3.9 解答

(a) $\int_{0}^{1}\frac{dx}{\sqrt{1-x}}$ の積分を行なう前に， $f(x) = \frac{1}{\sqrt{1-x}}$ は $[0,1)$

で連続であることに注意すると

$\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x}} = \int_{0}^{1-}\frac{dx}{\sqrt{1-x}}$

$\begin{displaymath}\begin{array}{l\vert lll} x&0 &\rightarrow & 1-0\ \hline t&1 &\rightarrow & 0+ \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{1-}\frac{dx}{\sqrt{1-x}}$	$\displaystyle =$	$\displaystyle \int_{1}^{0+}{\frac{-2tdt}{t}}$
	$\displaystyle =$	$\displaystyle \int_{1}^{0+}(-2) dt (t$ は0にならない $\displaystyle )$
	$\displaystyle =$	$\displaystyle -2t\mid_{1}^{0+} = -2(\lim_{t \to 0+}t - 1)= 2$

(b) $\int_{0}^{1}\frac{1}{x} dx$ の積分を行なう前に， $f(x) = \frac{1}{x}$ は $(0,1]$

で連続であることに注意すると

$\displaystyle \int_{0}^{1}\frac{1}{x} dx = \int_{0+}^{1}\frac{1}{x} dx$

$\displaystyle \int_{0+}^{1}\frac{1}{x} dx$

$\displaystyle =$

$\displaystyle \log\vert x\vert\mid_{0+}^{1} = \log{1} - \lim_{x \to 0+}\log\vert x\vert = \infty$

で連続であることに注意すると

$\displaystyle \int_{0}^{1}\log{x} dx = \int_{0+}^{1}\log{x} dx$

$\displaystyle \left(\begin{array}{ll} f(x) = \log{x} & g'(x) = 1\\ f'(x) = \frac{1}{x} & g(x) = x \end{array}\right)$

$\displaystyle \int_{0+}^{1}\log{x} dx$	$\displaystyle =$	$\displaystyle x\log{x}\mid_{0+}^{1} - \int_{0+}^{1}dx$
	$\displaystyle =$	$\displaystyle (x\log{x} - x)\mid_{0+}^{1}$
	$\displaystyle =$	$\displaystyle \log{1} - 1 - \lim_{x \to 0+}(x\log{x} - x)$
	$\displaystyle =$	$\displaystyle -1 - \lim_{x \to 0+}\frac{\log{x}}{\frac{1}{x}}$
	$\displaystyle =$	$\displaystyle -1 - \lim_{x \to 0+}\frac{1/x}{-1/x^2} ($ ロピタルの定理より $\displaystyle )$
	$\displaystyle =$	$\displaystyle -1 - \lim_{x \to 0+}(-x) = -1$

(a) $\int_{0}^{\infty}xe^{-x} dx$ の積分を行なう前に， $f(x) = xe^{-x}$ は $[0,\infty)$ で連続であることに注意すると

$\displaystyle \int_{0}^{\infty}xe^{-x} dx = \int_{0}^{\infty-}xe^{-x} dx$

$\displaystyle \int{udv} = uv - \int{vdu}$

$\displaystyle \left(\begin{array}{ll} u = x & dv = e^{-x}\\ du = dx & v = -e^{-x} \end{array}\right)$

$\displaystyle \int_{0}^{\infty-}xe^{-x} dx$	$\displaystyle =$	$\displaystyle -xe^{-x}\mid_{0}^{\infty-} + \int_{0}^{\infty-}e^{-x} dx$
	$\displaystyle =$	$\displaystyle \lim_{x \to \infty-}(-xe^{-x} - e^{-x})\mid_{0}^{\infty-}$
	$\displaystyle =$	$\displaystyle \lim_{x \to \infty-}(\frac{-x}{e^{x}})$
	$\displaystyle =$	$\displaystyle \lim_{x \to \infty}\frac{-1}{e^{x}} = 0 ($ ロピタルの定理 $\displaystyle )$

(b) $\int_{2}^{\infty}\frac{1}{x(\log{x})^{\alpha}} dx$ の積分を行なう前に， $f(x) = \frac{1}{x(\log{x})^{\alpha}}$ は $[2, \infty)$ で連続であることに注意すると

$\displaystyle \int_{2}^{\infty}\frac{1}{x(\log{x})^{\alpha}} dx = \int_{2}^{\infty-}\frac{1}{x(\log{x})^{\alpha}} dx$

$\begin{displaymath}\begin{array}{l\vert lll} x&2&\rightarrow&\infty-\ \hline t&\log{2}&\rightarrow&\infty- \end{array}\end{displaymath}$

$\displaystyle \int_{2}^{\infty-}\frac{1}{x(\log{x})^{\alpha}} dx$	$\displaystyle =$	$\displaystyle \int_{\log{2}}^{\infty-}\frac{1}{t^{-\alpha}}dt$
	$\displaystyle =$	$\displaystyle \left\{\begin{array}{ll} \frac{t^{-\alpha + 1}}{1 - \alpha} \mid_... ...\\ \log{\vert t\vert}\mid_{\log{2}}^{\infty-}, & \alpha = 1 \end{array}\right.$

$\displaystyle \lim_{x \to \infty-}{\frac{t^{-\alpha + 1}}{1 - \alpha}} = \left\{\begin{array}{ll} 0, & \alpha > 1\\ \infty, & \alpha < 1 \end{array}\right.$

$\displaystyle \int_{2}^{\infty-}\frac{1}{x(\log{x})^{\alpha}} dx = \left\{\begin{array}{ll} 0, & \alpha > 1\\ \infty, & \alpha \leq 1 \end{array}\right.$

(c) $\int_{2}^{\infty}\frac{1}{x^{\alpha}\log{x}} dx$ の積分を行なう前に， $f(x) = \frac{1}{x^{\alpha}\log{x}}$ は $[2, \infty)$ で連続であることに注意すると

$\displaystyle \int_{2}^{\infty}\frac{1}{x^{\alpha}\log{x}} dx = \int_{2}^{\infty-}\frac{1}{x^{\alpha}\log{x}} dx$

$\displaystyle x^{\alpha}f(x) = \frac{1}{\log{x}} \leq \frac{1}{\log{2}}$

$\displaystyle \int_{2}^{\infty-}f(x) dx \leq \frac{1}{\log{2}}\int_{2}^{\infty-}\frac{1}{x^{\alpha}} dx$

$\begin{displaymath}\begin{array}{l\vert lll} x&2&\rightarrow&\infty-\ \hline t&\log{2}&\rightarrow&\infty- \end{array}\end{displaymath}$

$\displaystyle \int_{2}^{\infty}\frac{1}{x\log{x}} dx$	$\displaystyle =$	$\displaystyle \int_{\log{2}}^{\infty-}\frac{dt}{t}$
	$\displaystyle =$	$\displaystyle \log\vert t\vert\mid_{\log{2}}^{\infty-} = \infty$

$\displaystyle \int_{2}^{\infty}\frac{1}{x^{\alpha}\log{x}} dx > \int_{2}^{\infty}\frac{1}{x\log{x}} dx$

(a) この問題は収束，発散について調べろという問題であり，何に収束するかという問題ではない．まず， $\cos{x}$ のテーラー展開を思いだすと

$\displaystyle \cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots +$

$\displaystyle \cos{x} \leq 1 - \frac{x^{2}}{2}$

$\displaystyle \frac{1}{\sqrt{\cos{x}}} < \frac{1}{\sqrt{1 - \frac{x^2}{2}}}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1 - \frac{x^2}{2}}} dx < \infty$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1 - \frac{x^2}{2}}} dx$	$\displaystyle =$	$\displaystyle \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{2 - x^2}} dx$
	$\displaystyle =$	$\displaystyle \sqrt{2}\sin{(\frac{x}{2})}\mid_{0}^{\frac{\pi}{2}} = \sqrt{2}\sin^{-1}{(\frac{\pi}{2\sqrt{2}})} < \infty$

$\begin{displaymath}\begin{array}{l\vert lll} x&0+&\rightarrow&1\ \hline t&0+&\rightarrow&1 \end{array}\end{displaymath}$

$\displaystyle \int_{0+}^{1}\frac{\log{x}}{\sqrt{x}} dx$	$\displaystyle =$	$\displaystyle \int_{0+}^{1}\frac{\log{t^2}}{t}(2tdt) = 4\int_{0+}^{1}\log{t} dt$
	$\displaystyle =$	$\displaystyle 4[t\log{t} - t\mid_{0+}^{1} = 4[\log{1} - 1 - \lim_{t \to 0+}(t\log{t} - t)]$
	$\displaystyle =$	$\displaystyle 4[-1 - \lim_{t \to 0+}(\frac{\log{t} -1}{\frac{1}{t}})]$
	$\displaystyle =$	$\displaystyle 4[-1 - \lim_{t \to 0+}\frac{1/t}{-1/t^2} ] = -4$

のとき，

. また， $K(1) = \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}} = [\sin^{-1}{x}]_{0}^{1} = \frac{\pi}{2}$ .