C.11 · 部分積分、順序の逆転および対称性
GRE Subject Test · GRE Mathematics トピック 29 3:16 英語ナレーション・英語＋中文字幕 burning-in
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This unit derives integration-by-parts reductions with valid endpoint limits, reverses a double integral by reconstructing its region, and uses reflection symmetry while checking removable endpoint behaviour.
本单元在端点极限有效的前提下推导分部积分的递推式，通过重建区域来交换二重积分的次序，并在检查可去端点的同时使用反射对称性。
Integration by parts comes from the product rule, and a definite integral keeps the boundary term at both ends.
分部积分来自乘积法则，定积分要保留两端的边界项。
For the integral from 1 to x of log t to the n, choose u as log t to the n: the lower boundary vanishes, giving I n equals x log x to the n minus n I n minus 1.
对从 1 到 x 的 (ln t)ⁿ 的积分，取 u = (ln t)ⁿ：下边界项为零，得 Iₙ = x(ln x)ⁿ − nIₙ₋₁。
For improper integrals, apply the identity on finite endpoints, then justify each limit.
对反常积分，先在有限端点上用恒等式，再逐个说明极限。
With u equals x and d v equals x e to the minus x squared, the boundary term vanishes at both infinities, so the second moment is root pi over 2, and evenness gives root pi over 4 on the half-line.
取 u = x，dv = xe^{−x²}dx，边界项在两端无穷远处都为零，所以二阶矩是 √π/2，由偶性得半直线上为 √π/4。
To reverse the order, describe the triangle: x from 0 to 1 and y from x to 1.
要交换次序，先描述三角形：x 从 0 到 1，y 从 x 到 1。
Read across instead: for each y, x from 0 to y.
再横着读：对每个 y，x 从 0 到 y。
With F equal to e to the y squared, the inner x-length supplies a factor y, and the integral becomes e minus 1 over 2.
当 F = e^{y²} 时，内层 x 的长度提供因子 y，积分变为 (e − 1)/2。
If reflecting the interval negates the integrand, the integral is zero.
若区间反射使被积函数变号，积分为零。
For sin 2m x over sin x on 0 to pi, reflection x to pi minus x reverses the numerator and keeps the denominator.
对 上的 sin(2mx)/sin x，反射 x ↦ π − x 使分子变号、分母不变。
The endpoints are removable: limits are 2m at zero and minus 2m at pi.
端点是可去的：在零处极限为 2m，在 π 处为 −2m。
The recurrence gives I 2 equals x log x squared minus 2 I 1, with I 1 equal to x log x minus x plus 1.
递推式给出 I₂ = x(ln x)² − 2I₁，其中 I₁ = x ln x − x + 1。
So I 2 is x times log squared minus 2 log plus 2, minus 2; the constant makes I 2 of 1 zero.
所以 I₂ = x − 2；常数保证 I₂(1) = 0。
In the reversed e to the y squared integral, dropping the factor y would change the value.
在交换次序后的 e^{y²} 积分中，漏掉因子 y 会改变结果。
Improper integration by parts needs endpoint limits.
反常积分的分部积分需要端点极限。
Reconstruct the region before reversing order.
交换次序前先重建区域。
And an odd-looking reflected integrand must be integrable, including any removable endpoints.
看起来关于中点反对称的被积函数必须可积，包括任何可去端点。
The region is x up to y.
区域是 x ≤ y。
Reversed, y runs from 0 to 1 and x from 0 to y; continuity on the compact triangle permits it, giving e minus 1 over 2.
交换后 y 从 0 到 1，x 从 0 到 y；在紧三角形上连续保证可以交换，得 (e − 1)/2。
The log formula follows by parts; differentiating checks the integrand and x equals 1 checks the constant.
对数公式由分部积分得出；求导检验被积函数，代入 x = 1 检验常数。
By the first-order sine limit, the endpoint values are 2m and minus 2m, so both singularities are removable and the integrand is integrable.
由一阶正弦极限，端点值分别是 2m 和 −2m，所以两个奇点都可去，被积函数可积。
Reflection flips the numerator and keeps the denominator, so the integral equals its negative: zero.
反射使分子变号、分母不变，所以积分等于它的相反数：零。
Check yourself.
自我检测。
The reduction has a minus sign.
递推式中是减号。
The half-line second moment is root pi over 4.
半直线上的二阶矩是 √π/4。
Reversed, x runs from 0 to y, and the integral is e minus 1 over 2.
交换后 x 从 0 到 y，积分为 (e − 1)/2。
And the symmetric sine integral is zero.
对称的正弦积分为零。