Evaluating Improper Integrals · 反常积分的评估
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| improper integral/ɪmˈprɒpə ˈɪntɪɡrəl/ | 反常积分 | fǎn cháng jī fēn |
| infinite limit/ˈɪnfɪnət ˈlɪmɪt/ | 无穷极限 | wú qióng jí xiàn |
| unbounded integrand/ʌnˈbaʊndɪd ˈɪntɪɡrænd/ | 无界被积函数 | wú jiè bèi jī hán shù |
Integrating over the infinite (or the unbounded)
- A normal definite integral has finite limits and a finite integrand. An improper integral 反常积分 breaks one of those.
- Two ways it goes improper: an infinite limit 无穷极限 of integration (like $\int_1^\infty$), or an unbounded integrand 无界被积函数 (a vertical asymptote inside the interval).
- We can't just "plug in $\infty$" or evaluate at a blow-up, so we use a limit.
- The integral converges if that limit is a finite number, and diverges otherwise.
在无穷(或无界)上积分
- 普通定积分有有限的积分限和有限的被积函数。反常积分打破其中之一。
- 它变反常有两种方式:积分的无穷限(如 $\int_1^\infty$),或无界被积函数(区间内有竖直渐近线)。
- 我们不能"代入 $\infty$"或在爆发处求值,所以用一个极限。
- 若该极限是有限数,积分收敛;否则发散。
An integral is improper when... · 当积分是反常的时...
Infinite limit or unbounded integrand → improper. · 无限极限或无界被积函数 → 反常。
You can evaluate an improper integral by directly substituting $\infty$ into the antiderivative. · 可以通过直接将 $\infty$ 代入原函数来求广义积分的值。
You must take a limit instead. · 必须取极限。
Infinite limits → a limit
- Replace the infinite limit with a variable, integrate, then let it run to infinity:
-
$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx$$
- Compute the ordinary integral up to $b$, then take the limit as $b\to\infty$.
- If the limit is finite, the improper integral converges to it.
无穷限 → 一个极限
- 把无穷限换成一个变量,积分,再让它跑向无穷:
-
$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx$$
- 算出到 $b$ 的普通积分,再取 $b\to\infty$ 的极限。
- 若极限有限,反常积分就收敛到它。
The infinite tail under 1/x² · 1/x² 下方的无限尾部
y = 1/x²
The area under $1/x^2$ from $1$ to · 到 $\infty$ is finite ($=1$) even though the region never ends — a convergent improper integral. · $1/x^2$从$1$到$\infty$下的面积是有限的（$=1$），尽管区域永无止境——这是一个收敛的反常积分。
$\displaystyle\int_a^\infty f\,dx$ is defined as... · $\displaystyle\int_a^\infty f\,dx$定义为...
It is the limit of the proper integral. · 它是正常积分的极限。
Unbounded integrand → a limit at the trouble point
- If $f$ blows up at an endpoint $c$, approach it with a limit instead of evaluating there:
- $\displaystyle\int_a^c f\,dx=\lim_{t\to c^-}\int_a^t f\,dx$ (and similarly from the other side).
- $\int_0^1 \tfrac{1}{\sqrt x}\,dx$ is improper at $0$ — take $\lim_{t\to0^+}\int_t^1$.
- Same principle: sneak up on the singularity with a limit.
无界被积函数 → 在麻烦点取极限
- 若 $f$ 在端点 $c$ 爆发,用极限靠近它,而非在那里求值:
- $\displaystyle\int_a^c f\,dx=\lim_{t\to c^-}\int_a^t f\,dx$(另一侧类似)。
- $\int_0^1 \tfrac{1}{\sqrt x}\,dx$ 在 $0$ 处反常——取 $\lim_{t\to0^+}\int_t^1$。
- 同一原理:用极限悄悄逼近奇点。
Evaluate · 评价 $\displaystyle\int_1^\infty \dfrac{1}{x^2}\,dx$. · 计算 $\displaystyle\int_1^\infty \dfrac{1}{x^2}\,dx$。
$\lim_{b\to\infty}(1-\tfrac1b)=1$.
Converge or diverge
- Converges: the limit is a finite number — that number is the value.
- Diverges: the limit is infinite or does not exist — the integral has no finite value.
- A famous pair: $\int_1^\infty\tfrac{1}{x}\,dx$ diverges, but $\int_1^\infty\tfrac{1}{x^2}\,dx$ converges (to $1$).
- The difference — same-looking integrands, opposite fates — is the whole point.
收敛还是发散
- 收敛: 极限是有限数——那个数就是值。
- 发散: 极限是无穷或不存在——积分没有有限值。
- 一对著名例子:$\int_1^\infty\tfrac{1}{x}\,dx$ 发散,而 $\int_1^\infty\tfrac{1}{x^2}\,dx$ 收敛(到 $1$)。
- 这个区别——看似相同的被积函数、相反的命运——正是要点。
What happens to $\displaystyle\int_1^\infty \dfrac{1}{x}\,dx$? · $\displaystyle\int_1^\infty \dfrac{1}{x}\,dx$会发生什么？
$\lim_{b\to\infty}\ln b=\infty$ → diverges. · $\lim_{b\to\infty}\ln b=\infty$ → 发散。
An improper integral ____ if its defining limit is a finite number. · 如果定义其极限为有限数，则反常积分____。
Finite limit → converges. · 有限极限 → 收敛。
You cannot treat an improper integral like a normal one — plugging in $\infty$ or a blow-up point is meaningless. Always rewrite it as a limit and evaluate that limit. And spot interior singularities: if the integrand blows up inside $[a,b]$, split the integral at that point and handle each piece separately.
你不能把反常积分当普通积分处理——代入 $\infty$ 或爆发点毫无意义。永远把它改写成极限并求那个极限。并要发现内部奇点:若被积函数在 $[a,b]$ 内部爆发,就在那点把积分分段,分别处理每块。
Evaluate $\displaystyle\int_1^\infty \dfrac{1}{x^2}\,dx$.
- Rewrite: $\displaystyle\lim_{b\to\infty}\int_1^b x^{-2}\,dx=\lim_{b\to\infty}\Big[-\tfrac1x\Big]_1^b$.
- $=\displaystyle\lim_{b\to\infty}\Big(-\tfrac1b+1\Big)=0+1=1$.
- The limit is finite, so the integral converges to $1$.
求 $\displaystyle\int_1^\infty \dfrac{1}{x^2}\,dx$。
- 改写:$\displaystyle\lim_{b\to\infty}\int_1^b x^{-2}\,dx=\lim_{b\to\infty}\Big[-\tfrac1x\Big]_1^b$。
- $=\displaystyle\lim_{b\to\infty}\Big(-\tfrac1b+1\Big)=0+1=1$。
- 极限有限,所以积分收敛到 $1$。
An improper integral has an infinite limit or an unbounded integrand. Rewrite it as a limit of a proper integral: $\int_a^\infty f=\lim_{b\to\infty}\int_a^b f$ (or a limit approaching a singularity). It converges if the limit is finite, and diverges otherwise.
反常积分有无穷限或无界被积函数。把它改写成一个普通积分的极限:$\int_a^\infty f=\lim_{b\to\infty}\int_a^b f$(或逼近奇点的极限)。若极限有限则收敛,否则发散。