发布时间 : 星期日 文章§19.2含参变量的反常积分更新完毕开始阅读
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例2 证明???0??cosxy1?x2dx在(??,??)上一致收敛. 证:??y?(??,??),有: 1而?dx收敛,??01?x2??cosxy1?x2?11?x2,由M?判别法,????0??cosxy1?x2dx在(??,??)内一致收敛. 幻灯片 14
例3 证明含参量广义积分??[0,d]上一致收敛. ??
e?xysinxxdx在?0sinx证:??dx收敛,?x?0(当然,对于参量y,它在[0,d]上一致收敛),??每个固定的y?[0,d],函数g(x,y)?e?xy为x的单调函数,且对任何0?y?d,x?0,都有g(x,y)?e?xy?1?xysinx由阿贝尔判别法,得?dx在[0,d]上一致收敛.?ex?0?? 幻灯片 15
例4 证明:设f(x,y)在[a,b]?[c,??)上连续， 又?则???c??c
f(x,y)dy在[a,b)上收敛，但在x?b处发散，f(x,y)dy在[a,b)上不一致收敛. ??证:(反证法)设?cA2A1f(x,y)dy在[a,b)上一致收敛,则???0,?M?c,?A1,A2?M,?x?[a,b),有:??f(x,y)dy??, (?)又?f(x,y)在[a,b]?[A1,A2]上连续,?A2A1f(x,y)dy作为x函数,在[a,b]上连续.(?)中,令x?b?,得:?A2A1f(b,y)dy??,????cf(x,y)dy在x?b处收敛.矛盾.// 143798332.doc Page 6 of 10
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分析:??An??,且An???,??????I(x)??f(x,y)dyc??An?1?n?1?Af(x,y)dy?un(x)??nn?1对[a,b]上一致收敛的级数?un(x),用性质定理,即得 定理19.9 (连续性) n?1设f(x,y)在[a,b]?[c,??)上连续. 若含参变量广义积分 I(x)????(cfx,y)dy 在[a,b]上一致收敛, 则I(x)在[a,b]上连续. 所以,极限运算与积分运算可交换幻灯片 定理19.10 (可微性) 17
设f(x,y)与fx(x,y)在[a,b]?[c,??)上连续. 若I(x)????f(x,y)dy在[a,b]上收敛, c???f)dy在[a,b]上一致收敛, cx(x,y则I(x)在[a,b]上可微,且 I?(x)????f(x,y)dy cx分析:A?I(x)??????f(x,y)dy???n?1f(x,y)dycn?1An??I?(x)????An?1?逐项求导?????f(x,y)dy?????An?1f(x,y)dyn?1An??n?1??An??定理19.3?????An?1x,y)dy?x(x,y)dyn?1?Afx(n???fc幻灯片 定理19.11(可积性) 18
设f(x,y)在[a,b]?[c,??)上连续. 若I(x)????f(x,y)dy在[a,b]上一致收敛, c则I(x)在[a,b]上可积,且 ?bdx??f(x,y)dy?dya?c???c?bf(x,y)dx a析:?bdxa???f(x,y)dyb??分c??adx?n?1?An?1Af(x,y)dyn逐项积分??????bdxAn?1aAx,y)dy定理19.6??An?1bn?1??f(n???n?1?Adyn?af(x,y)dx????dybx,y)dxc?f(a下面我们把一致收敛的函
数项级数的和函数性质，移植给含参变量广义积
分。给出一致收敛的含参变量广义积分的性质，它们的证明方法是通过化归的思想。只给证明思想.
只给证明思想.
只给证明思想.
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当定理19.11中x的取值
定理19.12 设f(x,y)在[a,??)?[c,??)上连续. 若(i) ?敛，???c??a范围为无限区间[a,+∞)时,有如下定理
f(x,y)dx关于y在任何闭区间[c,d]一致收f(x,y)dy关于x在任何闭区间[a,b]一致收敛; (ii) 积分 与 ????a??dx?dy???c??f(x,y)dy f(x,y)dx (18) ca中有一个收敛. 则(18)中另一个积分也收敛,且 ??a??c??c??a?dx?f(x,y)dy??dy?f(x,y)dx (19) 幻灯片 20
?证明:不妨设(18)中第一个积分收敛,??a??c
??dx?f(x,y)dy也收敛,当d?c时,Id??dcddy?dy?a??af(x,y)dx????adx?dc??cf(x,y)dy???cf(x,y)dx????adx?f(x,y)dy????adx???df(x,y)dy根据条件(i)及定理19.11,可得:Id??????aAdx?d??df(x,y)dyadx???f(x,y)dy????Adx???df(x,y)dy (20) 幻灯片 21
由条件(ii),对于???0,?G?a,?A?G,有:
??A??d?选定A后,由?dx?f(x,y)dy??2??cf(x,y)dy的一致收敛性,?M?c,?d?M,有:??2??df(x,y)dy??2(A?a)把这两个结果应用到(20)式得到Id?即limId?0,d????2???这就证明了(19)式.// 143798332.doc Page 8 of 10
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例5 计算: ?pxsinbx?sinaxdx, (p?0,b?a) I???ex?0??解:?sinbx?sinaxx????bacosxydy,?pxsinbx?sinax? I??dx?ex?0???px???e?0??cosxydy?dxba????0dx?e?pxcosxydyab 幻灯片 23
由于e?pxcosxy?e?px,广义积分???0
e?pxdx收敛根据M?判别法,知???0e?pxcosxydx在[a,b]上一致收敛,由于e?pxcosxy在[0,??)?[a,b]上连续,于是,可交换积分顺序,从而,I????babdy?2??0e?pxcosxydxdy?arctanbp?arctanap.//pp?y2a 幻灯片 24
??
sinaxxdx. 例6 计算: I???解:?0例5中,取b?0,得:??asinax?F(p)??e?pxdx?arctan, (p?0)px?0由阿贝尔判别法,可知:?pxsinax含参量p的广义积分:?dx?ex?0??在p?0上一致收敛.?F(p)在p?0上连续,sinax??dx?F(0)?limF(p)?p?0?x?0???lim?arctanp?0ap??2sgna.//