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sin nx在0到

∫(pi/0)sinxsin(nx)dx=-1/2∫(pi/0)[sin(x+nx)-sin(x-nx)]dx=(1/2)∫(pi/0)[sin(1-n)x-sin(1+n)]dx=(1/2)∫(pi/0)sin(1-n)xdx-(1/2)∫(pi/0)sin(1+n)xdx=(pi/0)cos(1-n)x/[2(1-n)]-(pi/0)cos(1+n)x/[2(1+n)]当n=奇数时,上式等于=2n/(n^2-1);当n=偶数时,上式等于=2/(n^2-1).

0到2π是一个周期,如果n为奇数,积分为零.

解:∵sinxsin(nx)=(1/2)[cos(n-1)x-cos(n+1)x].∴∫(x=0,π)sinxsin(nx)dx=(1/2)[(1/(n-1))sin(n-1)x-(1/(n+1))sin(n+1)x]丨(x=0,π)=0.供参考.

积化和差公式:sin(x/3)sin(nx) = (1/2)[cos(x/3 - nx) - cos(x/3 + nx)] ∫ sin(x/3)sin(nx) dx= (1/2)∫ cos(x/3 - nx) dx - (1/2)∫ cos(x/3 + nx) dx= (1/2)[1/(1/3 - n)]∫ cos[(1/3 - n)x] d[(1/3 + n)x] - (1/2)[1/(1/3 + n)]∫ cos[(1/3 + n)x] d[(1/3 + n)x]= 1/[2(1/3 - n)]sin(x/3 -

这有啥想不明白的,当 x = 3π 时,sin(3nπ) 恒为 0 ,原级数中每项都是 0 ,当然收敛于 0 .

设f(x)=cos[2cos(x)]sin(nx)f(-x)=cos[2cos(-x)]sin(-nx)=-cos[2cos(x)]sin(nx)=-f(x)所以f(x)为奇函数.f(x+2kπ)=cos[2cos(x+2kπ)2kπ+nx)=f(x),所以f(x)是以2π为周期的奇函数.f(x)从0到2π的积分为0.

由lim(sinx/x)=1得原式=lim(sin(sinx)/sinx*sinx/x)=lim(sin(sinx)/sinx)*lim(sinx/x)=1

=lim x->0 (sin mx/mx)*(mx/nx)令t=mx=(m/n)*lim t->0 sint/t=(m/n)*1=m/n

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