高数黎曼引理 第11题题目lim⁡n→∞∫01e−xsin⁡nx dx______ \lim_{n\rightarrow\infty}\int_{0}^{1} e^{-x}\sin nx \,dx \_\_\_\_\_\_n→∞lim​∫01​e−xsinnxdx______答案0\boldsymbol{0}0核心知识点黎曼引理结论1黎曼引理若函数f(x)f(x)f(x)在区间[a,b][a,b][a,b]上可积则lim⁡λ→∞∫abf(x)sin⁡λx dx0,lim⁡λ→∞∫abf(x)cos⁡λx dx0 \lim_{\lambda\rightarrow\infty}\int_{a}^{b} f(x)\sin\lambda x \,dx 0,\quad \lim_{\lambda\rightarrow\infty}\int_{a}^{b} f(x)\cos\lambda x \,dx 0λ→∞lim​∫ab​f(x)sinλxdx0,λ→∞lim​∫ab​f(x)cosλxdx0大白话当三角函数里面的系数趋向无穷大时sin⁡(λx)\sin(\lambda x)sin(λx)正负快速震荡一正一负相互抵消积分结果趋近于0。前提外面的f(x)f(x)f(x)在区间上可积有界、只有有限间断点就行连续函数一定满足。本题分析这里f(x)e−xf(x)e^{-x}f(x)e−x区间[0,1][0,1][0,1]e−xe^{-x}e−x在[0,1][0,1][0,1]上连续一定可积λn→∞\lambda n \to \inftyλn→∞直接套用黎曼引理lim⁡n→∞∫01e−xsin⁡nx dx0 \lim_{n\rightarrow\infty}\int_{0}^{1} e^{-x}\sin nx \,dx 0n→∞lim​∫01​e−xsinnxdx0补充分部积分严格推导考场想写过程可以用这个方法In∫01e−xsin⁡nx dx I_n\int_{0}^{1}e^{-x}\sin nx \,dxIn​∫01​e−xsinnxdx分部积分公式∫u dvuv−∫v du\int u\,dv uv-\int v\,du∫udvuv−∫vdu设ue−x, dvsin⁡nx dxue^{-x},\;dv\sin nx\,dxue−x,dvsinnxdxdu−e−xdx,v−1ncos⁡nx du-e^{-x}dx,\quad v-\frac1n\cos nxdu−e−xdx,v−n1​cosnxIn−1ne−xcos⁡nx∣01−1n∫01e−xcos⁡nx dx I_n \left.-\frac1n e^{-x}\cos nx\right|_{0}^{1}-\frac1n\int_{0}^{1}e^{-x}\cos nx\,dxIn​−n1​e−xcosnx​01​−n1​∫01​e−xcosnxdxIn1n(e0cos⁡0−e−1cos⁡n)−1n∫01e−xcos⁡nxdx I_n \frac1n\left(e^{0}\cos0 - e^{-1}\cos n\right)-\frac1n\int_{0}^{1}e^{-x}\cos nx dxIn​n1​(e0cos0−e−1cosn)−n1​∫01​e−xcosnxdx看两项1n(1−e−1cos⁡n)\dfrac1n(1 - e^{-1}\cos n)n1​(1−e−1cosn)cos⁡n\cos ncosn是有界函数绝对值≤1\le1≤1乘以1n\dfrac1nn1​n→∞n\to\inftyn→∞趋向0−1n∫01e−xcos⁡nxdx-\dfrac1n\int_{0}^{1}e^{-x}\cos nx dx−n1​∫01​e−xcosnxdx积分是有限值乘以1n\dfrac1nn1​n→∞n\to\inftyn→∞趋向0两项极限都是0lim⁡n→∞In0\boldsymbol{\lim\limits_{n\to\infty}I_n0}n→∞lim​In​0易错点总结黎曼引理条件有限闭区间f(x)f(x)f(x)可积❗ 无穷区间不能直接用黎曼引理sin⁡(nx)\sin(nx)sin(nx)高频震荡正负抵消积分趋于0选择题/填空题直接背结论秒出答案大题可以分部积分写推导。快速记忆有限区间f(x)f(x)f(x)可积sin⁡(λx)\sin(\lambda x)sin(λx)或者cos⁡(λx)\cos(\lambda x)cos(λx)λ→∞\lambda\to\inftyλ→∞积分极限0。