数学分析(复旦大学版)课后题答案40-45

243
1lÙ ¹ëCþ2ÂÈ©
1.
y²µeQ[ a, ∞; c, d ]S¤á|f (x, y)| F (x, y)§¿ 9uy ∈ [ c, d ]È© F (x, y) dxÂñ§u uy ∈ [ c, d ]½Âñ§ ýéÂñ. y²µÏÈ© F (x, y) dx9uy ∈ [ c, d ]Âñ§ud¹ëgþ'PÂÈ©' ÜÂñ¦n§& é∀ε > 0§QyÃ9'A (ε) > a§¨A, A A §éy ∈ [ c, d ]§k F (x, y) dx < ε
−∞
−∞
−(x−α)2
−x 4
2
+∞
−x 4
2
+∞
−x 4
2
−∞
0 +∞
−(x−α)2
−∞ +∞
−(x−α)2
+∞
−t2
α → +∞
A +∞
α → +∞
A−α
−(x−α)2
+∞
A
−(x−α)2
+∞
0
−(x−α)2
−∞ 1 0
(4) (i) |xp−1 ln2 x| = xp−1 ln2 x
È©
xp0 −1 ln2 x (p
1 0
§udÃF¼êPÂÈ©§y{'4Gª§& 1 ln xy dx9uy Q[ , b ](b > 1)þÂñ. b
+∞ a A
ln
0
b dx x
Âñ
#f (x, y)Q[ a, +∞; c, d ]ë§é[ c, d)þzy§ f (x, y) dxÂñ§¢È©Qy = duÑ. y²ùÈ©Q[ c, d ]Âñ. y²µd f (x, d) dxuѧ&∃ε > 0, ∀A > a, ∃A , A A §¦ f (x, d) dx ε
1 e
p
→ +∞ (p → +0)
ØÂñ.
245
(5)
^y{. b# e α > 0¤á
+∞ 0
−αx
sin x dx
A
9 uα > 0 Â ñ § u é∀ε > 0, ∃M
A A A +∞
> 0
§ ¨A
> A
> M
§é
e−αx sin x dx < ε e−αx sin x dx < ε
Âñ cos xy dx 9uy Q(−∞, +∞)SÂñ. x +1
+∞ 0
dx π = x2 + 1 2
(3) x = 0
y
b, b > 1, 0 < x
1
§| ln xy|
| ln x| + | ln y |
− ln x + ln b = ln
1
b x
Ï l d¼§y{§&
3.
+∞
b ln b x lim x ln = lim =0 x→+0 x x→+0 x− 1 4
(2)
0
√ −αx2 αe dx (0 < α < +∞)
244
+∞
(3)
−∞
e−(x−α) dx
2
(i) a < α < b (ii) −∞ < α < +∞
1
(4)
0
xp−1 ln2 x dx
(i) p p0 > 0 (ii) p > 0
+∞
(5)
e−αx sin x dx (α > 0)
)µ
π −1 1 p 2−p
1 π −1 π sin x sin x sin x sin x dx = dx + dx + dx p (π − x)2−p p (π − x)2−p p (π − x)2−p p (π − x)2−p x x x x 0 0 1 π −1 1 sin x dx p 2−p 0 x (π − x) sin x sin x (0 x 1, 0 < p1 p p2 < 2) p 2 − p p 2 x (π − x) x (π − x)2−p2 sin x 1 lim xp2 −1 p = 2−p 2 − p 2 2 2 x→+0 x (π − x) π 1 sin x p2 < 2 p2 − 1 < 1 dx p2 (π − x)2−p2 x 0 1 sin x dx p ∈ [ p1 , p2 ] p (π − x)2−p x 0 1 sin x sin x (0 , 1 ] × [ p , p ] dx [ p1 , p2 ] 1 2 p (π − x)2−p xp (π − x)2−p x 0 π
dx [ p1 , p2 ]
Q
ë
2−p
dx [ p1 , p2 ]
Q
ë
6.
α −x b −x +∞ 2 b −x b −x x→+∞ +∞ 1 α −x 1 +∞ −αx2 0 +∞ −αx2 +∞ −t2 +∞ −t2 α→+0 A α→+0 √ αA +∞ A 0 0 0 0 −α0 x2 +∞ 0 −α0 x2 0 +∞ A −αx
2
0
+∞
−(x−α)2
+∞
−(x−α)2
+∞
p0 > 0, 0
x
1)
x
p−1
ln x dx =
0
2
e
4
−p0 z 2
z dz
ud ܧy{'4Gª l d¼§y{§& x 1 (ii) Ϩx ∈ 0, , ln x 1 e
1 0 2
z →+∞
lim z 2 · e−p0 z z 2 = lim
z → +∞
z = 0 (p0 > 0) ep0 z
A 0 0 +∞ −αx 0 +∞ 0 2 2 π
+∞
sin x dx
0
uѧugñ§%
(2) F (p) =
0
y²µ α (1) #F (α) = dx. x +α é?Ûα = 0§Ø#α > 0§Vδ > 0§¦&α − δ > 0§eyF (α)Q[ α − δ, α + δ ]SÂñ α α +δ ¯¢þ§¨α ∈ [ α − δ, α + δ ]§ x + α x + (α − δ ) α +δ ÏÈ© (α − dxÂñ§ud¼§y{§&F (α)Q[ α − δ, α + δ ]þ9uαÂñ δ) + x u´dë½n§&F (α)QT«mþ´α'ë¼ê§AyQα Xë α duα = 0'?¿S§& x + dxé?Ûα = 0ë§ddF (α)Q?Ûعα = 0'«mþ α Ñë α π α π ¢d lim dx = , lim dx = − α +x 2 α +x 2 α &F (α)Qα = 0?Øë§u x + α dxQعα = 0'?Û«mþ´ë¼ê. (2) ?p ∈ (0, 2)§uQ0 < p , p < 2§¦0 < p p p <2 Ï0ÚπþU´ÛX§òÈ©©nã
2
2 +∞
x2
1 + a2
0
cos xy x2 + y 2 1 x2 + 1
2பைடு நூலகம்
Âñ dx 9uy Q[ a, +∞)(a > 0)SÂñ.
+∞ 0
x2
dx π = 2 +a 2a
(2)
Ï u´d¼§y{§& ´ÛX§¨ 1 b
1 4
cos xy y ∈ (−∞, +∞), 2 x +1
+∞ 0
Âñ
Q
ë
246
sin x x Q [ 1, π −1 ]×[ p , p ]ë§udëS½n§& Ï&ȼê x (πsin − x) x (π − x) x éu x (πsin dx − x) x sin(π − x) Ï x (πsin (π − 1 x π, 0 < p p p < 2) − x) x (π − x) π − x) 1 lim (π − x) x sin( = (π − x) π π − x) Ïp > 0§u1 − p < 1§u´d ܧy{'4Gª§& x sin( dxÂñ (π − x) x l d¼§y{§& x (πsin dx9up ∈ [ p , p ]Âñ − x) x x Q[ π−1, π)×[ p , p ]þë§udëS½n§& x (πsin q&ȼê x (πsin − x) − x) nܱþ§&F (p)Q[ p , p ]ë§l QÙþ?Xpë x dxQ(0, 2)Së.. qdp ∈ (0, 2)'?¿S§&F (p) = x (πsin − x)
+∞ a a +∞ a A 0 0 A A A A A A A +∞ +∞ a
+∞
f (x, y ) dx
9
F (x, y ) dx < εéy ∈ [ c, d ]Ѥá |f (x, y )| dx f (x, y) dx dá¹ëgþPÂÈ©' ÜÂñ¦n§ f (x, y) dx9uy ∈ [ c, d ]Âñ§ uy ∈ [ c, d ]Âñ u f (x, y) dx9uy ∈ [ c, d ]Âñ ýéÂñ. 2. y²eÈ©Q¤½'«mSÂñµ
π −1 p 2−p 1 2 1 p π π −1 p 2−p p 2−p p1 2−p1 1 2 1−p1 x→π −0 1 p1 2−p1 p1 π 1 π −1 p−1 2−p1 π π −1 p 2−p 1 2 π p 2−p 1 2 π −1 p 1 2 π 0 p 2−p +∞ +∞
2−p
+∞ 0 2 2 0 0 0 0 0 0 0 0 2 2 2 0 2 +∞ 0 0 0 2 2 0 0 0 +∞ 0 0 2 2 +∞ +∞ α→+0 0 2 2 α→−0 0 2 2 +∞ 0 2 2 1 2 1 2
sin x dx (0, 2) xp (π − x)2−p
Q
Së.
éu Ï Ï §u §u´d ܧy{'4Gª§& l d¼§y{§& 9u Âñ q&ȼê Q þë§udëS½n§& sin x dx´¹ëgþ'~ÂÈ© x (π − x)
0 0 0 0 a A A +∞ a 0
ùv²éy = d ∈ [ c, d ]k f (x, y) dx ε §`² 4. ?ØeÈ©Q½«m'ÂñSµ (1) x e dx (a α b; a, b?¿¢ê)
A +∞ α −x 1 +∞
f (x, y ) dx [ c, d ]
Q
Âñ.
A
l éuα ∈ (0, 1)½¤á
A
QØ1ªüb-α → 0§uk sin x dx ε§l sin x dxÂñ sin x dx = 1 − cos A§¨A → +∞§cos A'4ØQ§u´ b#ؤá l e sin x dx9uα > 0ØÂñ. 5. y²µ α (1) dxQعα = 0'?Û«mþ´ë¼ê¶ x +α
(1)
0
Ïα ∈ [ a, b ], x ∈ (1, +∞)§u0 < |x e | x e q lim x · x e = 0§uâáPÂÈ©' ܧy{'4Gª§& x e dxÂñ u´d¼§y{§& x e dx9uα ∈ [ a, b ](a, b?¿¢ê)Âñ. √ √ π (2) αe dx = Âñ§¢§Q(0, +∞)9uαÂñ 2 √ √ π é∀A > 0§Ï lim αe dx = lim e dt = e dt = 2 √ π √ √ % é u0 < ε < 2 § U Qα > 0§ ¦ & α e dx = α e dx > ε § = √αe dx9uαQ(0, +∞)þØÂñ. √ (3) é?¿'½'α ∈ (−∞, +∞)§È© e dxÑÂñ§ e dx = π (i) |x|¿©§éa < α < b§k0 < e < 2e Ï e dx = 2 e dxÂñ ud¼§y{§& e dxéa < α < bÂñ. √ (ii) é∀A > 0§k lim e dx = lim e dt = π √ π u¨α¿©§ e dx > 2 dd§& e dxQ−∞ < α < +∞þÂñ l e dxQ−∞ < α < +∞þÂñ.
+∞
e−p0 z z 2 dz
0
p−1
ln2
Âñ§u´ x ln x dxÂñ x dx9upQp p > 0þÂñ.
1 p0 −1 2 0 0 2
1 e
%k u´
1 0 1 0
x
p−1
ln x dx >
2
1 e
x
p−1
ln x dx >
0
x
p −1
xp−1 ln2 x dx p > 0
Q
0
1 dx = p
a +∞ a +∞
|f (x, y )| dx
9
(1)
0 +∞
cos xy dx (y x2 + y 2
a > 0)
(2)
0 1
cos xy dx (−∞ < y < +∞) x2 + 1 1 b y b, b > 1
(3)
0
ln xy dx
y²µ cos xy (1) Ïy a > 0§u x +y u´d¼§y{§&