Kenneth Kuttler

294 CHAPTER 12. SERIES AND TRANSFORMS

The justification for this interchange of integration follows as earlier. You can discount∫|y|>M | f (y)|dy for M large enough and then use Fubini’s theorem from Theorem 9.9.4 on

Page 214 to interchange the order. It is the same argument in Lemma 10.1.1.

=1π

∫∞

−∞

∫ R

0cos(t (x− y)) f (y)dtdy =

1π

∫∞

−∞

f (y)∫ R

0cos(t (x− y))dtdy

=1π

∫∞

−∞

f (y)sin(R(x− y))

x− ydy =

1π

∫∞

−∞

f (x−u)sin(Ru)

udy

=1π

∫∞

0f (x−u)

sin(Ru)u

dy+1π

∫ 0

−∞

f (x−u)sin(Ru)

udy

=1π

∫∞

0f (x−u)

sin(Ru)u

dy+1π

∫∞

0f (x+u)

sin(Ru)u

2π

∫∞

f (x−u)+ f (x+u)2u

sin(Ru)du

Also, from Theorem 10.2.5 or Corollary 10.2.8, the limit as R→ ∞ is f (x−)+ f (x+)2 . This

verifies 12.14.You can come up with lots of enigmatic integration formulas using this representation

theorem. You start with a function you want to represent and then use this formula to repre-sent it. The representation in terms of the formula yields strange and wonderful integrationfacts. Two cases which are nice to note are the case that f is even and the case that f isodd. If you are only interested in the function on the nonnegative real numbers, you couldconsider it either way as part of an odd function or part of an even function. This willchange what happens at 0. First suppose it is even. Then

limR→∞

1π

∫ R

∫∞

−∞

cos(t (x− y)) f (y)dydt

= limR→∞

1π

∫ R

∫∞

−∞

(cos txcos ty+ sin txsin ty) f (y)dydt

= limR→∞

1π

∫ R

0cos tx

∫∞

−∞

cos ty f (y)dydt

because y→ sin(ty) is odd. Thus you get

f (x+)+ f (x−)2

= limR→∞

1π

∫ R

0cos(tx)

∫∞

−∞

cos(ty) f (y)dydt

= limR→∞

2π

∫ R

0cos(tx)

∫∞

0cos(ty) f (y)dydt (12.15)

In case f is odd, you get

f (x+)+ f (x−)2

= limR→∞

1π

∫ R

0sin(tx)

∫∞

−∞

sin(ty) f (y)dydt

= limR→∞

2π

∫ R

0sin(tx)

∫∞

0sin(ty) f (y)dydt (12.16)

Lets see what this formula says for certain choices of f . We need to have f be in L1 (R)but this is easy to arrange. Just let it vanish off some interval. First suppose f (y) = y for

294 CHAPTER 12. SERIES AND TRANSFORMSThe justification for this interchange of integration follows as earlier. You can discountJiv|> u|f (v)|dy for M large enough and then use Fubini’s theorem from Theorem 9.9.4 onPage 214 to interchange the order. It is the same argument in Lemma 10.1.1.= if[- [ cos (1(x—y)) f(y fae aL mie ee_i +f 10 ) Sin (R=) =< f- P(x =) dy_ ‘f° f(x—u) ch ee,_ 2 [row EO ay tf ety nin)1 ss? f(x-u x+u= > | ‘At et + ) sin (Ru) duAlso, from Theorem 10.2.5 or Corollary 10.2.8, the limit as R — © is Pay fet) | Thisverifies 12.14. IJYou can come up with lots of enigmatic integration formulas using this representationtheorem. You start with a function you want to represent and then use this formula to repre-sent it. The representation in terms of the formula yields strange and wonderful integrationfacts. Two cases which are nice to note are the case that f is even and the case that f isodd. If you are only interested in the function on the nonnegative real numbers, you couldconsider it either way as part of an odd function or part of an even function. This willchange what happens at 0. First suppose it is even. Thenlim — LL [coste( x—y)) f (y) dydtRoo TT= lim — an [. (costxcosty + sintxsinty) f (y) dydtRow TT1= lim— Noose costyf (y) dydtRoo TT JO —oobecause y > sin (ty) is odd. Thus you get_ 1 /R rooPet) = £05) = jim = 008 (tx) [00s (ty) f (y) dydt2 pr -= ims ; cos (ix) | cos (ty) f (y) dydt (12.15)In case f is odd, you get_ R coPO) £00) = Jim = | sin(tx) [sin (ty) f ()dyat2 /R °°= jim = | sin(ex) [ sin(ty) f(y)dydt (12.16)Lets see what this formula says for certain choices of f. We need to have f be in L' (R)but this is easy to arrange. Just let it vanish off some interval. First suppose f(y) = y for