13.9. SPHERICAL COORDINATES IN p DIMENSIONS 365

Note that the above equals∫Ā×[0,∞)

f(

hp

(⃗φ ,θ ,ρ

))ρ

p−1Φ

(⃗φ ,θ

)dmp

and the iterated integral is also equal to∫[0,∞)

ρp−1

∫Ā

f(

hp

(⃗φ ,θ ,ρ

))Φ

(⃗φ ,θ

)dmp−1dm

because the difference is just a set of measure zero.

Notation 13.9.2 Often this is written differently. Note that from the spherical coordinateformulas, f

(h(⃗

φ ,θ ,ρ))

= f (ρω) where |ω| = 1. Letting Sp−1 denote the unit sphere,{ω ∈ Rp : |ω|= 1} , the inside integral in the above formula is sometimes written as∫

Sp−1f (ρω)dσ

where σ is a measure on Sp−1. See [83] for another description of this measure. It isn’t animportant issue here. Either 13.9.22 or the formula∫

∞

0ρ

p−1(∫

Sp−1f (ρω)dσ

)dρ

will be referred to as polar coordinates and is very useful in establishing estimates. Hereσ(Sp−1

)≡∫

A Φ

(⃗φ ,θ

)dmp−1.

Example 13.9.3 For what values of s is the integral∫

B(0,R)

(1+ |x|2

)sdy bounded inde-

pendent of R? Here B(0,R) is the ball, {x ∈ Rp : |x| ≤ R} .

I think you can see immediately that s must be negative but exactly how negative? Itturns out it depends on p and using polar coordinates, you can find just exactly what isneeded. From the polar coordinates formula above,∫

B(0,R)

(1+ |x|2

)sdy =

∫ R

0

∫Sp−1

(1+ρ

2)sρ

p−1dσdρ

= Cp

∫ R

0

(1+ρ

2)sρ

p−1dρ

Now the very hard problem has been reduced to considering an easy one variable problemof finding when ∫ R

0ρ

p−1 (1+ρ2)s

dρ

is bounded independent of R. You need 2s+(p−1)<−1 so you need s <−p/2.