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R∞ dx 0 1+xa u = xa 1 x = ua 1 dx = a1 u a −1du. R ∞ u a1 −1du 1 a 0 1+u 1 −1 −t(1+u) 1 R ∞ u a e du I(t) = a 0 1+u We want I(0). (Note, I(∞) = 0). R∞ 1 I’(t)= −1 a 0 u a −1e−t(1+u)du −1 −t ∞ a1 −1 −tu R I’(t)= a e 0 u e du −e−tΓ( a1 ) I’(t)= 1 at a 1 R∞ We have that 0 I 0(t)dt = I(∞) − I(0), I(∞) = 0 −Γ( a1 ) R ∞ e−t Then −I(0) = a 0 1 ta Γ( a1 ) R ∞ −1 −t Then I(0) = a 0 t a e Then I(0) = a1 Γ( a1 )Γ(1 − a1 ) Euler’s reflection formula: Γ(z)Γ(1 − z) = πcsc(πz) R ∞ dx π csc( π ). Then 0 1+x a = a a 2