证由定理条件,f(x)可展开为 Fourier级数。记∫(x)的 Fourier 系数为a和b,则有, f∫(x)dx=-[f(π)-f(-π)]=0, f∫(x) cos nxd x f(x)cos nx f(x)sin nxd x= nb n=1,2, f(x)sin ndx=-na n=1,2, 于是 f(x)~∑(- a. nsin nx+ b ncos nx)。证 由定理条件,f (x)可展开为 Fourier 级数。记 f (x)的 Fourier 系数为a b n和 n,则有, a0 = π π 1 ( )d π f x x − 1 [ (π) ( π)] 0 π = − − = f f , an = π π 1 ( )cos d π f x nx x − π -π ( )cos π f x nx = + π π ( )sin d π n f x nx x − = nbn, n = 1,2, , bn = π π 1 ( )sin d π f x nx x − = − nan, n = 1,2, 。 于是 f (x) ~ (− sin + cos ) = a n nx b n nx n n n 1