7.4 広義積分

1.

(a) $\Omega$を図示すると,

\begin{figure}\includegraphics[width=5cm]{CALCFIG/ren7-4-1a.eps}
\end{figure}
この領域上の$x = 0$の線上で被積分関数$e^{y/x}$は定義されていない.そこで,$x = 0$を避けるように,直線 $x = \frac{1}{n}$を引き,図のような領域を考える.この領域をv-simpleで表すと

$\displaystyle \Omega_{n} = \{(x,y) : \frac{1}{n} \leq x \leq 1, 0  leq y \leq x\}$

$\Omega$上での積分は
$\displaystyle \iint_{\Omega} e^{y/x}\;dxdy$ $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\iint_{\Omega_{n}}e^{y/x}\; dxdy$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{\frac{1}{n}}^{1}\int_{y=0}^{x}e^{y/x}\; dydx$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{\frac{1}{n}}^{1}[xe^{y/x}]_{0}^{x} \;dx$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{\frac{1}{n}}^{1}(x(e-1) \;dx$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}(e-1)\left[\frac{x^2}{2}\right]_{\frac{1}{n}}^{1} = (e-1)\lim_{n \to \infty}(\frac{1}{2} = \frac{1}{2n^2}) = \frac{e-1}{2}$  

(b) $\Omega$を図示すると,

\begin{figure}\includegraphics[width=5cm]{CALCFIG/ren7-4-1b.eps}
\end{figure}
この領域上の$y = 1-x$の線上で被積分関数 $\frac{x}{\sqrt{1 -x - y}}$は定義されていない.そこで,$y = 1-x$を避けるように,直線 $x = 1 - \frac{1}{n} - x$を引き,図のような領域を考える.この領域をv-simpleで表すと

$\displaystyle \Omega_{n} = \{(x,y) : 0 \leq x \leq 1-\frac{1}{n}, 0  leq y \leq 1- \frac{1}{n} -x\}$

$\Omega$上での積分は
$\displaystyle \iint_{\Omega} \frac{x}{\sqrt{1 - x- y}}\;dxdy$ $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\iint_{\Omega_{n}}\frac{x}{\sqrt{1-x-y}}\; dxdy$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1-\frac{1}{n}}^{1}\int_{y=0}^{1- \frac{1}{n} - x}\frac{x}{\sqrt{1-x-y}}\; dydx$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1-\frac{1}{n}}^{1}-\left[2x(1 - (x+y)^{1/2}\right]_{0}^{1 - \frac{1}{n} - x}\; dx$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1-\frac{1}{n}}^{1}\left(-2x(\frac{1}{n})^{1/2} + 2x(1-x)^{1/2}\right)\; dx$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\left[\frac{1}{\sqrt{n}}(-x^2) + \frac{4(1-x)^{5/2}}{5} - \frac{4(1-x)^{3/2}}{3}\right]_{0}^{1-\frac{1}{n}}$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\left(\frac{1}{\sqrt{n}}(-(1 - \frac{1}{n})^{2...
...}{n})^{5/2} - \frac{4}{3}(\frac{1}{n})^{3/2} - \frac{4}{5} + \frac{4}{3}\right)$  
  $\displaystyle =$ $\displaystyle \frac{8}{15}$  

(c) $\Omega$を図示すると,

\begin{figure}\includegraphics[width=5cm]{CALCFIG/ren7-4-1c.eps}
\end{figure}
この領域上の $x^2 + y^2 = 0$で被積分関数 $\log(x^2 + y^2)$は定義されていない.そこで, $x^2 + y^2 = 0$を避けるように,弧 $r = \frac{1}{n}$を描き,図のような領域を考える.この領域を平面の曲座標で表すと

$\displaystyle \Omega_{n} = \{(r,\theta) : 0 \leq \theta \leq \frac{\pi}{2}, \frac{1}{n}  leq r \leq 1\}$

$\Omega$上での積分は
$\displaystyle \iint_{\Omega} \log(x^2 + y^2)\;dxdy$ $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\iint_{\Omega_{n}}\log(x^2 + y^2)\; dxdy$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1}^{\frac{\pi}{2}}\int_{r=\frac{1}{n}}^{1}\log{r^2} \cdot r\; dr d\theta$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1}^{\frac{\pi}{2}}\int_{t=\frac{1}{n^2}}^{1}\log{t} \cdot \frac{dt}{2}\; d\theta$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1}^{\frac{\pi}{2}}\frac{1}{2}\left[t\log{t} - t\right]_{\frac{1}{n^2}} {1}\; d\theta$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\int_{1}^{\frac{\pi}{2}}\left(-1 - \frac{1}{n^2}\log(\frac{1}{n^2}) - \frac{1}{n^2}\right)\; d\theta$  
  $\displaystyle =$ $\displaystyle -\frac{1}{2}\cdot \frac{\pi}{2} = -\frac{\pi}{4}$