Advanced book on Mathematics Olympiad

Real Analysis 573

Consequently, the integral we are computing is equal toπ
2
32.
(American Mathematical Monthly, proposed by M. Hajja and P. Walker)

520.We have

I=

∫∫

D

ln|sin(x−y)|dxdy=

∫π

0

(∫y

0

ln|sin(y−x)|dx

)

dy

=

∫π

0

(∫y

0

ln sintdt

)

dy=y

∫y

0

ln sintdt

∣∣

∣∣

y=π

y= 0

−

∫π

0

yln sinydy

=πA−B,

whereA=

∫π

0 ln sintdt,B =

∫π
0 tln sintdt. Note that here we used integration by
parts! We compute further

A=

∫ π 2

0

ln sintdt+

∫π

π 2 ln sintdt=

∫ π 2

0

ln sintdt+

∫ π 2

0

ln costdt

=

∫ π 2

0

(ln sin 2t−ln 2)dt=−

π

2

ln 2+

1

2

A.

HenceA=−πln 2. ForBwe use the substitutiont=π−xto obtain

B=

∫π

0

(π−x)ln sinxdx=πA−B.

HenceB=π 2 A. Therefore,I=πA−B=−π
2
2 ln 2, and we are done.

Remark.The identity

∫ π 2

0

ln sintdt=−

π

2

ln 2

belongs to Euler.
(S. Radulescu, M. R ̆ adulescu, ̆ Teoreme ̧si Probleme de Analiza Matematic ̆ ̆a(Theorems
and Problems in Mathematical Analysis), Editura Didactica ̧ ̆si Pedagogic ̆a, Bucharest,
1982).

521.This problem applies the discrete version of Fubini’s theorem. Define

f(i,j)=

{

1 forj≤ai,

0 forj>ai.

The left-hand side is equal to

∑n

i= 1

∑m
∑m j=^1 f(i,j), while the right-hand side is equal to
j= 1

∑n

i= 1 f(i,j). The equality follows.