補(bǔ)全(-π,0)上,f(x)=-2x,使之為偶函數(shù)
a=1/π(-π,π)∫f(x)cosnxdx
=4/π(0,π)∫xcosnxdx
(0,π)∫xcosnxdx
=(0,π)[xsin(nx)/n]-(-π,π)∫sin(nx)/ndx]
=(0,π)[xsin(nx)/n]+cos(nx)/n^2]
=cos(nπ)/n^2=(-1)^n/n^2
故a=4(-1)^n/(πn^2)
a=1/π(-π,π)∫f(x)dx=2/π(0,π)∫2xdx=2/π*π^2=2π
所以f(x)=a/2+(n=1,∞)∑acosnx
=π+(n=1,∞)∑4(-1)^n/(πn^2)cosnx