Given that for each a∈(0,1), limh→0+∫1-h1 t-a(1-t)a-1 dt exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0,1), then The value of g'(1/2) is?

g(a) = limh→0+∫1-h1 t-a(1-t)a-1 dt
If h→0, then
g(a) = ∫01 t-a(1-t)a-1 dt
g'(a) = ∫01 [t-a(1-t)a-1 ln(1-t) - t-a(1-t)a-1ln(t)] dt
g'(1/2) = ∫01 [t-1/2(1-t)-1/2 ln(1-t) - t-1/2(1-t)-1/2ln(t)] dt
Let I = ∫01 t-1/2(1-t)-1/2 ln(1-t) dt
Let J = ∫01 t-1/2(1-t)-1/2 ln(t) dt
g'(1/2) = I - J
Let t = sin2θ. Then dt = 2sinθcosθ dθ
When t = 0, θ = 0. When t = 1, θ = π/2.
I = ∫0π/2 (sinθ)-1(cosθ)-1 ln(1-sin2θ) 2sinθcosθ dθ = ∫0π/2 2ln(cos2θ) dθ = 4∫0π/2 ln(cosθ) dθ = 4(-π/2) = -2π
J = ∫0π/2 (sinθ)-1(cosθ)-1 ln(sin2θ) 2sinθcosθ dθ = ∫0π/2 2ln(sin2θ) dθ = 4∫0π/2 ln(sinθ) dθ = 4(-π/2) = -2π
Therefore, g'(1/2) = I - J = -2π - (-2π) = 0