17
∞
Z
ln x
dx = lim
t→∞
x
1
Z
Z1 t
= lim
t→∞
t
ln x
dx
x
ln x(ln x)0 dx
1
Z
ln t
= lim
t→∞
udu (u = ln x)
0
1
(ln t)2
t→∞ 2
=∞.
= lim
R∞
So the improper integral
1
ln x
dx
x
is divergent.
22
Z
0
∞
ex
dx = lim
t→∞
e2x + 3
t
Z
0
ex
dx
e2x + 3
et
Z
1
du (u = ex )
t→∞ 1
u2 + 3
et
1
u
−1
√
= lim √ tan
t→∞
3
3 1
1
1
1
−1
−1
√
= √ lim tan (s) − √ tan
3 s→∞
3
3
1 π
1 π
=√ · − √ ·
3 2
3 6
π
= √ .
3 3
= lim
et
√ → ∞ as t → ∞
3
28
Z
0
5
Z 5
w
w
dw +
dw
0 w−2
2 w−2
Z t
Z 5
w
w
= lim−
dw + lim+
dw.
t→2
t→2
0 w−2
t w−2
w
dw =
w−2
Z
2
Consider
t
Z
lim
t→2−
0
Z t
2
w
dw = lim−
1+
dw
t→2
w−2
w−2
0
= lim− [w + 2 ln |w − 2|]t0
t→2
=2 + lim− ln(2 − t)
t→2
= − ∞.
Therefore the improper integral
R5
w
dw
0 w−2
is divergent.
41
Z
0
∞
x
dx =
3
x +1
Z
1
Z0 1
≤
Z0 1
=
0
Z ∞
x
dx +
x3 + 1
Z1 ∞
x
dx +
3
x +1
Z1 ∞
x
dx +
x3 + 1
1
1
x3
x
dx
+1
x
dx
x3
1
dx.
x2
Note that
R1
x
0
x3 +1
dx and
R∞
1
1
dx
x2
= 1 are both finite. So by comparison,
42
Z
1
∞
2 + e−x
dx ≥
x
Z
∞
2
dx
x
1
Z t
2
= lim
dx
t→∞ 1 x
= lim [2 ln |x|]t1
t→∞
=∞.
So by comparison,
R∞
1
2+e−x
dx
x
is divergent.
2
R∞
0
x
dx
x2 +1
is convergent.