含参常义积分和含参广义积分习题课教案 任课教师:赵萃魁辅导教师:贺飞 第十七章含参常义积分 基础知识 1.关于I()=f(x,y)dr (1)I(y)在可上连续的充分条件是∫(x,y)在[a,b;c,d上连续 (2)()在[c,小上可微且r(y)=mf(x,yd的充分条件是fr,y,f(x,y) 在[ab;c,④上连续 2.关于F(y) f(a, y)d (1)F(y)在[c,d上连续的充分条件是f(x,y)在[a,b;c,d上连续, a(y),b(y)在[cd上连续且a≤a(y)≤b,a≤b(y)≤b (2)F()在小上可微且F()=MmD(x,)dx+f(0),列b(o) fa(y),ya(y)的充分条件是f(x,y),f2(x,y)在a,b;c,可上连续,同时在 cd上b(y),a(y)存在且a≤a(y)≤b,a≤b(y)≤b. 3.关于积分可交换 dymf(x,y)dx=mdmf(x,y)d的充分条件是f(x,y)在bcd 上连续。 二、本章课后习题 求F(y) 解:因为e-y在RxR上连续,所以 F()=Jm(-2)-dx+c-n,2y-c-,1 2ye-y-ey- 2.F(y)=0(x+y)f(x)dx,其中f(x)是可微的,求F"(y)
1. I(y) = b a f(x, y)dx. (1)I(y) Æ [c, d] f(x, y) Æ [a, b; c, d] (2)I(y) Æ [c, d] I (y) = b a fy(x, y)dx f(x, y), fy(x, y) Æ [a, b; c, d] 2. F(y) = b(y) a(y) f(x, y)dx. (1)F(y) Æ [c, d] f(x, y) Æ [a, b; c, d] a(y), b(y) Æ [c, d] a a(y) b, a b(y) b. (2)F(y) Æ [c, d] F (y) = b(y) a(y) fy(x, y)dx + f[b(y), y]b (y) − f[a(y), y]a (y) f(x, y), fy(x, y) Æ [a, b; c, d] Æ [c, d] b (y), a (y) Æ a a(y) b, a b(y) b. 3. d c dy b a f(x, y)dx = b a dx d c f(x, y)dy f(x, y) Æ [a, b; c, d] !"#$% 1. F(y) = y2 y e−x2ydx, F (y). !"&# e−x2y Æ R × R $' F (y) = y2 y (−x2)e−x2ydx + e−y5 · 2y − e−y3 · 1 = 2ye−y5 − e−y3 − y2 y x2e−x2ydx. 2. F(y) = y 0 (x + y)f(x)dx, %( f(x) F(y). 1
解:因为(x+y)f(x)在R×R上连续,所以 F'(y)=Jo f(a)dar+(y+y)f(y) ∫f(x)dr+2yf(y F(y)=Jo odr+f(9)+2f(y)+2yf'(y) 3f(y)+2yf(y) 3.F(y)=bmv2+y2dx,直接计算积分求F(y,F(,用定理4计 算F(0),并验证结果正确性 解:当y≠0时, T2+y212=0-S √1+y2-(1 )d ln+y2-1+00(c+)d ln√1+y2-1+y F(0)=J alo-Jo da 0-limr-0rInr-1=-1 1 ( lim r In c= lim 所以 0 F y1+y2-1+ y arctan,y≠0 这里强调指出:利用左、右导数的定义,我们可计算 + arctan - F(y)-F(0 n(1+y2 2 故F()不存在设f(x,y)=ln(x2+y2),则f(x,y)=2y,f和f二 者均在[0,1-;-∞,+∞]上不连续,不满足定理4的条件。如果死套公式计算 F(0)=f(x,y)y=dr=Jd=0就会出现错误。即使是左右导数也不 能套用定理4中求导公式计算。因此在计算前必须先验证定理4的条件
!"&# (x + y)f(x) Æ R × R $' F (y) = y 0 f(x)dx + (y + y)f(y) = y 0 f(x)dx + 2yf(y). F(y) = y 0 0dx + f(y)+2f(y)+2yf (y) = 3f(y)+2yf(y). 3. F(y) = 1 0 ln x2 + y2dx, )&'( F(y), F (0)*)* 4 ' ( F (0), ++,,--./ !"0 y = 0 F(y) = xlnx2 + y2| x=1 x=0 − 1 0 x2 x2+y2 dx = ln 1 + y2 − 1 0 (1 − 1 x2+y2 )dx = ln 1 + y2 − 1 + y 1 0 ( 1 ( x y )2+1 )d x y = ln 1 + y2 − 1 + y arctan 1 y . F(0) = 1 0 ln xdx = x ln x| 1 0 − 1 0 dx = 0 − limx→0 x ln x − 1 = −1. (∵ limx→0 x ln x = limx→0 ln x 1 x = limx→0 1 x −1 x2 = 0) $' F(y) = ⎧ ⎨ ⎩ −1, y = 0, ln1 + y2 − 1 + y arctan 1 y , y = 0. .123/4"5*0 167)289'( F +(0) = lim y→+0 F(y) − F(0) y = lim y→+0[ ln(1 + y2) 2y + arctan 1 y ] = π 2 ; F −(0) = lim y→−0 F(y) − F(0) y = lim y→−0 [ ln(1 + y2) 2y + arctan 1 y ] = −π 2 . : F (0) ; Æ< f(x, y) = ln(x2 + y2), 3 fy(x, y) = 2y x2+y2 , f = fy > 4?Æ [0, 1; −∞, +∞] ; ;@5)* 4 A-BCDE'( F (0) = 1 0 fy(x, y)|y=0dx = 1 0 dx = 0 FG4HIJKL01676; MC*)* 4 ( 6DE'(&NÆ'(OPQR+,)* 4 2
6.(2)Jo In(1-2a cos +a2)d.c (a|< 1) fif: i f(a, a)=In(1-2a cos +a2), I(a)=o In(1-2a cos x+a2)da 因为1-2ac0sx+a2≥1+a2-2a=(1-|a)2>0,所以f(x,a)在 R×(-1,1)上连续,且fa(x,a)=1=2+在R×(-1,1)上连续利用 定理2可得: T'(a)=Jo 1-20 cos x+a24. aJ7(1+1=2+14)d Cs工 arctan(a tan 5)2=0 0. 于是I(a)=C(常数),而I(0)=0,故I(a)=C=0.口 7.利用xdy=求/sin(m)==d 解:不妨设0<a<b,因为sin(mn)在0,1;a,b上连续(注意在 {0}×{a,上定义sin(mn)xy为0).所以 f o sin(In -)mrdr sin(In )dx dyo sin(In i)r 作变换x=e-t可得 yd c-(+) sin tdt(注:这是一个含参广义积分 H++uF[-(y+1)sint-cos t]e-(+)1=t 1+(1+y 于是 Jo sin(In i)=ned x dy= arctan(1+yI arctan(1+b)-arctan(1 +a) rotan b-a +b)(1+a
6.(2) π 0 ln(1 − 2α cos x + α2)dx (|α| < 1). !"< f(x, α) = ln(1−2α cos x+α2), I(α) = π 0 ln(1−2α cos x+α2)dx. &# 1 − 2α cos x + α2 1 + α2 − 2|α| = (1 − |α|)2 > 0, $' f(x, α) Æ R × (−1, 1) fα(x, α) = −2 cos x+2α 1−2α cos x+α2 Æ R × (−1, 1) . 5* )* 2 S" I (α) = π 0 −2 cos x+2α 1−2α cos x+α2 dx = 1 α π 0 (1 + α2+1 1−2α cos x+α2 )dx = π α − 1−α2 α(1+α2) π 0 1 1+ −2α 1+α2 cos x dx = π α − 2 α arctan( 1+α 1−α tan x 2 )| x=π x=0 = π α − 2 α · π 2 = 0. I(α) = CTÆ7UV I(0) = 0, : I(α) = C = 0. 7. 5* b a xydy = xb−xa ln x 1 0 sin(ln 1 x ) xb−xa ln x dx. !";W< 0 <a<b, &# sin(ln 1 x )xy Æ [0, 1; a, b] T78Æ {0} × [a, b] )2 sin(ln 1 x )xy # 0U$' 1 0 sin(ln 1 x ) xb−xa ln x dx = 1 0 sin(ln 1 x )dx b a xydy = b a dy 1 0 sin(ln 1 x )xydx 9X x = e−t S 1 0 sin(ln 1 x )xydx = +∞ 0 e−(y+1)t sin tdt( 7".:YZ[\2 ) = 1 1+(1+y)2 [−(y + 1) sin t − cost]e−(y+1)t | t=+∞ t=0 = 1 1+(1+y)2 . 1 0 sin(ln 1 x ) xb−xa ln x dx = b a 1 1+(1+y)2 dy = arctan(1 + y)| b a = arctan(1 + b) − arctan(1 + a) = arctan b−a 1+(1+b)(1+a) . 3
的十不条件是设可积分 基时存积 本和二均满究无穷广强积分∫~f(x,g)dx的性如 故死收敛性。 称∫f(x,y)dx关于y∈cd为故死收敛的,如果ve>0,3A,当 A,A>A时,wy∈e,d,套|Af(x,y)d<e或Af(x,y)dal<e 2.故死收敛的判定 如果存在F(x),使得|f(x,列≤F(x),ⅶ≤x<+∞,Vc≤y≤d.且 ∫a~F(x)dx收敛,调∫f(x,y)dm关于y∈e,d为故死收敛的 3.连续性 +sf(x,y)dx是y∈cd上的连续函数的充分条件是f(x,y)在a 上连续且∫~f(x,y)dx关于y∈[e,d为故死收敛 4.积分交换次序 dyt+f(,yd=/td/mf(a,y)hy的充分条件是fa,y)在 a,+∞;c,d上连续且∫。f(x,y)dx关于y∈c,d为故死收敛 5.可导性 (v)=∫+f(x,y)dx在[ed上可导且r(y)=Jf(x,y)dx的 充分条件是f(x,y),f(x,y)在[a,+∞;cd上连续,∫。f(x,9)dx存在且 ∫+f(x,y)d关于y∈c,d为故死收敛 二、求解:为所题 4(1).J 解:当a≤a≤b且x≥1时0<xe-<xbe-x 接为 lim In 所计e-x=0(是)(x→+∞),而~dx收敛,故/rpe-ed收敛 从而积分x°e-d在区间a≤a≤b上故死收敛 a)d ia<a< b (i)-∞<
; < ]=>?@^_`\2 +∞ a f(x, y)dx /A 1. :Bab/ c +∞ a f(x, y)dx y ∈ [c, d] #:BabA- ∀ε > 0, ∃A0, 0 A ,A>A0 ∀y ∈ [c, d], C | A A f(x, y)dx| < ε d | +∞ A f(x, y)dx| < ε. 2. :Babe) A- Æ F(x), LS |f(x, y)| F(x), ∀a x < +∞, ∀c y d. +∞ a F(x)dx ab3 +∞ a f(x, y)dx y ∈ [c, d] #:Bab 3. / +∞ a f(x, y)dx y ∈ [c, d] f7 f(x, y) Æ [a, +∞; c, d] +∞ a f(x, y)dx y ∈ [c, d] #:Bab 4. gh d c dy +∞ a f(x, y)dx = +∞ a dx d c f(x, y)dy f(x, y) Æ [a, +∞; c, d] +∞ a f(x, y)dx y ∈ [c, d] #:Bab 5. 6/ I(y) = +∞ a f(x, y)dx Æ [c, d] 6 I (y) = +∞ a fy(x, y)dx f(x, y), fy(x, y) Æ [a, +∞; c, d] +∞ a f(x, y)dx Æ +∞ a fy(x, y)dx y ∈ [c, d] #:Bab !"#$% 4.(1). +∞ 1 xαe−xdx, a α b. !"0 a α b x 1 0 < xαe−x < xb e−x &# lim x→+∞ xb e−x 1 x2 = lim x→+∞ xb+2 ex = 0 $' xb e−x = o( 1 x2 ) (x → +∞), V +∞ 1 1 x2 dx ab, : +∞ 1 xb e−xdx ab. iV +∞ 1 xαe−xdx Æjk a α b :Bab. (3) +∞ −∞ e−(x−α)2 dx (i)a<α<b (ii)−∞ <α< +∞ 4
解:(i)当a<α<b时,取正数R,使得-R<a<b<R,于是对于 切a<a<b,有la|<R,从而有 0 vr≥R,va∈(a,b) 因为积分∫e-(-dx收敛,据致收敛判别法知,∫e-(to)d关于 a<a<b一致收敛,所以∫e-(-odx=0e-(a+o)dx+te-(-aodx 关于a<a<b一致收敛 ()当-∞<a<+∞,我们可以证明∫e-(x-o)dx关于-∞<a< +∞非一致收敛。对于任意给定的A>0有 lim lim lim 取o=√/2,VA>0,取A0=A+1,由上述极限可知存在ao>0.,使 得 由一致收敛定义的反叙述可得∫e-(-o)d关于-0<a<+∞非一致 收敛进而∫c-(-o)dx关于-∞<a<+∞非一致收敛口 (5)∫ ∞- a sin d (a>0) 解:对于任意A>0,有 lim eat sin rdr= lim e -aa(-a sin A-COs A) -COs A a2+1 特别地,当A=(2k+1)(k是非负整数)时 lim (2k+1)丌=1 所以取E0=>0,VA>0,取A=(24+1x>A,由上述极限的定义可 知,存在αo,使得 sIn.C dc-1 -oo sin r 由一致收敛定义的反叙述可得∫+~c- sin. cd关于a>0非一致收敛口
!"(i) 0 a<α<b l-7 R, LS −R<a<b<R, m : n a<α<b, C |α| < R, iVC 0 < e−(x±α)2 < e−(x−R)2 , ∀x R, ∀α ∈ (a, b). &# +∞ R e−(x−R)2 dx abo:BabepqD +∞ R e−(x±α)2 dx a<α<b :Bab$' +∞ −∞ e−(x−α)2 dx = +∞ 0 e−(x+α)2 dx+ +∞ 0 e−(x−α)2 dx a<α<b :Bab (ii) 0 −∞ <α< +∞, 89',r +∞ 0 e−(x−α)2 dx −∞ <α< +∞ s:Babm t8u) A > 0 C lim α→+∞ +∞ A e−(x−α)2 dx = lim α→+∞ +∞ A−α e−t2 dx = lim α→+∞ +∞ −∞ e−t2 dx = √π. l ε0 = √π/2, ∀A > 0, l A0 = A + 1, E vwxD Æ α0 > 0, L S | +∞ A0 e−(x−α0 )2 dx − √π| < ε0 ⇒ | +∞ A0 e−(x−α0 )2 dx| > √π 2 = ε0 E:Bab)2yzvS +∞ 0 e−(x−α)2 dx −∞ <α< +∞ s:B ab{V +∞ −∞ e−(x−α)2 dx −∞ <α< +∞ s:Bab (5) +∞ 0 e−αx sin xdx (α > 0) !"m t8 A > 0, C limα→0 +∞ A e−αx sin xdx = limα→0 e−αA(−α sin A − cos A) α2 + 1 = − cos A |p}0 A = (2k + 1)π(k s~F7) limα→0 +∞ (2k+1)π e−αx sin xdx = − cos(2k + 1)π = 1. $'l ε0 = 1 2 > 0, ∀A > 0, l A0 = (2[A] + 1)π > A, E vwx)2 D Æ α0, LS | +∞ A0 e−α0x sin xdx − 1| < 1 2 ⇒ | +∞ A0 e−α0x sin xdx| > 1 2 = ε0 E:Bab)2yzvS +∞ 0 e−αx sin xdx α > 0 s:Bab 5