54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1719
= π
(cos(
π
2(1−2α)
)+ isin
(π
2(1−2α)
))Then equating the imaginary parts yields
sin((1−2α)π)∫
∞
0
u2α−1
1+u2 du = π sin(
π
2(1−2α)
)and so using the trig identities for the sum of two angles,∫
∞
0
u2α−1
1+u2 du =π(sin(
π
2 (1−2α)))
2sin(
π
2 (1−2α))
cos(
π
2 (1−2α))
=π
2cos(
π
2 (1−2α)) = π
2sin(πα)
It remains to verify the last identity.
Γ(α)Γ(β ) ≡∫
∞
0
∫∞
0tα−1e−tsβ−1e−sdsdt
=∫
∞
0
∫∞
ttα−1e−u (u− t)β−1 dudt
=∫
∞
0e−u
∫ u
0tα−1 (u− t)β−1 dtdu
=∫ 1
0xα−1 (1− x)β−1 dx
∫∞
0e−uuα+β−1du
=
(∫ 1
0xα−1 (1− x)β−1 dx
)Γ(α +β )
This proves the lemma.If it is not stated otherwise, in all that follows α > 0.
Definition 54.3.16 Let A be a sectorial operator corresponding to the sector S−aφ where−a < 0. Then define for α > 0,
(−A)−α ≡ 1Γ(α)
∫∞
0tα−1S (t)dt
where S (t) is the analytic semigroup generated by A as in Corollary 54.3.8. Note thatfrom the estimate, ||S (t)|| ≤ Me−at of this corollary, the integral is well defined and is inL (H,H).
Theorem 54.3.17 For (−A)−α as defined in Definition 54.3.16
(−A)−α (−A)−β = (−A)−(α+β ) (54.3.17)
Also(−A)−1 (−A) = I, (−A)(−A)−1 = I (54.3.18)
and (−A)−α is one to one if α ≥ 0, defining A0 ≡ I.