Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
82 CHAPTER 12. THE RESIDUE THEOREM AND APPLICATIONS<br />
In particular, we see for α = 1/2 that Γ2 (1/2) = 2 � π/2<br />
dθ = π and 0<br />
� ∞<br />
e<br />
−∞<br />
−x2<br />
� ∞<br />
dx = 2 e<br />
0<br />
−x2<br />
� �<br />
1<br />
dx = Γ =<br />
2<br />
√ π.<br />
A substitution then leads to the important result � ∞<br />
a > 0.<br />
−∞ e−ax2<br />
For arbitrary α ∈ (0,1), we find, on substituting z = tan 2 θ,<br />
� π/2<br />
I(α) := Γ(α)Γ(1−α) = 2 tan 2α−1 θdθ =<br />
0 +<br />
� ∞<br />
0 +<br />
dx = � π/a for<br />
z α−1<br />
1+z dz.<br />
The integral here can be evaluated by a contour integration in the complex<br />
plane, noting that the function z α−1 = e (α−1)logz is holomorphic on the starshaped<br />
domain obtained by slicing the complex plane along the positive real<br />
axis. This branch cut is shown in red in the following figure. In other words we<br />
choose the antiderivative logz = log|z|+iargz <strong>of</strong> the function z ↦→ 1/z, where<br />
argz ∈ [0,2π).<br />
CR<br />
−1<br />
Imz<br />
ir<br />
Cr<br />
iR<br />
Rez<br />
Here the large circular contour CR is chosen to have radius R ≥ 2, so that<br />
|1+z| ≥ R/2 on CR, and the small semicircular contour Cr is chosen to have<br />
radius r ≤ 1/2, so that |1+z| ≥ 1/2 on Cr. On denoting<br />
f(z) := zα−1<br />
1+z<br />
e(α−1)logz<br />
= ,<br />
1+z<br />
we then see, accounting for the residue from the pole <strong>of</strong> f at z = −1, that<br />
2πie (α−1)iπ � R+ir �<br />
= f +<br />
ir<br />
CR<br />
f +<br />
� −ir<br />
R−ir<br />
�<br />
f +<br />
Cr<br />
f.