Improper Integrals: Infinite Limits and Unbounded Integrands | 反常积分：无穷限与无界被积函数

📚 Improper Integrals: Infinite Limits and Unbounded Integrands | 反常积分：无穷限与无界被积函数

An improper integral is a definite integral that cannot be evaluated directly: either one or both limits of integration are infinite, or the integrand becomes unbounded somewhere inside the interval. In AQA A-Level Mathematics the topic is treated as a direct application of limits, antiderivatives and the Fundamental Theorem of Calculus. The method never changes - rewrite the integral as a limit of proper integrals, evaluate the proper integral, then decide whether that limit exists as a finite number.

反常积分是指不能直接求值的定积分：积分限中有一个或两个为无穷大，或者被积函数在积分区间内某处无界。在 AQA A-Level 数学中，这一主题是极限、原函数与微积分基本定理的直接应用。方法始终不变——先把反常积分改写为正常积分的极限，求出这个正常积分，再判断该极限是否存在且为有限值。

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1. What Makes an Integral Improper? | 什么样的积分是反常积分

A proper definite integral requires two things: finite limits of integration, and an integrand that stays bounded on the closed interval. When both conditions hold, the Fundamental Theorem of Calculus applies immediately and the value of the integral is a single number obtained by substituting the limits into an antiderivative.

正常定积分需要满足两个条件：积分限有限，且被积函数在闭区间上有界。两个条件都成立时，可直接使用微积分基本定理，把积分限代入原函数即得一个确定的数值。

ere are exactly two families to recognise: integrals over an infinite interval, such as ∫₁^∞ f(x) dx, and integrals of an unbounded function, such as ∫₀¹ 1/√x dx. Both are called improper, but they are handled with the same limiting technique.

只要其中一个条件不满足，就会出现反常积分。需要识别的情况正好有两类：无穷区间上的积分，例如 ∫₁^∞ f(x) dx；以及无界函数的积分，例如 ∫₀¹ 1/√x dx。两者都称为反常积分，但都用同一套极限方法来处理。

The unifying idea is to remove the offending feature. Replace the infinite limit by a finite variable, or replace the singular endpoint by a point that has not yet reached the discontinuity. Then integrate normally and finally let that variable tend to its target value.

统一思路就是先把"出问题的部分"暂时移走：把无穷积分限换成有限的变量，或者把奇点端点换成一个尚未到达间断点的位置。然后正常积分，最后让这个变量趋向目标值。

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2. Type 1: Infinite Limits of Integration | 第一类：无穷积分限

For an integral whose upper limit is infinite, the definition is a limit from the right-hand end. The integral converges if this limit exists as a finite number, and diverges otherwise.

对于上限为无穷大的积分，定义就是一个趋向无穷的极限。若该极限存在且为有限值，则积分收敛；否则发散。

∫ₐ^∞ f(x) dx = lim_b→∞ ∫ₐ^b f(x) dx

infinite integral must be split at some convenient point c, and both pieces must converge separately before the whole integral can be given a value.

下限为负无穷时定义相同，只是取 a → -∞ 的极限。两端都无穷的积分必须在某个方便的点 c 处拆开，只有两部分分别收敛，整个积分才有值。

∫₋∞^∞ f(x) dx = ∫₋∞^c f(x) dx + ∫_c^∞ f(x) dx

antiderivative is -1/x, so the definite integral from 1 to b equals 1 - 1/b, which tends to 1. For 1/x, the antiderivative is ln x, and ln b grows without bound, so the integral diverges.

最经典的对照是 [1, ∞) 上的 1/x² 与 1/x。1/x² 的原函数是 -1/x，从 1 到 b 的定积分等于 1 - 1/b，趋于 1；而 1/x 的原函数是 ln x，ln b 无界增长，所以积分发散。

st as a finite number, which may happen because it grows without bound or because it oscillates.

要特别注意"发散"的含义：它并不总是指数值等于 +∞，而是指极限不存在为有限值，原因可能是无界增长，也可能是持续振荡。

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3. The p-Integral Test on Infinite Intervals | 无穷区间上的 p-积分判别法

Integrals of the form 1/x^p r depends only on the size of p. This result is worth memorising because it lets you decide convergence instantly, and it is the reference standard for the comparison test later.

形如 1/x^p 在 [1, ∞) 上的积分构成一个族，其敛散性只取决于 p 的大小。这个结论值得记住，因为它能让你立刻判断敛散性，也是后面比较判别法的参照标准。

∫₁^∞ 1/x^p dx converges when p > 1, and diverges when p ≤ 1

The intuition is practical. Larger p makes the integrand decay faster, so the tail contributes less and less. When p is exactly 1, the decay is too slow and the total area is infinite, even though the integrand tends to zero. Tending to zero is necessary for convergence but nowhere near sufficient.

直观理解很实际：p 越大，被积函数衰减越快，尾部贡献越来越小；当 p 恰好为 1 时衰减太慢，总面积无穷大，尽管被积函数趋于 0。趋于 0 是收敛的必要条件，但远远不是充分条件。

Integral | 积分	p value | p 值	Result | 结果
∫₁∞ x⁻² dx	p = 2	converges, value 1 | 收敛，值为 1
∫₁∞ x⁻³ᐟ² dx	p = 3/2	converges, value 2 | 收敛，值为 2
∫₁∞ x⁻¹ dx	p = 1	diverges | 发散
∫₁∞ x⁻¹ᐟ² dx	p = 1/2	diverges | 发散

always be checked separately, because the antiderivative is ln x rather than a power of x, and candidates frequently mis-handle this special case.

考试中 p = 1 这个临界情形必须单独讨论，因为此时原函数是 ln x 而不是 x 的幂函数，考生经常在这个特例上出错。

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4. Type 2: Unbounded Integrands | 第二类：无界被积函数

The second kind of improper integral occurs when the integrand has an infinite discontinuity somewhere in the interval. Typically the graph has a vertical asymptote at an endpoint, or inside the interval, and the function is not defined there at all.

第二类反常积分发生在被积函数在区间内存在无穷间断点的时候。通常图像在端点或区间内部有竖直渐近线，函数在该点根本没有定义。

If the singularity is at the lower endpoint a, the integral is defined by approaching a from the right with a positive variable that tends to zero. If the singularity is at the upper endpoint b, the limit comes from the left.

若奇点在下限 a，定义是让一个正变量从右侧趋向 a；若奇点在上限 b，则从左侧取极限。

∫ₐ^b f(x) dx = lim_c→a⁺ ∫_c^b f(x) dx

A clean convergent example is ∫₀¹ 1/√x dx. The antiderivative is 2√x, so the integral from c to 1 equals 2 - 2√c s c → 0⁺ this tends to 2. The region is infinitely tall at the left but has finite area 2.

典型的收敛例子是 ∫₀¹ 1/√x dx。原函数为 2√x，从 c 到 1 的积分等于 2 - 2√c，当 c → 0⁺ 时趋于 2。这个图形在左端无限高，但面积却是有限的 2。

o integrands look similar on a sketch, yet one produces finite area and the other does not.

相比之下，∫₀¹ 1/x dx 发散。原函数是 ln x，当 c → 0⁺ 时 ln c → -∞，极限无穷大。两个被积函数的图像看起来很像，但一个有有限面积，另一个没有。

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5. The p-Test Near a Singular Point | 奇点附近的 p-积分判别法

Near zero, the inequality is reversed compared with the infinite-interval case. This reversal is one of the most commonly confused facts in the whole topic, so it deserves a deliberate memory hook: at infinity, decay must be fast; near zero, the growth must be slow.

在零点附近，判别方向与无穷区间的情形正好相反。这是整个主题中最容易混淆的结论之一，值得用一句话记住：在无穷远处要求衰减足够快；在零点附近要求增长足够慢。

∫₀¹ 1/x^p dx converges when p < 1, and diverges when p ≥ 1

A worked case is ∫₀¹ x⁻²ᐟ³ dx. Here p = 2/3 < 1, so convergence is expected, and indeed the antiderivative 3x¹ᐟ³ gives 3 - 3c¹ᐟ³ → 3 as c → 0⁺. The value is finite, exactly as the test predicts.

例如 ∫₀¹ x⁻²ᐟ³ dx。此处 p = 2/3 < 1，应判为收敛；原函数为 3x¹ᐟ³，得 3 - 3c¹ᐟ³，当 c → 0⁺ 时趋于 3，与判别法完全一致。

When the integrand contains a linear expression such as 1/(x - 2)^p, the same rule applies once the singularity has been located. Identify the x-value at which the denominator vanishes, check whether it lies inside or at an endpoint of the interval, and only then compare p with 1.

若被积函数含线性表达，例如 1/(x - 2)^p，规则同样适用，但先要定位奇点：找出使分母为零的 x 值，判断它是否位于区间内部或端点，然后才比较 p 与 1。

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6. Splitting at an Interior Discontinuity | 在内部间断点处拆分

When the infinite discontinuity lies strictly inside the interval, split the integral at that point and treat each piece as a separate improper integral. Both pieces must converge. If either one diverges, the original integral diverges and no value can be assigned.

当无穷间断点位于区间内部时，必须在该点处拆开，把每一部分分别当作反常积分处理。两部分都必须收敛；只要有一部分发散，原积分就发散，不能赋予任何数值。

∫ₐ^b f(x) dx = ∫ₐ^c f(x) dx + ∫_c^b f(x) dx

, because the integrand is positive everywhere and an area cannot be negative. The reason is that the integrand is unbounded at x = 0, inside the interval, and both pieces diverge.

经典陷阱是 ∫₋₁¹ 1/x² dx。盲目使用基本定理会得到 [-1/x] 在 -1 到 1 的取值，即 -2。这显然荒谬，因为被积函数处处为正，面积不可能为负。原因是 x = 0 在区间内部且被积函数无界，两部分都发散。

Not every interior singularity causes divergence. For ∫₋₁¹ |x|⁻¹ᐟ uals 3/2 and the total is 3. Splitting is a necessary step, not a guarantee of divergence.

并非所有内部奇点都会导致发散。对 ∫₋₁¹ |x|⁻¹ᐟ³ dx 来说，奇点仍在 0，但每一半的 p = 1/3 < 1，各等于 3/2，总和为 3。拆开是必要步骤，但并不意味着一定发散。

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7. Mixed Improper Integrals and Substitution | 混合型反常积分与换元

Many exam integrals combine an infinite interval with a substitution or an integration by parts. The technique is unchanged: first find the antiderivative as a function of the original variable, then apply the limit process to the finished expression.

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