1718 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

Proof: First change variables to get rid of the σ . Let y = (t−σ)−1 (s−σ) . Then theintegral becomes ∫ 1

0(t− [(t−σ)y+σ ])α−1 (t−σ)−α y−α (t−σ)dy

=∫ 1

0((t−σ)(1− y))α−1 (t−σ)−α y−α (t−σ)dy

=∫ 1

0(1− y)α−1 y−α dy

Next let y = x2. The integral is

2∫ 1

0

(1− x2)α−1

x1−2α dx

Next let x = sinθ

2∫ 1

2 π

0(cos(θ))2α−1 sin(1−2α) (θ)dθ = 2

∫ 12 π

0

(cos(θ)sin(θ)

)2α−1

dθ

Now change the variable again. Let u = cot(θ) . Then this yields

2∫

∞

0

u2α−1

1+u2 du

This is fairly easy to evaluate using contour integrals. Consider the following contour calledΓR for large R. As R→ ∞, the integral over the little circle converges to 0 and so does theintegral over the big circle. There is one singularity at i.

−R R−R−1 R−1

Thus

limR→∞

∫ΓR

e(ln|z|+iarg(z))(1−2α)

1+ z2 dz =

= (1+ cos(1−2α)π)∫

∞

0

u2α−1

1+u2 du

+isin((1−2α)π)∫

∞

0

u2α−1

1+u2 du