📚 Improper Integrals: Infinite Limits and Unbounded Integrands | 反常积分：无穷限与无界被积函数
In A-Level Mathematics, a definite integral is usually evaluated over a finite interval with a continuous, bounded integrand. However, many important integrals involve an infinite limit of integration or a function that becomes unbounded within the interval. These are called improper integrals. In AQA exam questions, you will often be asked to decide whether an improper integral converges and, if possible, to find its exact value.
在 A-Level 数学中，定积分通常在有限区间上对连续、有界的被积函数进行求值。然而，许多重要积分涉及无穷积分限，或者被积函数在区间内趋于无界。这类积分称为反常积分。在 AQA 考试题中，你经常需要判断反常积分是否收敛，并在可能时求出其精确值。
1. What Is an Improper Integral? | 什么是反常积分？
A definite integral ∫ from a to b of f(x) dx is called proper when the interval [a, b] is finite and f is bounded on that interval. If either condition fails, the integral is improper. There are two main types: integrals with infinite limits of integration, and integrals whose integrand becomes unbounded at a point inside or at an endpoint of the interval.
当积分区间 [a, b] 有限，且被积函数 f 在该区间上有界时，定积分 ∫ 从 a 到 b 的 f(x) dx 称为正常积分。若任一条件不满足，该积分就是反常积分。主要有两类：积分限为无穷的积分，以及被积函数在区间内某点或端点处趋于无界的积分。
Recognising which type of improper integral you are dealing with is the first step, because the method of converting it into a limit is slightly different in each case.
识别你面对的是哪一类反常积分是第一步，因为将积分转化为极限的方法在两种情况下略有不同。
2. Type 1: Infinite Limits of Integration | 第一类：无穷积分限
If the upper limit is infinite, we define the improper integral as a limit of a proper integral with a finite upper limit. That is, ∫ from a to ∞ of f(x) dx = lim as b → ∞ of ∫ from a to b of f(x) dx. Similarly, if the lower limit is infinite, we use ∫ from -∞ to b of f(x) dx = lim as a → -∞ of ∫ from a to b of f(x) dx.
若上限为无穷，我们将反常积分定义为具有有限上限的正常积分的极限。即 ∫ 从 a 到 ∞ 的 f(x) dx = 当 b → ∞ 时 ∫ 从 a 到 b 的 f(x) dx 的极限。类似地，若下限为无穷，则使用 ∫ 从 -∞ 到 b 的 f(x) dx = 当 a → -∞ 时 ∫ 从 a 到 b 的 f(x) dx 的极限。
∫ from 1 to ∞ of 1/x² dx = lim as b → ∞ of ∫ from 1 to b of 1/x² dx
If this limit is a finite number, the integral converges. If the limit does not exist or is infinite, the integral diverges.
如果该极限是一个有限数，则该积分收敛。如果极限不存在或为无穷大，则该积分发散。
3. Type 2: Unbounded Integrands | 第二类：无界被积函数
An integral is also improper when the integrand has a vertical asymptote within the interval or at an endpoint. For example, ∫ from 0 to 1 of 1/√x dx is improper because 1/√x → ∞ as x → 0⁺. We handle this by replacing the problematic endpoint with a variable and taking a one-sided limit.
当被积函数在区间内或端点处具有垂直渐近线时，该积分也是反常积分。例如，∫ 从 0 到 1 的 1/√x dx 是反常的，因为当 x → 0⁺ 时，1/√x → ∞。我们通过用变量替换有问题的端点并取单侧极限来处理它。
∫ from 0 to 1 of 1/√x dx = lim as a → 0⁺ of ∫ from a to 1 of 1/√x dx
If the discontinuity occurs inside the interval, such as at x = c with a < c < b, you must split the integral at c and treat each side as a separate one-sided limit. Both sides must converge for the original integral to converge.
如果间断点出现在区间内部，例如在 a < c < b 中的 x = c，你必须将积分在 c 处拆开，并将每一边分别视为单侧极限。两边都必须收敛，原积分才收敛。
4. Convergence and Divergence | 收敛与发散
An improper integral converges if the defining limit exists as a finite number. It diverges if the limit is infinite, tends to positive or negative infinity, or fails to settle to any single value. For example, ∫ from 1 to ∞ of 1/x² dx converges, while ∫ from 1 to ∞ of 1/x dx diverges.
如果定义极限存在且为有限数，则反常积分收敛。如果极限为无穷大、趋于正无穷或负无穷，或不能稳定到任何单一值，则该积分发散。例如，∫ 从 1 到 ∞ 的 1/x² dx 收敛，而 ∫ 从 1 到 ∞ 的 1/x dx 发散。
It is important not to guess convergence from the graph alone. A function can tend to 0 as x → ∞, yet its integral may still diverge, as shown by 1/x. Always use a limit calculation or a recognised test.
不要仅根据图像猜测收敛性。一个函数在 x → ∞ 时可以趋于 0，但其积分仍可能发散，例如 1/x 就说明了这一点。务必使用极限计算或公认的检验方法。
5. The p-Integral Test | p 积分检验
For integrals of the form ∫ from 1 to ∞ of 1/x^p dx, the behaviour depends on the exponent p. By integrating x^(-p), we find that the limit is finite only when p > 1. When p ≤ 1, the integral diverges. This standard result is often used as a reference in comparison tests.
对于形如 ∫ 从 1 到 ∞ 的 1/x^p dx 的积分，其行为取决于指数 p。通过对 x^(-p) 积分，我们发现只有当 p > 1 时极限才有限。当 p ≤ 1 时，积分发散。这个标准结果常用作比较检验中的参考。
| Integral form | Converges if | Diverges if |
|---|---|---|
| ∫ from 1 to ∞ of 1/x^p dx | p > 1 | p ≤ 1 |
| ∫ from 0 to 1 of 1/x^p dx | p < 1 | p ≥ 1 |
The second row concerns unbounded integrands at the lower limit. Notice that the convergence conditions are reversed compared with the infinite upper limit case.
第二行涉及下限处无界的被积函数。注意与无穷上限情形相比，收敛条件正好相反。
6. Comparison Test for Improper Integrals | 反常积分的比较检验
Suppose 0 ≤ f(x) ≤ g(x) for all x ≥ a. If ∫ from a to ∞ of g(x) dx converges, then ∫ from a to ∞ of f(x) dx also converges. Conversely, if ∫ from a to ∞ of f(x) dx diverges, then ∫ from a to ∞ of g(x) dx also diverges.
假设对于所有 x ≥ a，都有 0 ≤ f(x) ≤ g(x)。如果 ∫ 从 a 到 ∞ 的 g(x) dx 收敛，那么 ∫ 从 a 到 ∞ 的 f(x) dx 也收敛。反过来，如果 ∫ 从 a 到 ∞ 的 f(x) dx 发散，那么 ∫ 从 a 到 ∞ 的 g(x) dx 也发散。
For example, since 1/(x² + 1) ≤ 1/x² for x ≥ 1, and ∫ from 1 to ∞ of 1/x² dx converges, the comparison test tells us that ∫ from 1 to ∞ of 1/(x² + 1) dx converges.
例如，由于当 x ≥ 1 时 1/(x² + 1) ≤ 1/x²，并且 ∫ 从 1 到 ∞ 的 1/x² dx 收敛，比较检验告诉我们 ∫ 从 1 到 ∞ 的 1/(x² + 1) dx 收敛。
This test only works for non-negative integrands. If the function changes sign, you may need to consider absolute values or split the interval.
该检验只适用于非负被积函数。如果函数变号，你可能需要考虑绝对值或拆分区间。
7. Limit Comparison Test | 极限比较检验
When f(x) and g(x) are positive for large x, and the limit of f(x)/g(x) as x → ∞ is a finite positive number L, then ∫ from a to ∞ of f(x) dx and ∫ from a to ∞ of g(x) dx either both converge or both diverge.
当 f(x) 和 g(x) 在 x 足够大时为正，且当 x → ∞ 时 f(x)/g(x) 的极限是一个有限正数 L，那么 ∫ 从 a 到 ∞ 的 f(x) dx 与 ∫ 从 a 到 ∞ 的 g(x) dx 要么都收敛，要么都发散。
For instance, to test ∫ from 2 to ∞ of 1/(x² – 1) dx, compare it with 1/x². The limit of [1/(x² – 1)] / [1/x²] = x²/(x² – 1) as x → ∞ is 1, which is finite and positive. Since ∫ from 2 to ∞ of 1/x² dx converges, the original integral converges.
例如，要检验 ∫ 从 2 到 ∞ 的 1/(x² – 1) dx，可将其与 1/x² 比较。当 x → ∞ 时，[1/(x² – 1)] / [1/x²] = x²/(x² – 1) 的极限为 1，这是有限正数。由于 ∫ 从 2 到 ∞ 的 1/x² dx 收敛，原积分收敛。
8. Evaluating Improper Integrals by Limits | 用极限求反常积分的值
To evaluate an improper integral, first write it as a limit of a proper integral. Then find an antiderivative of the integrand, substitute the finite and variable limits, and finally take the limit.
要求反常积分的值，首先将其写成正常积分的极限。然后求出被积函数的原函数，代入有限限和变量限，最后取极限。
Example 1: Evaluate ∫ from 1 to ∞ of 1/x² dx.
例 1：求 ∫ 从 1 到 ∞ 的 1/x² dx。
∫ from 1 to ∞ of 1/x² dx = lim as b → ∞ of [-1/x] from 1 to b = lim as b → ∞ of (-1/b + 1) = 1
Example 2: Evaluate ∫ from 0 to 1 of 1/√x dx.
例 2：求 ∫ 从 0 到 1 的 1/√x dx。
∫ from 0 to 1 of 1/√x dx = lim as a → 0⁺ of [2√x] from a to 1 = 2 – 0 = 2
Both examples converge and give finite values. Always check that the final limit is finite before stating the answer.
两个例子都收敛并得到有限值。在写出答案之前，务必检查最终极限是否有限。
9. Splitting Integrals Correctly | 正确拆分积分
An integral over the whole real line must be split at some finite point, often x = 0. For example, ∫ from -∞ to ∞ of f(x) dx is defined as ∫ from -∞ to 0 of f(x) dx + ∫ from 0 to ∞ of f(x) dx. Both separate integrals must converge.
在整个实数轴上的积分必须在某个有限点处拆分，通常取 x = 0。例如，∫ 从 -∞ 到 ∞ 的 f(x) dx 被定义为 ∫ 从 -∞ 到 0 的 f(x) dx + ∫ 从 0 到 ∞ 的 f(x) dx。这两个独立的积分都必须收敛。
Similarly, if there is a discontinuity at x = c inside [a, b], split as ∫ from a to c of f(x) dx + ∫ from c to b of f(x) dx, and treat both as one-sided limits. Do not try to cancel two infinite parts; this can lead to a Cauchy principal value, which is not the same as convergence.
类似地，如果 [a, b] 内 x = c 处有间断点，应拆分为 ∫ 从 a 到 c 的 f(x) dx + ∫ 从 c 到 b 的 f(x) dx，并将两者都视为单侧极限。不要试图抵消两个无穷部分；这可能导致柯西主值，而它与收敛不是一回事。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
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Forgetting to write the improper integral as a limit before substituting infinity or a discontinuous endpoint. This loses method marks.
在代入无穷或不连续端点之前，忘记将反常积分写成极限形式。这会失去方法分。
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Missing a hidden discontinuity inside the interval, for example in ∫ from 0 to 2 of 1/(x – 1)² dx. Always split at that point.
忽略区间内隐藏的间断点，例如 ∫ 从 0 到 2 的 1/(x – 1)² dx
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