이상적분improper_integral (rev. 1.7)
ex.
$\displaystyle \int\nolimits_1^{\infty}\frac1xdx$
pf.
$\displaystyle p=1\Rightarrow$
$\displaystyle \int\nolimits_1^{\infty}\frac1xdx$
$\displaystyle =\lim_{t\to\infty}\int\nolimits_1^t\frac1xdx$
$\displaystyle =\lim_{t\to\infty}\left[\ln|x|\right]_1^t$
$\displaystyle =\lim_{t\to\infty}\ln|t|$
$\displaystyle =\lim_{t\to\infty}\ln t$
$\displaystyle =\infty$ .....(1)
정리:$\displaystyle =\lim_{t\to\infty}\left[\ln|x|\right]_1^t$
$\displaystyle =\lim_{t\to\infty}\ln|t|$
$\displaystyle =\lim_{t\to\infty}\ln t$
$\displaystyle =\infty$ .....(1)
$\displaystyle \int\nolimits_1^{\infty}\frac1{x^p}dx$는 $\displaystyle p>1$ 이면 수렴, $\displaystyle p\le 1$ 이면 발산.
pf.
$\displaystyle p=1\Rightarrow$
위 (1)에 따라 발산.
$\displaystyle p\ne 1\Rightarrow$$\displaystyle =\int\nolimits_1^{\infty}x^{-p}dx$
$\displaystyle =\lim_{t\to\infty}\int\nolimits_1^t x^{-p}dx$
$\displaystyle =\lim_{t\to\infty}\left[\frac1{1-p}x^{1-p}\right]_1^t$
$\displaystyle =\lim_{t\to\infty}\left(\frac1{1-p}t^{1-p}-\frac1{1-p}\right)$
$\displaystyle =\begin{cases}\frac1{p-1}&(p>1)\\ \infty&(p<1)\end{cases}$ (첫번째는 수렴, 두번째는 발산)
Thm.$\displaystyle =\lim_{t\to\infty}\int\nolimits_1^t x^{-p}dx$
$\displaystyle =\lim_{t\to\infty}\left[\frac1{1-p}x^{1-p}\right]_1^t$
$\displaystyle =\lim_{t\to\infty}\left(\frac1{1-p}t^{1-p}-\frac1{1-p}\right)$
$\displaystyle =\begin{cases}\frac1{p-1}&(p>1)\\ \infty&(p<1)\end{cases}$ (첫번째는 수렴, 두번째는 발산)
$\displaystyle \int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^af(x)dx+\int_a^{\infty}f(x)dx$
ex.
$\displaystyle \int_{-\infty}^{\infty}\frac1{1+x^2}dx$ 의 값을 구하라.
sol.
$\displaystyle \int_0^{\infty}\frac1{1+x^2}dx$
$\displaystyle \int_{-\infty}^{\infty}\frac1{1+x^2}dx$ 의 값을 구하라.
sol.
$\displaystyle \int_0^{\infty}\frac1{1+x^2}dx$
$\displaystyle =\lim_{t\to\infty}\int_0^t\frac1{1+x^2}dx$
$\displaystyle =\lim_{t\to\infty}\left[\tan^{-1}x\right]_0^t$
$\displaystyle =\lim_{t\to\infty}\tan^{-1}t$
$\displaystyle =\frac{\pi}2$
$\displaystyle \int_{-\infty}^0\frac1{1+x^2}dx$$\displaystyle =\lim_{t\to\infty}\left[\tan^{-1}x\right]_0^t$
$\displaystyle =\lim_{t\to\infty}\tan^{-1}t$
$\displaystyle =\frac{\pi}2$
$\displaystyle =\lim_{t\to-\infty}\int_t^0\frac1{1+x^2}dx$
$\displaystyle =\lim_{t\to-\infty}\left[\tan^{-1}x\right]_t^0$
$\displaystyle =\lim_{t\to-\infty}(-\tan^{-1}t)$
$\displaystyle =-\left(-\frac{\pi}2\right)$
$\displaystyle =\frac{\pi}2$
답은 $\displaystyle \frac{\pi}2+\frac{\pi}2=\pi$$\displaystyle =\lim_{t\to-\infty}\left[\tan^{-1}x\right]_t^0$
$\displaystyle =\lim_{t\to-\infty}(-\tan^{-1}t)$
$\displaystyle =-\left(-\frac{\pi}2\right)$
$\displaystyle =\frac{\pi}2$