若将f(x)作偶延拓，则有bn=0 (n=1,2,…)
an?2π2πfxcosnxdx????x?1?cosnxdx ??00ππn?2,4,6??0,????4,n?1,3,5,??2?nπ1π2πfxdx????x?1?dx?π?2 ???π0ππa0?π?24?cos?2n?1?x??从而f?x?? (0≤x≤π) 22πn?1?2n?1?31.将f(x)=2+|x| (-1≤x≤1)展开成以2为周期的傅里叶级数，并由此求级数?1的和. 2n?1n?解：f(x)在(-∞,+∞)内连续，其傅里叶级数处处收敛，由f(x)是偶函数，故bn=0,(n=1,2,…)
a0??f?x?dx?2??2?x?dx?5
?11110an??f?x?cosnxdx?2??2?x?cosnxdx?101n?2,4,6??0,????4,n?1,3,5,???nπ?2?
所以
54f?x???22π?n?1??cos?2n?1?πx?2n?1?12，x∈[-1,1]
取x=0得，?n?1??2n?1?2π2?，故 8??1111?1π2??????2? ?2224n?1n8n?1nn?1?2n?1?n?1?2n?所以?1π? 26n?1n? 313
32.将函数f(x)=x-1(0≤x≤2)展开成周期为4的余弦级数.
解：将f(x)作偶延拓，作周期延拓后函数在(-∞,+∞)上连续,则有bn=0 (n=1,2,3,…)
212a0??f?x?dx???x?1?dx?0 02?2212nπxnπxfxcosdx?x?1cosdx???????202224n?22[??1??1] nπn?2,4,6,??0,???8?22,n?1,3,5,???nπan?8故f?x???2π?n?1?1?2n?1??cos2?2n?1?πx2 (0≤x≤2)
1?x,0?x?,?a0??2s?x????ancosnπx，-∞ ?5?1?s?????f?2?2???5???????2?1?1?3???1??2?2?4?5???1?f??????f?2???2???1???????2??1???f?????2??? 5234.设函数f(x)=x2(0≤x<1)，而s?x???b，-∞ 314 f(x)展开成正弦级数得到s(x)，延拓后f(x)在x??处连续，故. ?1?s?????f?2?1?1??1?. ??????????224????21235.将下列各周期函数展开成为傅里叶级数，它们在一个周期内的表达式分别为： 11?(1)f(x)=1-x2 ???x???； ?22??2x?1,?3?x?0,(2)f?x??? 1,0?x?3.?解：（1） f(x)在(-∞,+∞)上连续，故其傅里叶级数在每一点都收敛于f(x)，由于f(x)为偶函数，有bn=0 (n=1,2,3,…) a0?2?f?x?dx?4??1?x2?dx?121?2120121?212011， 6an?2?f?x?cos2nπxdx?4??1?x2?cos2nπxdx???1?2n?12 nπ?n?1,2,??所以 111???1?f?x???2?12πn?1n2n?1cos2nπx (-∞ 3131?0??1， (2) a0???3f?x?dx????3?2x?1?dx??0dx???33?an?13nπxfxcosdx????33310nπx13nπx???2x?1?cosdx??cosdx ?3033336n?22?1???1??,?n?1,2,3,???nπ? 315 bn?13nπxfxsindx????33310nπx13nπx???2x?1?sindx??sindx ?3033336n?1???1?,?n?1,2,??nπ而函数f(x)在x=3(2k+1)，k=0,±1,±2,…处间断，故 1??6?nπxnπx?nn?16k=0,f?x??????221???1??cos???1?sin? (x≠3(2k+1)，??2n?1?nπ3nπ3?±1,±2,…) 36.把宽为τ，高为h ，周期为T的矩形波（如图所示）展开成傅里叶级数的复数形式. 解：根据图形写出函数关系式 T??0,??t???22????u?t???h,??t? 22??T?0,?t??22?1l1T1?h?2 c0??u?t?dt??Tu?t?dt??2?hdt?2l?lT?2T?2TnπT2nπ?it?it1l1cn??u?t?eldt??2Tu?t?eTdt2l?lT?2nπnπtt?1?2?i2Th??T??2?i2T2nπ????hedt???ed?it? ??????T2T?2nπi?2T??h???i2nπt?2hnπ??????sinT????nπT?2nπi??e2? 316 故该矩形波的傅里叶级数的复数形式为 nπth?h?1nπ??i2T u?t????sineTπn???nTn?0(-∞ nc?1?lf?x?e?iπlxdx?1?1ne?xe?inπx2l?l2?1dx??12??e??1?nπi?x1?nπi??1?1e?e?1 ?2???1?n?11?nπi?sinh1???1?n?1?nπi1??nπ?2故f(x)的傅里叶级数的复数形式为 ?f?x??sinh1??1?n??1?inπ?inn?eπx (x≠2k+1，k???1??nπ?2=0,±1,±2,…) 38.求矩形脉冲函数f?t????A,0?t?T?0,其他的傅氏变换 解：F????????i?xdt??T?i?x??f?t?edt?A?1?e?i?x?0Aei? 39.求下列函数的傅里叶积分： (1)f?t????e?t,t?0?0,t?0 ??1,?1?t?0(2)f?t????1,0?t?1 ??0,其他 317