Unit B3: Improper Integrals and Numerical IntegrationB3 单元：反常积分与数值积分

Evaluate $\int_{1}^{\infty}\dfrac{dx}{x^{2}}$. By definition,

$$\int_{1}^{\infty}\frac{dx}{x^{2}}=\lim_{t\to\infty}\int_{1}^{t}x^{-2}\,dx=\lim_{t\to\infty}\Big[-\frac{1}{x}\Big]_{1}^{t}=\lim_{t\to\infty}\Big(1-\frac{1}{t}\Big)=1.$$

The integral converges to $1$. Contrast $\int_{1}^{\infty}\dfrac{dx}{x}$:

$$\lim_{t\to\infty}\int_{1}^{t}\frac{dx}{x}=\lim_{t\to\infty}\big[\ln x\big]_{1}^{t}=\lim_{t\to\infty}\ln t=\infty,$$

so this one diverges. Both integrands tend to $0$, yet only the first encloses finite area. Decay must be fast enough, and $1/x$ is exactly on the boundary $p=1$.

计算 $\int_{1}^{\infty}\dfrac{dx}{x^{2}}$。按定义：

$$\int_{1}^{\infty}\frac{dx}{x^{2}}=\lim_{t\to\infty}\int_{1}^{t}x^{-2}\,dx=\lim_{t\to\infty}\Big[-\frac{1}{x}\Big]_{1}^{t}=\lim_{t\to\infty}\Big(1-\frac{1}{t}\Big)=1.$$

积分收敛到 $1$。与之对比，$\int_{1}^{\infty}\dfrac{dx}{x}$：

$$\lim_{t\to\infty}\int_{1}^{t}\frac{dx}{x}=\lim_{t\to\infty}\big[\ln x\big]_{1}^{t}=\lim_{t\to\infty}\ln t=\infty,$$

发散。两个被积函数都趋向 $0$，但只有第一个围出有限面积。衰减速度必须足够快，而 $1/x$ 恰好处于临界点 $p=1$ 处。

Evaluate $\int_{0}^{\infty} e^{-x}\,dx$.

$$\int_{0}^{\infty} e^{-x}\,dx=\lim_{t\to\infty}\big[-e^{-x}\big]_{0}^{t}=\lim_{t\to\infty}\big(1-e^{-t}\big)=1.$$

Exponential decay always wins: $\int_{0}^{\infty}e^{-kx}\,dx=1/k$ for any $k>0$.

计算 $\int_{0}^{\infty} e^{-x}\,dx$：

$$\int_{0}^{\infty} e^{-x}\,dx=\lim_{t\to\infty}\big[-e^{-x}\big]_{0}^{t}=\lim_{t\to\infty}\big(1-e^{-t}\big)=1.$$

指数衰减总能保证收敛：对任意 $k>0$，$\int_{0}^{\infty}e^{-kx}\,dx=1/k$。

Evaluate $\int_{-\infty}^{\infty}\dfrac{dx}{1+x^{2}}$. Split at the convenient point $c=0$ and require each half to converge on its own:

$$\int_{0}^{\infty}\frac{dx}{1+x^{2}}=\lim_{t\to\infty}\big[\arctan x\big]_{0}^{t}=\lim_{t\to\infty}\arctan t=\frac{\pi}{2}.$$

By the evenness of the integrand the left half contributes the same amount,

$$\int_{-\infty}^{0}\frac{dx}{1+x^{2}}=\lim_{t\to-\infty}\big[\arctan x\big]_{t}^{0}=0-\Big(-\frac{\pi}{2}\Big)=\frac{\pi}{2}.$$

Both halves are finite, so the doubly infinite integral converges to $\tfrac{\pi}{2}+\tfrac{\pi}{2}=\pi$. The key discipline is that we evaluated two independent one-sided limits rather than a single symmetric limit.

计算 $\int_{-\infty}^{\infty}\dfrac{dx}{1+x^{2}}$。在 $c=0$ 处拆分，要求各段独立收敛：

$$\int_{0}^{\infty}\frac{dx}{1+x^{2}}=\lim_{t\to\infty}\big[\arctan x\big]_{0}^{t}=\lim_{t\to\infty}\arctan t=\frac{\pi}{2}.$$

由被积函数的偶函数性，左侧贡献相同：

$$\int_{-\infty}^{0}\frac{dx}{1+x^{2}}=\lim_{t\to-\infty}\big[\arctan x\big]_{t}^{0}=0-\Big(-\frac{\pi}{2}\Big)=\frac{\pi}{2}.$$

两段均有限，故积分收敛到 $\tfrac{\pi}{2}+\tfrac{\pi}{2}=\pi$。关键在于：我们计算了两个独立的单侧极限，而非一个对称极限。

Test $\int_{2}^{\infty}\dfrac{dx}{x\ln x}$. The integrand tends to $0$, yet the substitution $u=\ln x$, $du=dx/x$ exposes the truth:

$$\int_{2}^{\infty}\frac{dx}{x\ln x}=\lim_{t\to\infty}\int_{\ln 2}^{\ln t}\frac{du}{u}=\lim_{t\to\infty}\big[\ln u\big]_{\ln 2}^{\ln t}=\lim_{t\to\infty}\big(\ln\ln t-\ln\ln 2\big)=\infty.$$

The integral diverges. The lesson of Example 1.1 generalizes: an integrand vanishing at infinity is necessary but never sufficient for convergence. Here the decay is slower than every $1/x^{p}$ with $p>1$, so the area still accumulates without bound.

判断 $\int_{2}^{\infty}\dfrac{dx}{x\ln x}$。被积函数趋向 $0$，但换元 $u=\ln x$，$du=dx/x$ 揭示真相：

$$\int_{2}^{\infty}\frac{dx}{x\ln x}=\lim_{t\to\infty}\int_{\ln 2}^{\ln t}\frac{du}{u}=\lim_{t\to\infty}\big[\ln u\big]_{\ln 2}^{\ln t}=\lim_{t\to\infty}\big(\ln\ln t-\ln\ln 2\big)=\infty.$$

积分发散。这推广了例题 1.1 的结论：被积函数在无穷处趋零是收敛的必要条件，但远非充分条件。此处衰减比任何 $p>1$ 的 $1/x^p$ 都慢，故面积仍无限累积。

We prove that $\int_{1}^{\infty} x^{-p}\,dx$ converges if and only if $p>1$. For $p\ne 1$, the antiderivative of $x^{-p}$ is $\dfrac{x^{1-p}}{1-p}$, so

$$\int_{1}^{t} x^{-p}\,dx=\frac{x^{1-p}}{1-p}\Big|_{1}^{t}=\frac{1}{1-p}\big(t^{1-p}-1\big).$$

As $t\to\infty$, the term $t^{1-p}$ tends to $0$ when the exponent $1-p<0$, that is $p>1$, and tends to $\infty$ when $1-p>0$, that is $p<1$. Hence

$$\int_{1}^{\infty} x^{-p}\,dx=\frac{1}{p-1}\quad(p>1),\qquad \text{and diverges for }p<1.$$

The remaining case $p=1$ is exactly Example 1.1: the antiderivative is $\ln t$, which diverges. So convergence holds precisely for $p>1$, and the boundary case $p=1$ falls on the divergent side. The mirror test at $0$ in the next section is proved the same way, with the limit taken as $t\to 0^{+}$ instead, which is why the inequality flips.

证明：$\int_{1}^{\infty} x^{-p}\,dx$ 收敛当且仅当 $p>1$。当 $p\ne 1$ 时，$x^{-p}$ 的原函数（antiderivative）为 $\dfrac{x^{1-p}}{1-p}$，故

$$\int_{1}^{t} x^{-p}\,dx=\frac{x^{1-p}}{1-p}\Big|_{1}^{t}=\frac{1}{1-p}\big(t^{1-p}-1\big).$$

令 $t\to\infty$：当指数 $1-p<0$（即 $p>1$）时，$t^{1-p}\to 0$；当 $1-p>0$（即 $p<1$）时，$t^{1-p}\to\infty$。故

$$\int_{1}^{\infty} x^{-p}\,dx=\frac{1}{p-1}\quad(p>1),\qquad \text{当 }p<1\text{ 时发散。}$$

剩余情形 $p=1$ 即例题 1.1：原函数为 $\ln t$，发散。因此收敛恰在 $p>1$ 时成立，临界情形 $p=1$ 属于发散一侧。下一节中零点处的镜像判别法采用完全相同的推导，只是极限方向改为 $t\to 0^{+}$，因此不等号方向翻转。

Evaluate $\int_{0}^{1}\dfrac{dx}{\sqrt{x}}$. The integrand is unbounded as $x\to 0^{+}$. Here $p=\tfrac12<1$, so we expect convergence.

$$\int_{0}^{1}x^{-1/2}\,dx=\lim_{t\to 0^{+}}\big[2\sqrt{x}\big]_{t}^{1}=\lim_{t\to 0^{+}}\big(2-2\sqrt{t}\big)=2.$$

The area is finite even though the curve climbs to infinity at the left edge.

计算 $\int_{0}^{1}\dfrac{dx}{\sqrt{x}}$。被积函数在 $x\to 0^{+}$ 时无界，此处 $p=\tfrac12<1$，预期收敛。

$$\int_{0}^{1}x^{-1/2}\,dx=\lim_{t\to 0^{+}}\big[2\sqrt{x}\big]_{t}^{1}=\lim_{t\to 0^{+}}\big(2-2\sqrt{t}\big)=2.$$

尽管曲线在左端点趋向无穷大，面积仍然有限。

Consider $\int_{-1}^{1}\dfrac{dx}{x^{2}}$. The integrand blows up at the interior point $x=0$, so we must split:

$$\int_{-1}^{1}\frac{dx}{x^{2}}=\int_{-1}^{0}\frac{dx}{x^{2}}+\int_{0}^{1}\frac{dx}{x^{2}}.$$

Each half is a $p=2$ integral at $0$, hence diverges:

$$\int_{0}^{1}x^{-2}\,dx=\lim_{t\to 0^{+}}\Big[-\frac{1}{x}\Big]_{t}^{1}=\lim_{t\to 0^{+}}\Big(-1+\frac{1}{t}\Big)=\infty.$$

The integral diverges. Naively applying the power rule across $x=0$ gives the absurd answer $-2$; this is exactly the trap that the split is designed to expose.

考虑 $\int_{-1}^{1}\dfrac{dx}{x^{2}}$。被积函数在内部点 $x=0$ 处爆炸，必须拆分：

$$\int_{-1}^{1}\frac{dx}{x^{2}}=\int_{-1}^{0}\frac{dx}{x^{2}}+\int_{0}^{1}\frac{dx}{x^{2}}.$$

每段都是零点处 $p=2$ 的反常积分（improper integral），故发散：

$$\int_{0}^{1}x^{-2}\,dx=\lim_{t\to 0^{+}}\Big[-\frac{1}{x}\Big]_{t}^{1}=\lim_{t\to 0^{+}}\Big(-1+\frac{1}{t}\Big)=\infty.$$

整个积分发散。直接跨越 $x=0$ 套用幂函数求导公式会得到荒谬答案 $-2$；拆分恰好是为了避开这一陷阱。

Evaluate $\int_{0}^{1}\ln x\,dx$. The integrand is unbounded as $x\to 0^{+}$ (it tends to $-\infty$), so this is improper at the left endpoint. Integrate by parts with $u=\ln x$, $dv=dx$:

$$\int_{t}^{1}\ln x\,dx=\big[x\ln x-x\big]_{t}^{1}=(0-1)-(t\ln t-t)=-1-t\ln t+t.$$

As $t\to 0^{+}$ the product $t\ln t\to 0$ (the linear factor beats the logarithm), so

$$\int_{0}^{1}\ln x\,dx=\lim_{t\to 0^{+}}\big(-1-t\ln t+t\big)=-1.$$

The integral converges even though the integrand is unbounded, because the singularity is only logarithmic, far milder than any power $x^{-p}$ with $p\ge 1$.

计算 $\int_{0}^{1}\ln x\,dx$。被积函数在 $x\to 0^{+}$ 时趋向 $-\infty$，故在左端点处为反常积分。用分部积分法，令 $u=\ln x$，$dv=dx$：

$$\int_{t}^{1}\ln x\,dx=\big[x\ln x-x\big]_{t}^{1}=(0-1)-(t\ln t-t)=-1-t\ln t+t.$$

当 $t\to 0^{+}$ 时，$t\ln t\to 0$（线性因子压过对数），故

$$\int_{0}^{1}\ln x\,dx=\lim_{t\to 0^{+}}\big(-1-t\ln t+t\big)=-1.$$

虽然被积函数无界，积分仍收敛，因为此处奇性仅为对数级，远弱于任何 $p\ge 1$ 的 $x^{-p}$。

Some integrals are improper for both reasons. Consider $\int_{0}^{\infty}\dfrac{dx}{\sqrt{x}\,(1+x)}$, unbounded at $x=0$ and over an infinite interval. Split at $x=1$ to isolate the two defects:

$$\int_{0}^{\infty}\frac{dx}{\sqrt{x}\,(1+x)}=\int_{0}^{1}\frac{dx}{\sqrt{x}\,(1+x)}+\int_{1}^{\infty}\frac{dx}{\sqrt{x}\,(1+x)}.$$

Near $0$ the integrand behaves like $x^{-1/2}$, an integrable singularity ($p=\tfrac12<1$); near infinity it behaves like $x^{-3/2}$, a convergent tail ($p=\tfrac32>1$). Both pieces converge, so the whole does. The substitution $x=u^{2}$, $dx=2u\,du$ even gives the exact value:

$$\int_{0}^{\infty}\frac{2u\,du}{u\,(1+u^{2})}=2\int_{0}^{\infty}\frac{du}{1+u^{2}}=2\cdot\frac{\pi}{2}=\pi.$$

有些积分同时具有两种反常性。考虑 $\int_{0}^{\infty}\dfrac{dx}{\sqrt{x}\,(1+x)}$，在 $x=0$ 处无界且区间无穷。在 $x=1$ 处拆分，分离两处"缺陷"：

$$\int_{0}^{\infty}\frac{dx}{\sqrt{x}\,(1+x)}=\int_{0}^{1}\frac{dx}{\sqrt{x}\,(1+x)}+\int_{1}^{\infty}\frac{dx}{\sqrt{x}\,(1+x)}.$$

在 $0$ 附近被积函数行为如 $x^{-1/2}$，可积奇性（$p=\tfrac12<1$）；在无穷远处行为如 $x^{-3/2}$，收敛尾部（$p=\tfrac32>1$）。两段均收敛，整体收敛。令 $x=u^{2}$，$dx=2u\,du$ 还可得到精确值：

$$\int_{0}^{\infty}\frac{2u\,du}{u\,(1+u^{2})}=2\int_{0}^{\infty}\frac{du}{1+u^{2}}=2\cdot\frac{\pi}{2}=\pi.$$

We show $\int_{0}^{1} x^{-p}\,dx$ converges if and only if $p<1$. For $p\ne 1$,

$$\int_{t}^{1} x^{-p}\,dx=\frac{x^{1-p}}{1-p}\Big|_{t}^{1}=\frac{1}{1-p}\big(1-t^{1-p}\big).$$

Now let $t\to 0^{+}$. The term $t^{1-p}$ tends to $0$ when the exponent $1-p>0$, that is $p<1$, and blows up when $1-p<0$, that is $p>1$. Therefore

$$\int_{0}^{1} x^{-p}\,dx=\frac{1}{1-p}\quad(p<1),\qquad\text{and diverges for }p>1.$$

The case $p=1$ gives $[\ln x]_{t}^{1}=-\ln t\to\infty$, divergent. Compare with the test at infinity: there we needed $t^{1-p}\to 0$ as $t\to\infty$, which forced $1-p<0$. Near $0$ we need $t^{1-p}\to 0$ as $t\to 0^{+}$, which forces $1-p>0$. The same algebra, evaluated at opposite ends of the real line, flips the inequality. This duality is worth memorizing: small $p$ tames a singularity at $0$, large $p$ tames a tail at infinity.

证明：$\int_{0}^{1} x^{-p}\,dx$ 收敛当且仅当 $p<1$。当 $p\ne 1$ 时，

$$\int_{t}^{1} x^{-p}\,dx=\frac{x^{1-p}}{1-p}\Big|_{t}^{1}=\frac{1}{1-p}\big(1-t^{1-p}\big).$$

令 $t\to 0^{+}$：当 $1-p>0$（即 $p<1$）时，$t^{1-p}\to 0$；当 $1-p<0$（即 $p>1$）时，$t^{1-p}$ 爆炸。因此

$$\int_{0}^{1} x^{-p}\,dx=\frac{1}{1-p}\quad(p<1),\qquad p>1\text{ 时发散。}$$

$p=1$ 时得 $[\ln x]_{t}^{1}=-\ln t\to\infty$，发散。与无穷处判别法对比：在那里需要 $t^{1-p}\to 0$（$t\to\infty$），迫使 $1-p<0$；在零点处需要 $t^{1-p}\to 0$（$t\to 0^{+}$），迫使 $1-p>0$。同一套代数，在数轴两端求极限，不等号方向翻转。值得记住：小 $p$ 驯服零点处的奇性，大 $p$ 驯服无穷处的尾部。

This integrand has no elementary antiderivative, so numerical methods are essential. With $n=4$, $h=0.25$ and nodes $0,0.25,0.5,0.75,1$:

$$f(0)=1,\ f(0.25)=0.9394,\ f(0.5)=0.7788,\ f(0.75)=0.5698,\ f(1)=0.3679.$$ $$T_4=\frac{0.25}{2}\big[1+2(0.9394)+2(0.7788)+2(0.5698)+0.3679\big]=0.125\,(5.9439)=0.7430.$$

The true value is about $0.7468$, so the error is roughly $0.0038$. Doubling $n$ would cut the error by about a factor of four.

此被积函数无初等原函数，数值方法必不可少。取 $n=4$，$h=0.25$，节点 $0,0.25,0.5,0.75,1$：

$$f(0)=1,\ f(0.25)=0.9394,\ f(0.5)=0.7788,\ f(0.75)=0.5698,\ f(1)=0.3679.$$ $$T_4=\frac{0.25}{2}\big[1+2(0.9394)+2(0.7788)+2(0.5698)+0.3679\big]=0.125\,(5.9439)=0.7430.$$

真实值约为 $0.7468$，误差约 $0.0038$。将 $n$ 加倍，误差大约减小四倍。

Numerical rules shine when $f$ is known only at sample points. A sensor reports a flow rate $f(t)$ (in liters per second) at five equally spaced times over $[0,4]$ seconds, $h=1$:

$$f(0)=0,\ f(1)=3,\ f(2)=5,\ f(3)=4,\ f(4)=2.$$

The total volume is $\int_0^4 f\,dt$, estimated by

$$T_4=\frac{1}{2}\big[f(0)+2f(1)+2f(2)+2f(3)+f(4)\big]=\frac12\big[0+6+10+8+2\big]=\frac12(26)=13\ \text{liters}.$$

No formula for $f$ is needed; the trapezoidal rule only consumes the sampled values, which is why it is the workhorse of experimental data reduction.

当 $f$ 只在采样点处已知时，数值法大显身手。一个传感器在 $[0,4]$ 秒内五个等间距时刻报告流量 $f(t)$（升/秒），$h=1$：

$$f(0)=0,\ f(1)=3,\ f(2)=5,\ f(3)=4,\ f(4)=2.$$

总体积为 $\int_0^4 f\,dt$，估计为

$$T_4=\frac{1}{2}\big[f(0)+2f(1)+2f(2)+2f(3)+f(4)\big]=\frac12\big[0+6+10+8+2\big]=\frac12(26)=13\ \text{升}.$$

无需知道 $f$ 的解析式；梯形法则只消耗采样值，这正是它成为实验数据处理主力工具的原因。

Estimate $\int_0^1 e^{x}\,dx$ with $n=2$, $h=0.5$, and decide whether the estimate is high or low. Nodes $0,0.5,1$ give $f=1,\ e^{0.5}\approx1.6487,\ e\approx2.7183$:

$$T_2=\frac{0.5}{2}\big[1+2(1.6487)+2.7183\big]=0.25\,(7.0157)=1.7539.$$

The exact value is $e-1\approx1.7183$, so $T_2$ overestimates. That is expected: $e^{x}$ is concave up, so each chord lies above the curve and every trapezoid overshoots. For a concave-up integrand the trapezoidal rule always overestimates, and for concave-down it always underestimates, a fact worth using as a sanity check.

用 $n=2$，$h=0.5$ 估计 $\int_0^1 e^{x}\,dx$，并判断结果偏高还是偏低。节点 $0,0.5,1$ 给出 $f=1,\ e^{0.5}\approx1.6487,\ e\approx2.7183$：

$$T_2=\frac{0.5}{2}\big[1+2(1.6487)+2.7183\big]=0.25\,(7.0157)=1.7539.$$

精确值为 $e-1\approx1.7183$，$T_2$ 高估。这是预料之中的：$e^{x}$ 下凸（函数图像向上弯曲），每段弦在曲线上方，梯形偏大。对下凸函数，梯形法则总是高估；对上凸函数，总是低估，此规律可用于合理性验证。

On the subinterval $[x_{i-1},x_i]$, replace $f$ by the chord joining $(x_{i-1},f(x_{i-1}))$ and $(x_i,f(x_i))$. The region under that chord is a trapezoid of parallel sides $f(x_{i-1})$ and $f(x_i)$ and width $h$, so its area is $\tfrac{h}{2}\big(f(x_{i-1})+f(x_i)\big)$. Summing over all $n$ subintervals,

$$T_n=\sum_{i=1}^{n}\frac{h}{2}\big(f(x_{i-1})+f(x_i)\big)=\frac{h}{2}\sum_{i=1}^{n}\big(f(x_{i-1})+f(x_i)\big).$$

In this telescoped sum, $f(x_0)$ and $f(x_n)$ each appear once (only one trapezoid touches each endpoint), whereas every interior $f(x_k)$ appears twice (it is the right end of one trapezoid and the left end of the next). Collecting like terms,

$$T_n=\frac{h}{2}\Big[f(x_0)+2f(x_1)+\cdots+2f(x_{n-1})+f(x_n)\Big],$$

which is precisely the boxed formula. The $1,2,2,\dots,2,1$ weight pattern is not a convention to memorize; it is forced by how many trapezoids share each node.

在子区间 $[x_{i-1},x_i]$ 上，用连接 $(x_{i-1},f(x_{i-1}))$ 和 $(x_i,f(x_i))$ 的弦替代 $f$。弦下方的区域是宽为 $h$、两平行边分别为 $f(x_{i-1})$ 和 $f(x_i)$ 的梯形，面积为 $\tfrac{h}{2}\big(f(x_{i-1})+f(x_i)\big)$。对所有 $n$ 段求和，

$$T_n=\sum_{i=1}^{n}\frac{h}{2}\big(f(x_{i-1})+f(x_i)\big)=\frac{h}{2}\sum_{i=1}^{n}\big(f(x_{i-1})+f(x_i)\big).$$

在这个叠缩求和中，$f(x_0)$ 和 $f(x_n)$ 各出现一次（各端点只属于一个梯形），而每个内部 $f(x_k)$ 出现两次（它既是一个梯形的右端，也是下一个梯形的左端）。合并同类项，

$$T_n=\frac{h}{2}\Big[f(x_0)+2f(x_1)+\cdots+2f(x_{n-1})+f(x_n)\Big],$$

正是方框中的公式。$1,2,2,\dots,2,1$ 权重模式不是需要记忆的约定，而是由每个节点被多少个梯形共享所决定的。

Reusing the nodes from Example 4.1 with $h=0.25$:

$$S_4=\frac{0.25}{3}\big[1+4(0.9394)+2(0.7788)+4(0.5698)+0.3679\big].$$ $$S_4=\frac{0.25}{3}\big[1+3.7576+1.5576+2.2792+0.3679\big]=\frac{0.25}{3}(8.9623)=0.74686.$$

This matches the true value $0.74682$ to four decimals, far better than the trapezoidal estimate $0.7430$ at the same cost in function evaluations. Parabolic pieces capture the curvature that chords miss.

复用例题 4.1 的节点，$h=0.25$：

$$S_4=\frac{0.25}{3}\big[1+4(0.9394)+2(0.7788)+4(0.5698)+0.3679\big].$$ $$S_4=\frac{0.25}{3}\big[1+3.7576+1.5576+2.2792+0.3679\big]=\frac{0.25}{3}(8.9623)=0.74686.$$

与真实值 $0.74682$ 吻合到四位小数，远优于相同函数评估代价下梯形估计的 $0.7430$。抛物线能捕捉到弦所忽略的曲率。

On $[-h,h]$ Simpson's rule reads $\tfrac{h}{3}\big(f(-h)+4f(0)+f(h)\big)$. Test it on $f(x)=x^3$. The exact integral is $\int_{-h}^{h}x^3\,dx=0$ by oddness. The rule gives

$$\frac{h}{3}\big((-h)^3+4\cdot 0+h^3\big)=\frac{h}{3}(0)=0,$$

which is exact. More generally the rule is built to be exact on quadratics (three nodes determine a parabola), and the cubic term integrates to zero on the symmetric interval and is also reproduced exactly. This extra degree of exactness is why Simpson's error scales like $h^4$ rather than $h^2$.

在 $[-h,h]$ 上，辛普森法则为 $\tfrac{h}{3}\big(f(-h)+4f(0)+f(h)\big)$。用 $f(x)=x^3$ 验证。精确积分由奇函数性质得 $\int_{-h}^{h}x^3\,dx=0$。法则给出

$$\frac{h}{3}\big((-h)^3+4\cdot 0+h^3\big)=\frac{h}{3}(0)=0,$$

精确。更一般地，辛普森法则对二次多项式精确（三点确定一条抛物线），而三次项在对称区间上积分为零，同样被精确再现。这一额外精度的度是辛普森误差按 $h^4$ 而非 $h^2$ 缩减的原因。

Approximate $\int_1^3\dfrac{dx}{x}$, whose exact value is $\ln 3\approx1.0986$. With $n=2$, $h=1$, nodes $1,2,3$ and $f=1,\tfrac12,\tfrac13$:

$$S_2=\frac{1}{3}\Big[f(1)+4f(2)+f(3)\Big]=\frac13\Big[1+4\cdot\tfrac12+\tfrac13\Big]=\frac13\Big[1+2+\tfrac13\Big]=\frac13\cdot\frac{10}{3}=\frac{10}{9}\approx1.1111.$$

Even with a single parabola the estimate is within about $0.013$ of $\ln 3$. The trapezoidal rule at the same two nodes would give $\tfrac12[1+2\cdot\tfrac12+\tfrac13]\cdot\dots$ a coarser $1.1667$, so Simpson's curvature correction already buys an order of magnitude.

近似 $\int_1^3\dfrac{dx}{x}$，精确值为 $\ln 3\approx1.0986$。取 $n=2$，$h=1$，节点 $1,2,3$，函数值 $f=1,\tfrac12,\tfrac13$：

$$S_2=\frac{1}{3}\Big[f(1)+4f(2)+f(3)\Big]=\frac13\Big[1+4\cdot\tfrac12+\tfrac13\Big]=\frac13\Big[1+2+\tfrac13\Big]=\frac13\cdot\frac{10}{3}=\frac{10}{9}\approx1.1111.$$

仅用一条抛物线，估计值与 $\ln 3$ 的误差约为 $0.013$。相同两段的梯形法则给出粗糙的 $1.1667$，辛普森的曲率修正已带来一个量级的改善。

Apply Simpson's rule to $\int_0^2 x^{2}\,dx$ with $n=2$, $h=1$. Nodes $0,1,2$ give $f=0,1,4$:

$$S_2=\frac{1}{3}\big[0+4(1)+4\big]=\frac{1}{3}(8)=\frac{8}{3}.$$

The exact value is $\int_0^2 x^2\,dx=\tfrac{8}{3}$, so Simpson's rule is exact, as it must be: the rule integrates the interpolating parabola exactly, and for a quadratic integrand the parabola is the function itself. This is the qualitative content of the error bound to come, whose $f^{(4)}$ factor vanishes for any cubic.

用 $n=2$，$h=1$ 对 $\int_0^2 x^{2}\,dx$ 应用辛普森法则。节点 $0,1,2$ 给出 $f=0,1,4$：

$$S_2=\frac{1}{3}\big[0+4(1)+4\big]=\frac{1}{3}(8)=\frac{8}{3}.$$

精确值为 $\int_0^2 x^2\,dx=\tfrac{8}{3}$，辛普森法则精确，这是必然的：该法则对插值抛物线精确积分，对二次被积函数，抛物线就是函数本身。这是下文误差界的定性内容，其 $f^{(4)}$ 因子对任意三次多项式均为零。

Fit a parabola $p(x)=Ax^{2}+Bx+C$ through three equally spaced nodes, taken as $-h,0,h$ for convenience. Then $p(-h)=Ah^{2}-Bh+C$, $p(0)=C$, $p(h)=Ah^{2}+Bh+C$. Integrate the parabola exactly:

$$\int_{-h}^{h} p(x)\,dx=\Big[\frac{A}{3}x^{3}+\frac{B}{2}x^{2}+Cx\Big]_{-h}^{h}=\frac{2A}{3}h^{3}+2Ch.$$

Now express this in terms of the three sampled values. A short computation gives

$$\frac{h}{3}\big[p(-h)+4p(0)+p(h)\big]=\frac{h}{3}\big[(Ah^{2}-Bh+C)+4C+(Ah^{2}+Bh+C)\big]=\frac{h}{3}\big[2Ah^{2}+6C\big]=\frac{2A}{3}h^{3}+2Ch.$$

The two expressions agree exactly, so $\tfrac{h}{3}(1,4,1)$ reproduces the integral of any parabola through the three nodes. Patching one such formula over each consecutive pair of subintervals and adding, the shared interior nodes (counted once on each side) acquire weight $2$, while the strictly interior odd nodes keep weight $4$. That assembly produces the global pattern $1,4,2,4,2,\dots,4,1$.

用三个等间距节点（取为 $-h,0,h$ 方便起见）拟合抛物线 $p(x)=Ax^{2}+Bx+C$。则 $p(-h)=Ah^{2}-Bh+C$，$p(0)=C$，$p(h)=Ah^{2}+Bh+C$。精确积分抛物线：

$$\int_{-h}^{h} p(x)\,dx=\Big[\frac{A}{3}x^{3}+\frac{B}{2}x^{2}+Cx\Big]_{-h}^{h}=\frac{2A}{3}h^{3}+2Ch.$$

用三个采样值表示这一结果，简单计算给出

$$\frac{h}{3}\big[p(-h)+4p(0)+p(h)\big]=\frac{h}{3}\big[2Ah^{2}+6C\big]=\frac{2A}{3}h^{3}+2Ch.$$

两式完全吻合，$\tfrac{h}{3}(1,4,1)$ 精确再现了三节点抛物线的积分。将该公式覆盖每两段连续子区间并求和，共享的内部节点（两侧各计一次）权重变为 $2$，而严格内部的奇数索引节点保持权重 $4$。组合后产生全局模式 $1,4,2,4,2,\dots,4,1$。

How large must $n$ be so the trapezoidal rule approximates $\int_1^2 \tfrac{1}{x}\,dx$ within $0.0001$? Here $f(x)=1/x$, $f''(x)=2/x^3$, and on $[1,2]$ the maximum of $|f''|$ is at $x=1$, giving $K_2=2$. With $b-a=1$:

$$|E_T|\le \frac{2\cdot 1}{12\,n^2}=\frac{1}{6n^2}\le 0.0001 \iff n^2\ge \frac{1}{0.0006}\approx 1666.7 \iff n\ge 41.$$

So $n=41$ subintervals suffice (round up). Simpson's rule, with its $n^4$ denominator, would need only a handful of subintervals for the same accuracy.

梯形法则估计 $\int_1^2 \tfrac{1}{x}\,dx$ 误差在 $0.0001$ 以内，$n$ 至少需要多大？$f(x)=1/x$，$f''(x)=2/x^3$，在 $[1,2]$ 上 $|f''|$ 的最大值在 $x=1$ 处取到，$K_2=2$。$b-a=1$：

$$|E_T|\le \frac{2\cdot 1}{12\,n^2}=\frac{1}{6n^2}\le 0.0001 \iff n^2\ge \frac{1}{0.0006}\approx 1666.7 \iff n\ge 41.$$

$n=41$ 段已足够（向上取整）。辛普森法则的误差界含 $n^4$，达到相同精度只需几段。

On a single subinterval $[x_{i-1},x_i]$ of width $h$, Taylor-expanding $f$ about the midpoint and integrating shows the local trapezoidal error is $-\tfrac{h^3}{12}f''(\xi_i)$ for some $\xi_i$ in the subinterval. Summing over the $n$ subintervals and using $nh=b-a$:

$$E_T=-\frac{h^3}{12}\sum_{i=1}^{n} f''(\xi_i)=-\frac{h^2}{12}\,(b-a)\,\overline{f''},$$

where $\overline{f''}$ is an average of second derivatives. Bounding $|\overline{f''}|\le K_2$ and writing $h=(b-a)/n$ gives the stated bound with its $1/12$ and $1/n^2$. The same Taylor bookkeeping, carried to fourth order with the cubic terms cancelling, produces Simpson's $1/180$ and $1/n^4$.

在宽度为 $h$ 的单段子区间 $[x_{i-1},x_i]$ 上，对 $f$ 在中点展开泰勒级数并积分，可得局部梯形误差为 $-\tfrac{h^3}{12}f''(\xi_i)$，其中 $\xi_i$ 在子区间内。对 $n$ 段求和，用 $nh=b-a$：

$$E_T=-\frac{h^3}{12}\sum_{i=1}^{n} f''(\xi_i)=-\frac{h^2}{12}\,(b-a)\,\overline{f''},$$

其中 $\overline{f''}$ 是二阶导数的平均值。用 $|\overline{f''}|\le K_2$ 估计，代入 $h=(b-a)/n$，即得含 $1/12$ 和 $1/n^2$ 的误差界。同样的泰勒计账法推至四阶（三次项相消），产生辛普森的 $1/180$ 和 $1/n^4$。

Bound the error of Simpson's rule with $n=4$ for $\int_0^1 e^{x}\,dx$. Here $f(x)=e^{x}$ so $f^{(4)}(x)=e^{x}$, and on $[0,1]$ the maximum is $K_4=e^{1}=e\approx2.7183$. With $b-a=1$,

$$|E_S|\le\frac{K_4\,(b-a)^{5}}{180\,n^{4}}=\frac{e\cdot 1}{180\cdot 4^{4}}=\frac{2.7183}{180\cdot 256}=\frac{2.7183}{46080}\approx 5.9\times 10^{-5}.$$

So Simpson's rule with just four subintervals is guaranteed accurate to within about $0.00006$. The actual error is even smaller; the bound is a worst-case guarantee, not the realized error.

给出 $n=4$ 时辛普森法则估计 $\int_0^1 e^{x}\,dx$ 的误差界。$f(x)=e^{x}$，故 $f^{(4)}(x)=e^{x}$，在 $[0,1]$ 上最大值 $K_4=e\approx2.7183$。$b-a=1$，

$$|E_S|\le\frac{K_4\,(b-a)^{5}}{180\,n^{4}}=\frac{e\cdot 1}{180\cdot 4^{4}}=\frac{2.7183}{46080}\approx 5.9\times 10^{-5}.$$

仅四段的辛普森法则精度有保证，误差在约 $0.00006$ 以内。实际误差更小；该界是最坏情况的保证，不是实际误差。

For $\int_0^{\pi} \sin x\,dx=2$, compare the worst-case errors of the two rules with $n=10$. Here $|f''|\le 1$ and $|f^{(4)}|\le 1$, with $b-a=\pi$. The trapezoidal bound is

$$|E_T|\le\frac{1\cdot\pi^{3}}{12\cdot 10^{2}}=\frac{\pi^{3}}{1200}\approx 0.0258,$$

while Simpson's bound is

$$|E_S|\le\frac{1\cdot\pi^{5}}{180\cdot 10^{4}}=\frac{\pi^{5}}{1.8\times 10^{6}}\approx 1.70\times 10^{-4}.$$

At identical cost (eleven function evaluations) Simpson's bound is roughly $150$ times smaller. The gap widens as $n$ grows, because the bounds scale like $1/n^{2}$ versus $1/n^{4}$.

对 $\int_0^{\pi} \sin x\,dx=2$，比较 $n=10$ 时两种法则的最坏误差界。$|f''|\le 1$，$|f^{(4)}|\le 1$，$b-a=\pi$。梯形误差界：

$$|E_T|\le\frac{1\cdot\pi^{3}}{12\cdot 10^{2}}=\frac{\pi^{3}}{1200}\approx 0.0258,$$

辛普森误差界：

$$|E_S|\le\frac{1\cdot\pi^{5}}{180\cdot 10^{4}}=\frac{\pi^{5}}{1.8\times 10^{6}}\approx 1.70\times 10^{-4}.$$

相同计算量（11 次函数求值）下，辛普森误差界约小 $150$ 倍。随着 $n$ 增大，差距继续扩大，因为两个界分别按 $1/n^{2}$ 和 $1/n^{4}$ 缩减。

On one subinterval $[0,h]$, compare the exact integral with one trapezoid. Write the antiderivative $F$ with $F'=f$ and Taylor-expand about $0$. The exact integral is $\int_0^h f=F(h)-F(0)$. Expanding $F(h)=F(0)+hf(0)+\tfrac{h^{2}}{2}f'(0)+\tfrac{h^{3}}{6}f''(0)+O(h^{4})$ gives

$$\int_0^h f=hf(0)+\frac{h^{2}}{2}f'(0)+\frac{h^{3}}{6}f''(0)+O(h^{4}).$$

The trapezoid uses $\tfrac{h}{2}\big(f(0)+f(h)\big)$; expanding $f(h)=f(0)+hf'(0)+\tfrac{h^{2}}{2}f''(0)+O(h^{3})$,

$$\frac{h}{2}\big(f(0)+f(h)\big)=hf(0)+\frac{h^{2}}{2}f'(0)+\frac{h^{3}}{4}f''(0)+O(h^{4}).$$

Subtracting, the $f(0)$ and $f'(0)$ terms cancel and the local error is

$$\int_0^h f-\frac{h}{2}\big(f(0)+f(h)\big)=\Big(\frac16-\frac14\Big)h^{3}f''(0)+O(h^{4})=-\frac{h^{3}}{12}f''(0)+O(h^{4}).$$

So each subinterval contributes a local error of size $\tfrac{h^{3}}{12}|f''|$. Summing $n$ of these and using $nh=b-a$ converts the local $h^{3}$ into the global $\tfrac{(b-a)^{3}}{12n^{2}}K_2$ of the boxed bound. The factor $\tfrac16-\tfrac14=-\tfrac{1}{12}$ is the origin of the mysterious $1/12$.

在一段子区间 $[0,h]$ 上，比较精确积分与单个梯形。设原函数 $F$，$F'=f$，在 $0$ 处做泰勒展开。精确积分为 $\int_0^h f=F(h)-F(0)$。展开 $F(h)=F(0)+hf(0)+\tfrac{h^{2}}{2}f'(0)+\tfrac{h^{3}}{6}f''(0)+O(h^{4})$，得

$$\int_0^h f=hf(0)+\frac{h^{2}}{2}f'(0)+\frac{h^{3}}{6}f''(0)+O(h^{4}).$$

梯形使用 $\tfrac{h}{2}\big(f(0)+f(h)\big)$；展开 $f(h)=f(0)+hf'(0)+\tfrac{h^{2}}{2}f''(0)+O(h^{3})$，

$$\frac{h}{2}\big(f(0)+f(h)\big)=hf(0)+\frac{h^{2}}{2}f'(0)+\frac{h^{3}}{4}f''(0)+O(h^{4}).$$

相减，$f(0)$ 和 $f'(0)$ 项消去，局部误差为

$$\int_0^h f-\frac{h}{2}\big(f(0)+f(h)\big)=\Big(\frac16-\frac14\Big)h^{3}f''(0)+O(h^{4})=-\frac{h^{3}}{12}f''(0)+O(h^{4}).$$

每段子区间贡献局部误差 $\tfrac{h^{3}}{12}|f''|$。对 $n$ 段求和，用 $nh=b-a$，将局部 $h^{3}$ 转化为方框公式中全局的 $\tfrac{(b-a)^{3}}{12n^{2}}K_2$。因子 $\tfrac16-\tfrac14=-\tfrac{1}{12}$ 正是神秘 $1/12$ 的来源。

The gamma function $\Gamma(s)=\int_{0}^{\infty} x^{s-1}e^{-x}\,dx$ converges for $s>0$: near $0$ the factor $x^{s-1}$ is an integrable singularity when $s>0$, and at infinity $e^{-x}$ forces convergence. The substitution $x=u^2$ connects it to the Gaussian:

$$\Gamma\!\Big(\tfrac12\Big)=\int_{0}^{\infty} x^{-1/2}e^{-x}\,dx=\int_{0}^{\infty}\frac{e^{-u^{2}}}{u}\,2u\,du=2\int_{0}^{\infty} e^{-u^{2}}\,du=\sqrt{\pi}.$$

So a single substitution turns the half-integer gamma value into the Gaussian integral above.

伽玛函数 $\Gamma(s)=\int_{0}^{\infty} x^{s-1}e^{-x}\,dx$ 在 $s>0$ 时收敛：当 $s>0$ 时，因子 $x^{s-1}$ 在 $0$ 附近是可积奇点，而在无穷远处 $e^{-x}$ 保证收敛（convergence）。换元 $x=u^2$ 将其与高斯积分联系起来：

$$\Gamma\!\Big(\tfrac12\Big)=\int_{0}^{\infty} x^{-1/2}e^{-x}\,dx=\int_{0}^{\infty}\frac{e^{-u^{2}}}{u}\,2u\,du=2\int_{0}^{\infty} e^{-u^{2}}\,du=\sqrt{\pi}.$$

一次换元即将半整数伽玛值转化为上述高斯积分。

Consider $\int_{0}^{1}\dfrac{\ln x}{\sqrt{x}}\,dx$. The integrand is singular at $0$ and negative there, so a crude comparison is awkward. Substitute $x=t^2$, $dx=2t\,dt$:

$$\int_{0}^{1}\frac{\ln x}{\sqrt{x}}\,dx=\int_{0}^{1}\frac{2\ln t}{t}\,(2t)\,dt=4\int_{0}^{1}\ln t\,dt.$$

Now $\int_0^1 \ln t\,dt=\big[t\ln t-t\big]_0^1=-1$ (the boundary term $t\ln t\to 0$ as $t\to 0^+$), so the original integral equals $-4$. The substitution converted a singular integrand into one with a removable boundary term, after which an elementary antiderivative finished the job.

考虑 $\int_{0}^{1}\dfrac{\ln x}{\sqrt{x}}\,dx$。被积函数在 $0$ 处奇异且为负，直接比较法不便操作。令 $x=t^2$，$dx=2t\,dt$：

$$\int_{0}^{1}\frac{\ln x}{\sqrt{x}}\,dx=\int_{0}^{1}\frac{2\ln t}{t}\,(2t)\,dt=4\int_{0}^{1}\ln t\,dt.$$

现在 $\int_0^1 \ln t\,dt=\big[t\ln t-t\big]_0^1=-1$（边界项 $t\ln t\to 0$，当 $t\to 0^+$），故原积分等于 $-4$。换元将奇异被积函数转化为只含可去边界项的形式，再用初等原函数（antiderivative）完成计算。

Many transforms used in engineering are improper integrals. Compute the Laplace transform of $f(t)=1$, namely $\int_0^{\infty} e^{-st}\,dt$ for $s>0$:

$$\int_0^{\infty} e^{-st}\,dt=\lim_{b\to\infty}\Big[-\frac{1}{s}e^{-st}\Big]_0^{b}=\lim_{b\to\infty}\frac{1}{s}\big(1-e^{-sb}\big)=\frac{1}{s}.$$

The integral converges precisely because $s>0$ makes the exponential decay; for $s\le 0$ it diverges. The convergence condition $s>0$ is exactly the region of definition of the transform, a recurring theme: an improper integral is a function of its parameters, defined only where it converges.

工程中许多变换都是反常积分（improper integral）。计算 $f(t)=1$ 的拉普拉斯变换，即 $\int_0^{\infty} e^{-st}\,dt$（$s>0$）：

$$\int_0^{\infty} e^{-st}\,dt=\lim_{b\to\infty}\Big[-\frac{1}{s}e^{-st}\Big]_0^{b}=\lim_{b\to\infty}\frac{1}{s}\big(1-e^{-sb}\big)=\frac{1}{s}.$$

积分收敛（converges）恰因 $s>0$ 使指数衰减；$s\le 0$ 时发散（diverges）。收敛条件 $s>0$ 正是变换的定义域，这是一个反复出现的主题：反常积分是其参数的函数，仅在收敛之处有定义。

To evaluate $\int_1^{\infty}\dfrac{dx}{x^{2}\sqrt{1+x}}$ numerically, the infinite interval is inconvenient. The substitution $x=1/u$, $dx=-u^{-2}\,du$ maps $[1,\infty)$ to the finite interval $(0,1]$:

$$\int_1^{\infty}\frac{dx}{x^{2}\sqrt{1+x}}=\int_{1}^{0}\frac{-u^{-2}\,du}{u^{-2}\sqrt{1+1/u}}=\int_0^{1}\frac{du}{\sqrt{1+1/u}}=\int_0^{1}\frac{\sqrt{u}\,du}{\sqrt{u+1}}.$$

The transformed integrand is bounded and smooth on $[0,1]$, so Simpson's rule converges rapidly on it. This is the standard recipe for numerical work with improper integrals: change variables so the infinite tail or the singularity becomes a tame finite interval, then apply a quadrature rule whose error bounds you can trust.

对 $\int_1^{\infty}\dfrac{dx}{x^{2}\sqrt{1+x}}$ 进行数值计算时，无穷区间不便处理。换元 $x=1/u$，$dx=-u^{-2}\,du$ 将 $[1,\infty)$ 映射到有限区间 $(0,1]$：

$$\int_1^{\infty}\frac{dx}{x^{2}\sqrt{1+x}}=\int_{1}^{0}\frac{-u^{-2}\,du}{u^{-2}\sqrt{1+1/u}}=\int_0^{1}\frac{du}{\sqrt{1+1/u}}=\int_0^{1}\frac{\sqrt{u}\,du}{\sqrt{u+1}}.$$

变换后的被积函数在 $[0,1]$ 上有界光滑，辛普森法则快速收敛。这是处理反常积分（improper integral）数值计算的标准方法：换元使无穷尾部或奇异性变为可控的有限区间，再应用误差界可信赖的数值积分规则。

The same machinery decides convergence of series. Let $f$ be positive, continuous, and decreasing on $[1,\infty)$ with $a_n=f(n)$. Because $f$ decreases, on each subinterval $[n,n+1]$ we have $f(n+1)\le f(x)\le f(n)$, so integrating across $[n,n+1]$ traps the term:

$$f(n+1)\le\int_n^{n+1} f(x)\,dx\le f(n).$$

Summing the right inequality from $n=1$ to $N-1$ and the left from $n=1$ to $N-1$ gives

$$\sum_{n=2}^{N} a_n\le\int_1^{N} f(x)\,dx\le\sum_{n=1}^{N-1} a_n.$$

Letting $N\to\infty$, the partial sums and the improper integral are squeezed together: $\sum a_n$ converges if and only if $\int_1^{\infty} f$ converges. Applied to $f(x)=x^{-p}$ this recovers the $p$ series test from the $p$ integral test, the cleanest possible bridge between the two halves of this unit. The harmonic series $\sum 1/n$ diverges for exactly the reason $\int_1^{\infty}dx/x$ does.

同样的机制也能判断级数收敛（convergence）。设 $f$ 在 $[1,\infty)$ 上正值、连续、递减，且 $a_n=f(n)$。由于 $f$ 递减，在每段 $[n,n+1]$ 上有 $f(n+1)\le f(x)\le f(n)$，对 $[n,n+1]$ 积分可夹住各项：

$$f(n+1)\le\int_n^{n+1} f(x)\,dx\le f(n).$$

将右侧不等式从 $n=1$ 到 $N-1$ 求和，左侧同理：

$$\sum_{n=2}^{N} a_n\le\int_1^{N} f(x)\,dx\le\sum_{n=1}^{N-1} a_n.$$

令 $N\to\infty$，部分和与反常积分被夹紧：$\sum a_n$ 收敛当且仅当 $\int_1^{\infty} f$ 收敛。对 $f(x)=x^{-p}$ 应用此结论，可从 $p$ 积分判别法导出 $p$ 级数判别法，这是本单元两个主题之间最简洁的桥梁。调和级数 $\sum 1/n$ 发散，恰因 $\int_1^{\infty}dx/x$ 发散。