一道微积分证明题……求救啊 积分上限π/2,下限0 函数是f（x）绝对值sin nx n趋向正无穷 还有一个是上限和下限
1个回答
(0,π/2)∫f(x)|sinnx|dx
换元nx=t
=1/n*(0,nπ/2)∫f(t/n)|sint|dt
=1/n*[(0,π)∫f(x/n)sinxdx-(π,2π)∫f(x/n)sinxdx+...+(-1)^k(kπ,(k+1)π))∫f(x/n)sinxdx+...]
对负项,换元x-kπ=t,从而-((k-1)π,kπ))∫f(x/n)sinxdx=(0,π))∫f[(t+k)/n]sintdt
所以原式
=1/n*[(0,π)∫f(x/n)sinxdx+(0,π)∫f[(x+π)/n]sinxdx+...+(0,π))∫f[(x+kπ)/n)sinxdx+...]
k和n之间的关系
当nπ/2为偶数时,2k=n
当nπ/2为奇数时,2k+1=n
当n→∞,有限项皆可略去,方便起见,就取2k=n
原式
=1/(2k)*{(0,π)∫f[x/(2k)]sinxdx+(0,π)∫f[(x+π)/(2k)]sinxdx+...+(0,π))∫f[(x+kπ)/(2k)sinxdx+...}
=1/(2k)*(0,π)∫sinx{f[x/(2k)]+∫f[(x+π)/(2k)]+...+f[(x+kπ)/(2k)]+...}dx
根据定积分的定义k→∞
结束项（上限）lim (x+kπ)/(2k)=t,开始项（下限）=lim x/2k=0
积分区间长=lim (kπ+x-x)/(2k)=π/2,所以lim π/2/k=lim π/(2k)=dt
所以原式
=1/π*(0,π)∫sinx[(0,π/2)∫f(t)dt]dx
=1/π*(0,π)∫sinxdx(0,π/2)∫f(t)dt
=2/π*(0,π/2)∫f(t)dt
=2/π*(0,π/2)∫f(x)dx
证毕
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