Chapter 5 Techniques of Integrals(3/3) - 微積分甲 (一) - 自修課程 (大學)

Improper integrals

# Improper Integrals (瑕積分)

Case1: 積分範圍無限，但積分函數有限
Case2: 積分範圍有限，但積分函數無限
Case3: 積分範圍無限，且積分函數無限

# Case 1

If $f$ is continuous on $[a, \infty)$ ,then $\int^{\infty}_a f(x) dx = \lim_{b \to \infty} \int^b_a f(x)dx$
If $f$ is continuous on $(-\infty, b]$ ,then $\int^{b}_{-\infty} f(x) dx = \lim_{a \to -\infty} \int^b_a f(x)dx$
If $f$ is continuous on $(-\infty, \infty)$ ,then $\int^{\infty}_{-\infty} f(x) dx = \int^{a}_{-\infty} f(x) dx + \int^{\infty}_{a} f(x) dx = \lim_{t \to \infty} \int^t_a f(x) dx + \lim_{t \to -\infty} \int^a_t f(x) dx$

$\int^{\infty}_0 \frac{1}{x^2+1}dx$
$\int^{-2}_{-\infty} \frac{1}{x}dx$

$\int^{\infty}_0 \frac{1}{x^2+1}dx = \lim_{b\to \infty} \int^b_a \frac{1}{x^2+1}dx = \lim_{b \to \infty} \{ tan^{-1}x|^{x=b}_{x=0} \} = \lim_{b \to \infty} \{ tan^{-1}b - tan^{-1}0 \} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ .

$\Rightarrow \int^{\infty}_0 \frac{1}{x^2+1}dx$ converge to $\frac{\pi}{2}$ .

$\int^{-2}_{-\infty} \frac{1}{x} dx = \lim_{a\to -\infty}\frac{1}{x} dx = \lim_{a \to -\infty} \{ ln|x| |^{x = -2}_{a} \} = \lim_{a \to -\infty} \{ ln2 - ln|a| \} = -\infty$
$\Rightarrow \int^{-2}_{-\infty} \frac{1}{x} dx$ diverges.

# Case 2

If $f$ is continuous on $[a,b)$ and $\lim_{t \to b^-} f(x) = \infty / -\infty$ ,then $\int^b_af(x)dx = \lim_{t \to b^-} \int^t_a f(x)dx$ .
If $f$ is continuous on $(a,b]$ and $\lim_{t \to a^+} f(x) = \infty / -\infty$ ,then $\int^b_af(x)dx = \lim_{t \to a^+} \int^b_t f(x)dx$ .
If $f$ is continuous on $(a,b)$ and both $\lim_{t \to b^-} f(x)$ , $\lim_{t \to a^+} f(x)$ are infinite,then $\int^b_af(x)dx = \int^c_af(x)dx + \int^b_cf(x)dx$ .

$\int^1_0 \frac{1}{\sqrt[3]{x}}dx$
$\int^2_{-1} \frac{1}{x^3}dx$

$\int^1_0 \frac{1}{\sqrt[3]{x}}dx = \int^1_0 x^{-\frac{1}{3}} dx = \lim_{a \to 0^+} \int^1_a x^{-\frac{1}{3}}dx = \lim_{a \to 0^+} \frac{3}{2} x^{\frac{2}{3}} |^{x=1}_{x=a} = \lim_{a \to 0^+} {\frac{3}{2} - \frac{3}{2}a^{\frac{2}{3}}} = \frac{3}{2}$ .

$\Rightarrow \int^1_0 \frac{1}{\sqrt[3]{x}}dx$ converges to $\frac{3}{2}$ .

$\int^2_{-1} \frac{1}{x^3}dx = \int^0_{-1} \frac{1}{x^3} dx + \int^2_0 \frac{1}{x^3} dx = \lim_{b \to 0^-} \int^b_{-1} x^{-3} dx + \lim_{a \to 0^+} \int^2_a x^{-3}dx = \lim_{b \to 0^-} \frac{-1}{2}x^{-2} |^{x=b}_{x=-1} + \lim_{a \to 0^+} \frac{-1}{2}x^{-2} |^{x=2}_{x=a} = \lim_{b \to 0^-} (-\frac{1}{2} \cdot \frac{1}{b^2} + \frac{1}{2}) + \lim_{a \to 0^+} (-\frac{1}{8} + \frac{1}{2} \cdot \frac{1}{a^2})$

$\Rightarrow \int^2_{-1} \frac{1}{x^3}dx$ diverges.( $\int^2_{-1} \frac{1}{x^3}dx$ does not exist.)

# Case 3

# Example

$\int^{\infty}_0 \frac{1}{x(x+1)}dx$

solution

$\int^{\infty}_0 \frac{1}{x(x+1)}dx= \int^1_0\frac{1}{x(x+1)}dx + \int^{\infty}_1 \frac{1}{x(x+1)}dx = \lim_{a \to 0^+}\int^1_a \frac{1}{x(x+1)}dx + \lim_{b \to \infty}\int^b_1 \frac{1}{x(x+1)}dx \\= \lim_{a \to 0^+}ln|\frac{x}{x+1}||^1_a + \lim_{b \to \infty}ln|\frac{x}{x+1}||^b_1 = \lim_{a \to 0^+}(ln\frac{1}{2} - ln|\frac{a}{a+1}|) + \lim_{b \to \infty}(ln|\frac{b}{b+1}| - ln\frac{1}{2}) \\= -\lim_{a \to 0^+}ln|\frac{a}{a+1}| + ln1$

$\Rightarrow \int^{\infty}_0\frac{1}{x(x+1)}dx$ diverges.

# Comparative method

若無法計算瑕積分的值，可以看出瑕積分的收斂或發散

Disuss the improper integral $\int^{\infty}_1 \frac{1}{x^p}dx$ .

If $p \neq 1$ $p \neq = 1$ , $\int^b_1 \frac{1}{x^p}dx = \frac{1}{-p+1}x^{-p+1}|^{x=b}_{x=1} = \frac{1}{-p+1}(\frac{1}{b^{p-1}}-1)$ $\int_{1}^{b} \frac{1}{x ^{p}} d x = \frac{1}{- p + 1} x^{- p + 1} ∣_{x = 1}^{x = b} = \frac{1}{- p + 1} (\frac{1}{b ^{p - 1}} - 1)$ .
1. 當 $p>1$ ， $\lim_{b \to \infty}(\frac{1}{b^{p-1}}-1) = -1$ .
2. 當 $p<1$ ， $\lim_{b \to \infty}(\frac{1}{b^{p-1}}-1) = \infty$ .
  $\therefore \int^{\infty}_1 \frac{1}{x^p}dx$ is convergent if $p > 1$ .
  $\therefore \int^{\infty}_1 \frac{1}{x^p}dx$ is divergent if $p < 1$ .
If $p =1$ ,then $\int^{\infty}_1 \frac{1}{x}dx = \lim_{b \to \infty}\int^{b}_1 \frac{1}{x}dx = \lim_{b \to \infty}(ln|b| - ln|1|) = \infty$
$\Rightarrow \int^{b}_1\frac{1}{x}dx$ is divergent.

結果很重要

$\int^{\infty}_1 \frac{1}{x^p}dx$ is convergent if $p > 1$ .
$\int^{\infty}_1 \frac{1}{x^p}dx$ is divergent if $p < 1$ .
$\int^{\infty}_1 \frac{1}{x^p}dx$ is divergent if $p = 1$ .

$\int^{\infty}_2 \frac{1}{\sqrt{x-1}}dx$ 收斂 or 發散？
$\int^{\infty}_1 \frac{1}{\sqrt{x^2+3}}dx$ 收斂 or 發散？
Show that $\int^{\infty}_1 e^{-x}dx$ is convergent.

$\because 0 < \frac{1}{\sqrt{x}} < \frac{1}{\sqrt{x-1}}$ on $[2,\infty)$ .
$\therefore \int^{\infty}_2 \frac{1}{\sqrt{x}}dx < \int^{\infty}_2 \frac{1}{\sqrt{x-1}}dx$
$\Rightarrow \int^{\infty}_2 \frac{1}{\sqrt{x-1}}dx$ diverges since $\int^{\infty}_2 \frac{1}{\sqrt{x}}dx$ diverges.

$\because 0 < \frac{1}{x^2+3} < \frac{1}{x^2}$ on $[1,\infty)$ .
$\therefore \int^{\infty}_1 \frac{1}{x^2+3}dx \le \int^{\infty}_1 \frac{1}{x^2}dx$
$\Rightarrow \int^{\infty}_1 \frac{1}{x^2+3}dx$ converges since $\int^{\infty}_1 \frac{1}{x^2}dx$ converges.

$\int^{\infty}_0 e^{-x^2} dx =\int^1_0 e^{-x^2} dx + \int^{\infty}_1 e^{-x^2} dx$
$\because e^{-x} \ge e^{-x^2} \ge 0$ on $[1,\infty)$ .
$\Rightarrow \int^{b}_1 e^{-x^2} dx \le \int^{b}_1 e^{-x} dx = \frac{1}{e} - \frac{1}{e^b}$ .
$\because \lim_{b \to \infty} \int^b_1 e^{-x}dx = \frac{1}{e} \therefore \int^{\infty}_1 e^{-x}dx$ converges.
$\because \int^1_0 e^{-x^2}dx$ is a definite integral and $\int^{\infty}_1 e^{-x}dx$ is convergent.
$\therefore \int^{\infty}_0 e^{-x}dx$ is convergent.

# Reference

蘇承芳老師 - 微積分甲（一）109 學年度 - Calculus (I) Academic Year 109

數學微積分大學自修