Advanced High-School Mathematics

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

1. 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

2. 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

.)

3. Compute
   xlim→∞

∫∞

0 e

−tcos(xt)dt.

4. LetA, B be constants,A >0. Show that
   ∫∞
   0 e

−AtsinBtdt= B

A^2 +B^2

.

5. Define the function

Π(x) =







1 if|x|≤^12 ,

0 otherwise.