Improper Integrals
Definition 1. Integrals with infinite limits of integration are improper integrals of Type I.
(1) If $f(x)$ is continuous on $[0, \infty)$, then $$\int_a^{\infty} f(x) dx = \lim_{b \rightarrow \infty} \int_a^b f(x) dx.$$ (2) If $f(x)$ is continuous on $(- \infty, b]$, then $$\int_{- \infty}^b f(x) dx = \lim_{a \rightarrow - \infty} \int_a^b f(x) dx.$$ (3) If $f(x)$ is continuous on $(- \infty, \infty)$, then $$\int_{- \infty}^{\infty} f(x) dx = \int_{- \infty}^c f(x) dx + \int_c^{\infty} f(x) dx,$$ where $c$ is any real number.
In each case, if the limit exists and is finite we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges.
Definition 2. Integrals of functions that become infinite at a point within the interval of integration are improper integrals of Type II.
(1) If $f(x)$ is continuous on $(a, b]$ and discontinuous at $a$, then $$\int_a^b f(x) dx = \lim_{c \rightarrow a^+} \int_c^b f(x) dx.$$ (2) If $f(x)$ is continuous on $[a, b)$ and discontinuous at $b$, then $$\int_a^b f(x) dx = \lim_{c \rightarrow b^-} \int_a^c f(x) dx.$$ (3) If $f(x)$ is discontinuous at $c$, where $a < c < b$, and continuous on $[a, c) \cup (c, b]$, then $$\int_{a}^{b} f(x) dx = \int_{a}^c f(x) dx + \int_c^{b} f(x) dx.$$ In each case, if the limit exists and is finite we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges.
적분의 구간을 무한대로 늘리고자 할 때, 혹은 세로 점근선이 존재하는 함수를 적분하고자 할 때 위와 같은 정의가 합당함을 쉽게 동의할 수 있을 것이다.
The Integral $\int_1^{\infty} \frac{dx}{x^p}$
Theorem 1. The integral $\int_1^{\infty} \frac{dx}{x^p}$ converges if $p > 1$ and diverges if $p \leq 1$.
Proof. If $p \neq 1$, $$\int_1^{\infty} \frac{dx}{x^p} = \lim_{c \rightarrow \infty} \int_1^c x^{-p} dx = \lim_{c \rightarrow \infty} \left[ \frac{1}{1-p} x^{1-p} \right]_1^c \\ = \lim_{c \rightarrow \infty} \left( \frac{1}{1-p} c^{1-p} - \frac{1}{1-p} \right) = \begin{cases} \infty & \text{if } 1-p > 0 \\ \frac{1}{p-1} & \text{if } 1 - p <0 \end{cases}.$$ If $p=1$, then $$\int_1^{\infty} \frac{dx}{x^p} = \lim_{c \rightarrow \infty} \int_1^c \frac{1}{x} dx = \lim_{c \rightarrow \infty} (\text{ln } c) = \infty.$$ Thus if $p > 1$ the integral converges, and if $p \leq 1$ the integral diverges. $\blacksquare$
The Comparison Test
Theorem 2. Let $f$ and $g$ be continuous on $[a, \infty)$ with $0 \leq f(x) \leq g(x)$ for all $x \geq a$. Then
(1) If $\int_a^{\infty} g(x) dx$ converges, then $\int_a^{\infty} f(x) dx$ also converges.
(2) If $\int_a^{\infty} f(x) dx$ diverges, then $\int_a^{\infty} g(x) dx$ also diverges.
Proof. (1) By the property of definite integral, $$\int_a^b f(x) dx \leq \int_a^b g(x) dx, \text{ } b > a.$$ Since $\int_a^{\infty} g(x) dx$ converges, $\int_a^{\infty} f(x) dx$ also converge.
(2) From the contrapositive of (1), it is valid. $\blacksquare$
Limit Comparison Test
Theorem 3. If the positive functions $f$ and $g$ are continuous on $[a, \infty)$, and if $$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = L, \text{ } 0 < L < \infty,$$ then $$\int_{a}^{\infty} f(x) dx \text{ and } \int_{a}^{\infty} g(x) dx$$ either both converge or both diverge.