Subject: Mathematics — Definite Integration
Question: Given that for each
$a \in (0,1),{\lim_{h \rightarrow 0^{+}}\int_{h}^{1 - h}t^{- a}}{(1 - t)^{a - 1}dt}$
exists. Let this limit be g(a). In addition, it is given that the
function g(a) is differentiable on (0,1).
The value of $g' = \frac{1}{2}$.
Answer: 0
Solution: $g(a) = \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}t^{- a}(1 - t)^{a - 1}dt$
$$g(a) = \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}(1 - t)^{- a}t^{a - 1}dt$$
Using,
$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b - x)dx$$
Also,
$$g(1 - a) = \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}t^{a - 1}(1 - t)^{- a}dt$$
Thus,
$$g(a) = g(1 - a)$$
$$\Rightarrow \quad g'(a) = - g'(1 - a) \Rightarrow g'(a) + g'(1 - a) = 0$$
Putting $a = \frac{1}{2}$, we get
$$g'\left( \frac{1}{2} \right) + g'\left( \frac{1}{2} \right) = 0$$
or,
$$g'\left( \frac{1}{2} \right) = 0$$