19.13. FRACTIONAL POWERS OF OPERATORS 571

Finally consider the claim about e−At being Hermitian. For Φ(t)≡ e−At

Φ′ (t)+AΦ(t) = 0,Φ(0) = I

Φ∗′ (t)+Φ

∗ (t)A = 0, Φ∗ (0) = I

and so, from what was just shown about commuting,

Φ∗′ (t)+Φ

∗ (t)A = 0, Φ∗ (0) = I

Φ′ (t)+Φ(t)A = 0,Φ(0) = I

Thus Φ(t) and Φ∗ (t) satisfy the same initial value problem and so they are the same Thanksto Theorem 19.12.3.

Next it follows that A−α is Hermitian because(A−α x0,y0

)≡

(∫∞

0tα−1e−Atx0dt,y0

)=∫

∞

0

(tα−1e−Atx0,y0

)dt

=∫

∞

0

(x0, tα−1e−Aty0

)dt =

(x0,∫

∞

0tα−1e−At (y0)dt

)=

(x0,A−α y0

)Note that Lemma 19.13.2 shows that AA−α = A−α A. Also A−α A−β = A−β A−α and

in fact, A−α commutes with every operator which commutes with A. Next is a technicallemma which will prove useful.

Lemma 19.13.5 For α,β > 0,Γ(α)Γ(β ) = Γ(α +β )∫ 1

0 (1− v)α−1 vβ−1dv

Proof:

Γ(α)Γ(β )≡∫

∞

0

∫∞

0e−(t+s)tα−1sβ−1dtds =

∫∞

0

∫∞

se−u (u− s)α−1 sβ−1duds

=∫

∞

0e−u

∫ u

0(u− s)α−1 sβ−1dsdu =

∫∞

0e−u

∫ u

0(u− s)α−1 sβ−1dsdu

=∫

∞

0e−u

∫ 1

0(u−uv)α−1 (uv)β−1 udvdu

=∫

∞

0e−u

∫ 1

0uα−1uβ (1− v)α−1 vβ−1dvdu

=∫ 1

0(1− v)α−1 vβ−1dv

∫∞

0uα+β−1e−udu

= Γ(α +β )∫ 1

0(1− v)α−1 vβ−1dv

Now consider whether A−α acts like it should. In particular, is A−α A−(1−α) = A−1?