Evaluate Derivative of Integral Function at Half

To find the value of $g'\left(\frac{1}{2}\right)$, we start with the definition of the function $g(a)$ for $a \in (0, 1)$: $g(a) = \lim_{h \to 0^+} \int_{h}^{1-h} t^{-a} (1 - t)^{a-1} \, dt$

Let the integral inside the limit be denoted as: $I_h(a) = \int_{h}^{1-h} t^{-a} (1 - t)^{a-1} \, dt$

Substitute $t = 1 - u$, which gives $dt = -du$. When $t = h$, $u = 1 - h$, and when $t = 1 - h$, $u = h$. The integral becomes: $I_h(a) = \int_{1-h}^{h} (1 - u)^{-a} u^{a-1} (-du) = \int_{h}^{1-h} u^{a-1} (1 - u)^{-a} \, du$

Rewriting the exponents in the form corresponding to the definition: $I_h(a) = \int_{h}^{1-h} u^{-(1-a)} (1 - u)^{(1-a)-1} \, du = I_h(1-a)$

Taking the limit as $h \to 0^+$ on both sides: $g(a) = g(1 - a) \quad \text{for all } a \in (0, 1)$

Since $g(a)$ is differentiable on the interval $(0, 1)$, we can differentiate both sides of the identity $g(a) = g(1 - a)$ with respect to $a$ using the chain rule: $g'(a) = \frac{d}{da} [g(1 - a)] = -g'(1 - a)$

Now, evaluating this derivative at $a = \frac{1}{2}$: $g'\left(\frac{1}{2}\right) = -g'\left(1 - \frac{1}{2}\right)$ $g'\left(\frac{1}{2}\right) = -g'\left(\frac{1}{2}\right)$ $2g'\left(\frac{1}{2}\right) = 0$ $g'\left(\frac{1}{2}\right) = 0$

Therefore, the correct option is (D).