1傅里叶级数 sin(-y)sin(-ny)d nnrdx= 0 因此,在区间(-x,π)上 √1-∞x=x 22_42Snx 当x=±π时,上式右端收敛于 f(r-0)+(x+0)√2+√2 2=f(±π) 所以,在区间[-x,π1上 2√24√2 ASn cosT (3)f(x)=ar+ br+c (ax2+ bx +c)dx 3 T 2bT + 2c 1(ax br t c)cos a r conde t rcosnxdx t oNair (ax2+ bx+c)sinnar = 因此,在区间(0,2x)上 r2+b+c+∑(42csnx innr () 0=x,(ax2+bx+c)d=1(E+ )|x are 2