Value of $g'(\frac{1}{2})$

Question

Given than for each $a\in (0,1)$

$$\lim_{h \to 0^+} \int_{h}^{1-h} t^{-a} (1-t)^{a-1} dt$$

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 $g'(\frac{1}{2})$ is ?

How should I go about? Need hints

Answer 1

$$g(a) = \rm{B}(1-a, 2-a) = {\frac {\Gamma \left( 1-a \right) \Gamma \left( 2-a \right) }{\Gamma \left( 3-2\,a \right) }}$$ where $\rm{B}$ is the beta function. Then

$$ g'(a) = -{\frac {\Psi \left( 1-a \right) \Gamma \left( 1-a \right) \Gamma \left( 2-a \right) }{\Gamma \left( 3-2\,a \right) }}-{\frac {\Gamma \left( 1-a \right) \Psi \left( 2-a \right) \Gamma \left( 2-a \right) }{\Gamma \left( 3-2\,a \right) }}+2\,{\frac {\Gamma \left( 1-a \right) \Gamma \left( 2-a \right) \Psi \left( 3-2\,a \right) }{\Gamma \left( 3-2\,a \right) }} $$ and $$ \eqalign{g'(1/2) &= \frac{\left( \gamma+2\,\ln \left( 2 \right) \right) \pi}{2}+\frac{\pi\, \left(- 2+\gamma+2\,\ln \left( 2 \right) \right)}{2} +\pi\, \left( 1- \gamma \right) \cr &= 2 \pi \ln(2)} $$

Answer 2

Since $g(a)=g(1-a)$, we can say that $g'(a)=-g'(1-a)$. Putting $a=\frac{1}{2}$, we get $g'(\frac{1}{2}) = -g'(\frac{1}{2})$. This implies $g'(\frac{1}{2})=0$

EDIT: $$g(a)= \lim_{h \to 0} \int_{h}^{1-h} t^{-a} (1-t)^{a-1} dt$$ $$g(1-a)=\lim_{h \to 0} \int_{h}^{1-h} t^{a-1} (1-t)^{-a} dt$$

Using $\int_{a}^{b}f(x) dx=\int_{a}^{b}f(a+b-x) dx$

$$g(1-a)=\lim_{h \to 0} \int_{h}^{1-h} (1-t)^{a-1} t^{-a} dt$$

This proves $g(a)=g(1-a).$