5.1 解答

5.1

1.

(a)

$${\cal L}\{f(t)\} = \int_{0}^{4}2e^{-st}dt + \int_{4}^{\infty}\underbrace{t}_{u}\unde... ...u = t & dv = e^{-st}dt \\ du = dt & v = -\frac{1}{s}e^{-st} \end{array}\right)$$

$$= -\frac{2}{s}e^{-st}\mid_{0}^{4} - \frac{t}{s}e^{-st}\mid_{4}^{\infty -} + \frac{1}{s}\int_{4}^{\infty}e^{-st}dt$$

$$= -\frac{2}{s}(e^{-4s} - 1) + \frac{4}{s}e^{-4s} - \frac{1}{s^2}e^{-st}\mid_{4}^{\infty -}$$

$$= \frac{2}{s}(e^{-4s} + 1) + \frac{1}{s^2}e^{-4s} \ \framebox{終}$$

(b)

$${\cal L}\{f(t)\} = \int_{0}^{\infty}(t^{2} - 2t - 5)e^{-st}dt$$

$$= \int_{0}^{\infty}\underbrace{ t^{2}}_{u}\underbrace{e^{-st}dt}_{d... ...{2} & dv = e^{-st}dt\\ du = 2t dt & v = -\frac{1}{s}e^{-st} \end{array}\right)$$

$$- 2\int_{0}^{\infty}te^{-st}dt - 5\int_{0}^{\infty}e^{-st}dt$$

$$= -\frac{t^2}{s}e^{-st}\mid_{0}^{\infty -} + \frac{2}{s}\int_{0}^{\infty}te^{-st}dt$$

$$- 2\int_{0}^{\infty}te^{-st}dt + \frac{5}{s}e^{-st}\mid_{0}^{\infty -}$$

$$= \frac{5}{s} + (\frac{2}{s} - 2)\int_{0}^{\infty}\underbrace{t}_{u... ... = t & dv = e^{-st}dt\\ du = dt & v = -\frac{1}{s}e^{-st}dt \end{array}\right)$$

$$= (\frac{2}{s} - 2)( \frac{1}{s}\int_{0}^{\infty}e^{-st}dt) - \frac{5}{s}$$

$$= (\frac{2}{s} - 2)(-\frac{1}{s^2})e^{-st}\mid_{0}^{\infty-} + \frac{5}{s} = (\frac{2}{s} - 2)\frac{1}{s^2} - \frac{5}{s} \ \framebox{終}$$

(c)

$${\cal L}\{f(t)\} = \int_{0}^{\infty}te^{-t}e^{-st}dt = \int_{0}^{\infty}\underbrace{... ... e^{-(1 + s)t}dt\\ du = dt & v = - \frac{1}{s+1}e^{-(1+s)t} \end{array}\right)$$

$$= -\frac{t}{1+s}e^{-(1+s)t}\mid_{0}^{\infty -} + \frac{1}{1+s}\int_{0}^{\infty}e^{-(1+s)t}dt$$

$$= -\frac{e^{-(1+s)t}}{(1+s)^{2}}\mid_{0}^{\infty-} = \frac{1}{(1+s)^2} \ \ \ \framebox{終}$$

(d)

$${\cal L}\{e^{2t}\cos{t}\} = \int_{0}^{\infty}e^{2t}\cos{t}e^{-st}dt = \int_{0}^{\infty}e^{(2-s)t}\cos{t}dt$$

$$= \int_{0}^{\infty}e^{(2-s)t}(\frac{e^{it} + e^{-it}}{2})dt$$

$$= \frac{1}{2}\int_{0}^{\infty}(e^{(2-s+i)t} + e^{(2-s-i)t})dt$$

$$= \frac{1}{2}(\frac{e^{(2-s-i)t}}{2-s+i} + \frac{e^{(2-s-i)t}}{2-s-i})\mid_{0}^{\infty-}$$

$$= -\frac{1}{2}(\frac{1}{2-s+i} + \frac{1}{2-s-i})$$

$$= -\frac{1}{2}(\frac{2(2-s)}{(2-s)^{2} + 1}) = \frac{s-2}{(s-2)^{2} + 1} \ \ \framebox{終}$$

(e)

${\cal L}\{t^{n}\} = \int_{0}^{\infty}t^{n}e^{-st}dt = I_{n}$とおくと

$$I_{n} = -\frac{t^{n}}{s}e^{-st}\mid_{0}^{\infty-} + \frac{n}{s}\underbrace{\int_{0}^{\infty}t^{n-1}e^{-st}dt}_{I_{n-1}} = \frac{n}{s}I_{n-1}$$

また $I_{0} = {\cal L}\{1\} = \frac{1}{s}$より

$$\frac{I_{n}}{I_{0}} = \frac{I_{n}}{I_{n-1}}\frac{I_{n-1}}{I_{n-2}... ...}}\frac{I_{1}}{I_{0}} = \frac{n}{s}\frac{n-1}{s}\frac{n-2}{s}\cdots\frac{1}{s}$$

よって

$$I_{n} = \frac{n!}{s^{n+1}} \ \ \ \framebox{終}$$