数学分析(复旦大学版)课后题答案40-45

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
dx [ p1 , p2 ]
Q
ë
2−p
dx [ p1 , p2 ]
Q
ë
6.
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
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
−∞
−∞
−(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
+∞
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
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.
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
+∞ 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
+∞ 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)
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 > 1y²µ cos xy (1) Ïy a > 0§u x +y u´d¼§y{§&
(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.
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
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)
2
2 +∞