Math 411: Honours Complex Variables - University of Alberta

82 CHAPTER 12. THE RESIDUE THEOREM AND APPLICATIONS
In particular, we see for α = 1/2 that Γ2 (1/2) = 2 � π/2
dθ = π and 0
� ∞
e
−∞
−x2
� ∞
dx = 2 e
0
−x2
� �
1
dx = Γ =
2
√ π.
A substitution then leads to the important result � ∞
a > 0.
−∞ e−ax2
For arbitrary α ∈ (0,1), we find, on substituting z = tan 2 θ,
� π/2
I(α) := Γ(α)Γ(1−α) = 2 tan 2α−1 θdθ =
0 +
� ∞
0 +
dx = � π/a for
z α−1
1+z dz.
The integral here can be evaluated by a contour integration in the complex
plane, noting that the function z α−1 = e (α−1)logz is holomorphic on the starshaped
domain obtained by slicing the complex plane along the positive real
axis. This branch cut is shown in red in the following figure. In other words we
choose the antiderivative logz = log|z|+iargz of the function z ↦→ 1/z, where
argz ∈ [0,2π).
CR
−1
Imz
ir
Cr
iR
Rez
Here the large circular contour CR is chosen to have radius R ≥ 2, so that
|1+z| ≥ R/2 on CR, and the small semicircular contour Cr is chosen to have
radius r ≤ 1/2, so that |1+z| ≥ 1/2 on Cr. On denoting
f(z) := zα−1
1+z
e(α−1)logz
= ,
1+z
we then see, accounting for the residue from the pole of f at z = −1, that
2πie (α−1)iπ � R+ir �
= f +
ir
CR
f +
� −ir
R−ir
�
f +
Cr
f.