Advanced book on Mathematics Olympiad

574 Real Analysis

522.First, note that forx>0,

e−sxx−^1 |sinx|<e−sx,

so the integral that we are computing is finite.
Now consider the two-variable function

f (x, y)=e−sxysinx.

We have
∫∞

0

∫∞

1

|f (x, y)|dydx=

∫∞

0

∫∞

1

e−sxy|sinx|dydx=

1

s

∫∞

0

e−sxx−^1 |sinx|dx,

and we just saw that this is finite. Hence we can apply Fubini’s theorem, to conclude that
on the one hand,
∫∞

0

∫∞

1

f (x, y)dydx=

1

s

∫∞

0

e−sxx−^1 sinxdx,

and on the other hand,
∫∞

0

∫∞

1

f (x, y)dydx=

∫∞

1

1

s^2 y^2 + 1

dy.

Here of course we used the fact that
∫∞

0

e−axsinxdx=

1

a^2 + 1

,a> 0 ,

a formula that can be proved by integrating by parts. Equating the two expressions that
we obtained for the double integral, we obtain
∫∞

0

e−sxx−^1 sinxdx=

π

2

−arctans=arctan(s−^1 ),

as desired.
(G.B. Folland,Real Analysis, Modern Techniques and Their Applications, Wiley,
1999)

523.Applying Tonelli’s theorem to the functionf (x, y)=e−xy, we can write
∫∞

0

e−ax−e−bx

x

dx=

∫∞

0

∫b

a

e−xydydx=

∫b

a

∫∞

0

e−xydxdy

=

∫b

a

1

y

dy=ln

b

a

.