256 CHAPTER 5 Series and Differential Equations
This is a fairly simple integration: using the substitution (u= lnx) one
first computes the indefinite integral
∫ dx
xlnpx
=
1
(1−p)xlnpx
.
Therefore,
∫∞
2
dx
xlnpx
= lima→∞
Ñ
1
(1−p)xlnpx
é∣
∣∣
∣∣
a
2
=
1
2(1−p) lnp 2
.
Exercises
- For each improper integral below, compute its value (which might
be±∞) or determine that the integral does not exist.
(a)
∫∞
2
dx
√
x− 2
(b)
∫ 1
− 1
dx
√
1 −x^2
(c)
∫∞
2
√ dx
x^2 − 4
(d)
∫ 1
0 lnxdx
- Letk≥1 andp >1 and prove that
∫∞
1 sin
k
( 2 π
xp
)
dx <∞.(Hint:
note that ifxp>4 then sink
(
2 π
xp
)
≤sin
(
2 π
xp
)
<
2 π
xp
.)
- Compute
xlim→∞
∫∞
0 e
−tcos(xt)dt.
- LetA, B be constants,A >0. Show that
∫∞
0 e
−AtsinBtdt= B
A^2 +B^2
.
- Define the function
Π(x) =
1 if|x|≤^12 ,
0 otherwise.