Improper Integrals | 反常积分

📚 Improper Integrals | 反常积分

In A-Level Mathematics, standard integrals are defined on a closed interval with a continuous integrand. An improper integral extends this idea to intervals that are infinite or to functions that become unbounded within the integration interval. Understanding these integrals is essential for tackling advanced problems in calculus, statistics and physics.

在 A-Level 数学中，标准积分定义在闭区间上且被积函数连续。反常积分将这一概念推广到无穷区间或函数在积分区间内无界的情形。理解反常积分对于解决微积分、统计学和物理学中的高级问题至关重要。

1. Definition and Motivation | 定义与动机

An improper integral occurs when one or both limits of integration are infinite, or when the integrand has a vertical asymptote within the interval. For example, ∫₁^∞ 1/x² dx and ∫₀¹ 1/√x dx are both improper. In the first case the interval is infinite; in the second the integrand tends to infinity at x = 0.

当积分区间上限或下限为无穷大，或者被积函数在积分区间内存在垂直渐近线时，就出现了反常积分。例如，∫₁^∞ 1/x² dx 和 ∫₀¹ 1/√x dx 都是反常积分。前者区间无穷，后者被积函数在 x = 0 处趋于无穷。

To give such an integral a value, we replace the troublesome point or infinity with a variable limit, evaluate the resulting ordinary integral, and then take a limit. If this limit exists and is finite, the improper integral converges; otherwise it diverges.

为了给这类积分赋值，我们用变量替换“问题点”或无穷大，计算由此得到的普通积分，再取极限。如果极限存在且有限，则反常积分收敛；否则发散。

2. Type 1: Infinite Intervals | 第一类：无穷区间

For an integral over an infinite interval, define:

对于无穷区间上的积分，定义如下：

∫ₐ^∞ f(x) dx = lim_{t→∞} ∫ₐ^t f(x) dx

Similarly, ∫_{-∞}^b f(x) dx = lim_{t→-∞} ∫_t^b f(x) dx. For an integral over the whole real line, split at any point c: ∫_{-∞}^∞ f(x) dx = ∫_{-∞}^c f(x) dx + ∫_c^∞ f(x) dx, and both parts must converge.

类似地，∫_{-∞}^b f(x) dx = lim_{t→-∞} ∫_t^b f(x) dx。对于整个实数轴上的积分，可以在任意点 c 处拆分：∫_{-∞}^∞ f(x) dx = ∫_{-∞}^c f(x) dx + ∫_c^∞ f(x) dx，且两部分都必须收敛。

Example: ∫₁^∞ 1/x² dx. Compute lim_{t→∞} [-1/x]₁^t = lim_{t→∞} (1 – 1/t) = 1. This integral converges to 1.

例：∫₁^∞ 1/x² dx。计算 lim_{t→∞} [-1/x]₁^t = lim_{t→∞} (1 – 1/t) = 1。该积分收敛于 1。

3. Type 2: Discontinuous Integrands | 第二类：被积函数不连续

If f(x) becomes unbounded at x = a, then define ∫_a^b f(x) dx = lim_{t→a⁺} ∫_t^b f(x) dx. If the discontinuity is at the upper limit x = b, use lim_{t→b⁻} ∫_a^t f(x) dx. If the discontinuity is interior at c, split the integral at c and evaluate both one-sided improper integrals.

若 f(x) 在 x = a 处无界，则定义 ∫_a^b f(x) dx = lim_{t→a⁺} ∫_t^b f(x) dx。若间断点在上限 x = b，则使用 lim_{t→b⁻} ∫_a^t f(x) dx。若间断点在内部 c 处，则必须在 c 处拆分积分，分别计算两个单侧反常积分。

Example: ∫₀¹ 1/√x dx = lim_{t→0⁺} [2√x]_t¹ = lim_{t→0⁺} (2 – 2√t) = 2. Despite the vertical asymptote, the integral converges.

例：∫₀¹ 1/√x dx = lim_{t→0⁺} [2√x]_t¹ = lim_{t→0⁺} (2 – 2√t) = 2。尽管存在垂直渐近线，该积分仍然收敛。

4. Convergence and Divergence | 收敛与发散

An improper integral is said to converge if the defining limit exists as a finite number. If the limit is infinite or does not exist, the integral diverges. It is important to note that a divergent integral may approach +∞, -∞, or oscillate indefinitely.

如果定义中的极限存在且为有限数，则称反常积分收敛。如果极限为无穷或不存在，则积分发散。注意，发散积分可能趋于 +∞、-∞ 或无限振荡。

For example, ∫₁^∞ 1/x dx diverges to infinity, while ∫₀^∞ sin x dx diverges because the partial integrals do not settle to a single value.

例如，∫₁^∞ 1/x dx 发散到无穷大；而 ∫₀^∞ sin x dx 发散，因为部分积分无法趋于一个确定值。

In AQA marking, you must show the limiting process explicitly. Simply plugging in infinity as if it were a number will lose marks. Write “as t → ∞” or “as t → 0⁺” and evaluate the limit.

在 AQA 评分中，必须明确写出极限过程。直接将无穷大当数代入会失分。应写出“当 t → ∞”或“当 t → 0⁺”，并计算极限。

5. Evaluating by Limits | 利用极限计算

The core technique is to rewrite the improper integral with a dummy variable and then apply standard integration rules. After integrating, substitute the limits and take the limit of the resulting expression.

核心技巧是用一个变量改写反常积分，然后应用标准积分法则。积分后代入上、下限，并对所得表达式取极限。

Example: Evaluate ∫₀^∞ e^{-3x} dx.

例：计算 ∫₀^∞ e^{-3x} dx。

lim_{t→∞} ∫₀^t e^{-3x} dx = lim_{t→∞} [-(1/3)e^{-3x}]₀^t = lim_{t→∞} (1/3 – (1/3)e^{-3t}) = 1/3

Since e^{-3t} → 0 as t → ∞, the value is 1/3. Always check whether the exponential term decays before writing the final answer.

因为当 t → ∞ 时 e^{-3t} → 0，所以值为 1/3。在写出最终答案前，务必检查指数项是否衰减为零。

6. The p-Integral Test | p-积分判别法

Integrals of the form ∫₁^∞ 1/x^p dx converge if p > 1 and diverge if p ≤ 1. Similarly, ∫₀¹ 1/x^p dx converges if p < 1 and diverges if p ≥ 1. These results are extremely useful as comparison benchmarks.

形如 ∫₁^∞ 1/x^p dx 的积分在 p > 1 时收敛，在 p ≤ 1 时发散。类似地，∫₀¹ 1/x^p dx 在 p < 1 时收敛，在 p ≥ 1 时发散。这些结论作为比较基准非常有用。

Integral	Condition	Result
∫₁^∞ 1/x^p dx	p > 1	Converges to 1/(p-1)
∫₁^∞ 1/x^p dx	p ≤ 1	Diverges
∫₀¹ 1/x^p dx	p < 1	Converges to 1/(1-p)
∫₀¹ 1/x^p dx	p ≥ 1	Diverges

For example, ∫₁^∞ 1/x^{1.5} dx converges because 1.5 > 1. In contrast, ∫₁^∞ 1/√x dx diverges because p = 0.5 ≤ 1.

例如，∫₁^∞ 1/x^{1.5} dx 因 1.5 > 1 而收敛。相反，∫₁^∞ 1/√x dx 因 p = 0.5 ≤ 1 而发散。

7. Comparison Test | 比较判别法

When an integrand is too complicated to integrate directly, compare it with a known integral. If 0 ≤ f(x) ≤ g(x) for all x in the interval, and ∫ g(x) dx converges, then ∫ f(x) dx also converges. If f(x) ≥ g(x) ≥ 0 and ∫ g(x) dx diverges, then ∫ f(x) dx diverges.

当被积函数过于复杂而无法直接积分时，可将其与已知积分比较。若在区间上 0 ≤ f(x) ≤ g(x)，且 ∫ g(x) dx 收敛，则 ∫ f(x) dx 也收敛。若 f(x) ≥ g(x) ≥ 0 且 ∫ g(x) dx 发散，则 ∫ f(x) dx 发散。

Example: Does ∫₁^∞ e^{-x²} dx converge? Since 0 ≤ e^{-x²} ≤ e^{-x} for x ≥ 1, and ∫₁^∞ e^{-x} dx = 1/e converges, the original integral converges.

例：∫₁^∞ e^{-x²} dx 是否收敛？因为当 x ≥ 1 时 0 ≤ e^{-x²} ≤ e^{-x}，且 ∫₁^∞ e^{-x} dx = 1/e 收敛，所以原积分收敛。

For examinations, clearly state the comparison function and justify the inequality. A vague sentence like “it is less than something” without specifying the interval will not receive full credit.

在考试中，要清楚地说明比较函数并证明不等式。若只笼统地说“它小于某个函数”而不说明区间，则不能得到满分。

8. Absolute Convergence | 绝对收敛

An integral ∫ f(x) dx converges absolutely if ∫ |f(x)| dx converges. If the absolute value integral diverges but the original integral converges, we call it conditionally convergent. In A-Level questions, absolute convergence is used mainly to justify switching limits or applying tests.

若 ∫ |f(x)| dx 收敛，则称积分 ∫ f(x) dx 绝对收敛。若绝对值积分发散而原积分收敛，则称其为条件收敛。在 A-Level 问题中，绝对收敛主要用于证明极限交换或应用判别法时的合理性。

For example, ∫₁^∞ sin(x)/x² dx converges absolutely because |sin(x)/x²| ≤ 1/x² and ∫₁^∞ 1/x² dx converges. Meanwhile, ∫₁^∞ sin(x)/x dx converges conditionally, but this is beyond the core A-Level syllabus.

例如，∫₁^∞ sin(x)/x² dx 绝对收敛，因为 |sin(x)/x²| ≤ 1/x² 且 ∫₁^∞ 1/x² dx 收敛。另外，∫₁^∞ sin(x)/x dx 条件收敛，但这超出了 A-Level 核心大纲。

9. Applications in Probability | 在概率中的应用

Many probability density functions are defined on infinite intervals. For a continuous random variable X, the total probability is ∫_{-∞}^∞ f(x) dx = 1. Thus the integral must converge to 1. The exponential distribution has density f(x) = λe^{-λx} for x ≥ 0, and ∫₀^∞ λe^{-λx} dx = 1.

许多概率密度函数定义在无穷区间上。对于连续随机变量 X，总概率为 ∫_{-∞}^∞ f(x) dx = 1。因此该积分必须收敛到 1。指数分布的密度为 f(x) = λe^{-λx}（x ≥ 0），且 ∫₀^∞ λe^{-λx} dx = 1。

The normal distribution involves ∫_{-∞}^∞ e^{-x²/2} dx = √(2π). This is a classic improper integral that cannot be expressed in elementary functions, but its convergence can be proved by comparison with e^{-x} for large |x|.

正态分布涉及 ∫_{-∞}^∞ e^{-x²/2} dx = √(2π)。这是一个经典反常积分，无法用初等函数表示，但可以通过与大 |x| 时的 e^{-x} 比较来证明其收敛性。

10. Common Pitfalls and Exam Tips | 常见误区与考试技巧

Pitfall 1: Treating infinity as a number. Example: writing ∫₁^∞ 1/x dx = [ln x]₁^∞ = ∞ is acceptable only after stating the limit.

误区一：把无穷大当数。例如 ∫₁^∞ 1/x dx = [ln x]₁^∞ = ∞，只有在写出极限过程后才算合理。

Pitfall 2: Ignoring a discontinuity inside the interval. For ∫₋₁¹ 1/x² dx, one might compute [-1/x]₋₁¹ = -2, which is wrong because the integrand is unbounded at x = 0. The integral actually diverges.

误区二：忽略区间内部的间断点。对于 ∫₋₁¹ 1/x² dx，若直接计算 [-1/x]₋₁¹ = -2，则是错误的，因为被积函数在 x = 0 处无界。实际上该积分发散。

Tip: Always check the domain of the integrand before integrating. If there is any vertical asymptote, split the integral. Use a dummy variable t and write “lim” explicitly.

技巧：积分前务必检查被积函数的定义域。如果存在垂直渐近线，必须拆分积分。使用变量 t 并明确写出“lim”。

11. Worked Examples | 典型例题

Example A: Evaluate ∫₂^∞ 1/(x ln² x) dx.

例 A：计算 ∫₂^∞ 1/(x ln² x) dx。

Substitute u = ln x, du = 1/x dx. Then the integral becomes ∫_{ln 2}^∞ 1/u² du = [-1/u]_{ln 2}^∞ = 1/ln 2.

令 u = ln x，du = 1/x dx。则积分变为 ∫_{ln 2}^∞ 1/u² du = [-1/u]_{ln 2}^∞ = 1/ln 2。

Example B: Determine whether ∫₀^∞ xe^{-x} dx converges.

例 B：判断 ∫₀^∞ xe^{-x} dx 是否收敛。

Use integration by parts: lim_{t→∞} [-(x+1)e^{-x}]₀^t = lim_{t→∞} (1 – (t+1)e^{-t}) = 1. It converges to 1.

使用分部积分：lim_{t→∞} [-(x+1)e^{-x}]₀^t = lim_{t→∞} (1 – (t+1)e^{-t}) = 1。积分收敛于 1。

12. Summary | 总结

Improper integrals are evaluated by taking limits, either because the interval is infinite or because the integrand is unbounded. There are two types: infinite limits and discontinuous integrands. Use the p-integral test and comparison test to determine convergence quickly. Always show the limiting process explicitly to earn full marks.

反常积分通过取极限来求值，原因要么是区间无穷，要么是被积函数无界。共有两类：无穷限和间断被积函数。利用 p-积分判别法和比较判别法可快速判断收敛性。务必明确写出极限过程以获得满分。

Practise recognising improper integrals quickly: look for ∞ in the limits, or for denominators that become zero inside the interval. With careful limit evaluation, you can tackle any AQA-style improper integral question.

练习快速识别反常积分：寻找积分限中的 ∞，或积分区间内分母为零的情况。只要仔细求极限，就能应对任何 AQA 风格的反常积分问题。

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Improper Integrals | 反常积分