}\sin{\omega x}dx$$

$$= \pi \int_{0}^{\pi}\frac{\cos{(\omega - 1)x} - \cos{(\omega +1)x }}{2} dx$$

$$= \frac{\pi}{2}[\frac{\sin{(\omega - 1)x}}{\omega -1} - \frac{\sin{(\omega + 1)x}}{\omega + 1} \mid_{0}^{\pi}$$

$$= \frac{\pi}{2}[\frac{\sin{(\omega - 1)\pi}}{\omega -1} - \frac{\sin{(\omega + 1)\pi}}{\omega + 1}]$$

$$= \frac{\pi}{2}[\frac{-\sin{\omega \pi}}{\omega -1} + \frac{\sin{\omega \pi}}{\omega + 1}]$$

$$= \frac{\pi}{2}[\frac{-2\sin{\omega \pi}}{\omega^2 - 1}] = \frac{\pi \sin{\omega \pi}}{1 - \omega^2}$$

よって

$$f(x) = \frac{1}{\pi}\int_{0}^{\infty}\frac{\pi \sin{\omega \pi}}{1 - \omega^2}\sin{\omega x}d \omega$$

$$= \int_{0}^{\infty}\frac{\sin{\pi \omega}\sin{\omega x}}{1 - \omega^2} d\omega$$