Advanced book on Mathematics Olympiad

538 Real Analysis

arctan

(

1 −cosα

sinα

)

−arctan

(

− 1 −cosα

sinα

)

,

where the angles are to be taken between−π 2 andπ 2. But

1 −cosα

sinα

×

− 1 −cosα

sinα

=− 1.

Hence the difference between these angles is±π 2. Notice that the sign of the integral
is the same as the sign ofα. HenceI(α)= π 2 ifα ∈( 2 kπ, ( 2 k+ 1 )π )and−π 2 if
α∈(( 2 k+ 1 )π, ( 2 k+ 2 )π )for some integerk.

Remark.This is an example of an integral with parameter that does not depend continu-
ously on the parameter.
(E. Goursat,A Course in Mathematical Analysis, Dover, NY, 1904)

463.First, note that 1/

√

xhas this property forp>2. We will alter slightly this function
to make the integral finite forp=2. Since we know that logarithms grow much slower
than power functions, a possible choice might be

f(x)=

1

√

xlnx

.

Then
∫∞

2

f^2 (x)dx=

∫∞

2

1

xln^2 x

=−

1

lnx

∣∣

∣∣

∞

2

=

1

ln 2

<∞.

Consequently, the integral offpis finite for all real numbersp≥2.
Let us see what happens forp<2. An easy application of L’Hôpital’s theorem gives

xlim→∞

f(x)p

x−^1

=xlim→∞

x−

p

(^2) ln−px
x−^1
=xlim→∞
x^1 −
p
2
lnpx

=∞,

and hence the comparison test implies that forp<2 the integral is infinite. Therefore,
f(x)=√x^1 lnxsatisfies the required condition.

Remark.Examples like the above are used in measure theory to prove that inclusions
betweenLpspaces are strict.

464.Letnbe the degree ofP(x). Integrating successively by parts, we obtain
∫t

0

e−xP(x)dt=−e−xP(x)|t 0 +

∫t

0

e−xP′(x)dx

=−e−xP(x)|t 0 −e−xP′(x)|t 0 +

∫t

0

e−xP′(x)dx= ···