ImproperDoubleIntegral

解答 $\Omega = \{(x,y) : -\infty < x, y < \infty\}$ は $xy$

平面全体を表わすので有界でない．そこで

$\displaystyle \Omega_{n} = \{(x,y) : x^2 + y^2 \leq n^2\} \ (n \geq 1)$

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

$\displaystyle I_{n}$	$\displaystyle =$	$\displaystyle \iint_{\Omega_{n}}\frac{1}{1 + (x^2 + y^2)^2}dxdy = \iint_{\Gamma_{n}}\frac{1}{1 + r^4}\vert J\vert dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{n}\frac{r}{1 + r^4}dr d\theta \ \left(\b... ...rac{1}{2}\tan^{-1}{t} + c\\ = \frac{1}{2}\tan^{-1}{r^2} + c \end{array}\right)$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi} d\theta\left[\frac{1}{2}\tan^{-1}{r^2}\right]_{0}^{2\pi}$
	$\displaystyle =$	$\displaystyle 2\pi \cdot \frac{1}{2}(\tan^{-1}n^2 - \tan^{-1}{0})$

$\displaystyle I = \iint_{\Omega}\frac{1}{1 + (x^2 + y^2)^2}dxdy = \lim_{n \to \infty}I_{n} = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}.$