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: $\pi$
Solution: $g(a) = \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}t^{- a}(1 - t)^{a - 1}dt$
Putting $a = \frac{1}{2}$,
$$\therefore\quad g\left( \frac{1}{2} \right) = \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}t^{- 1/2}(1 - t)^{- 1/2}dt$$
$$= \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}\frac{1}{\sqrt{t(1 - t)}}dt = \lim_{h \rightarrow 0^{\text{+}}}\int_{h}^{1 - h}\frac{1}{\sqrt{\left( \frac{1}{2} \right)^{2} - \left( t - \frac{1}{2} \right)^{2}}}dt$$
$$= \lim_{h \rightarrow 0^{\text{+}}}\left\lbrack \sin^{- 1}\left( \frac{t - \frac{1}{2}}{\frac{1}{2}} \right) \right\rbrack_{h}^{1 - h} = \lim_{h \rightarrow 0^{\text{+}}}\left\lbrack \sin^{- 1}(2t - 1) \right\rbrack_{h}^{1 - h}$$
$$= \lim_{h \rightarrow 0^{\text{+}}}\left\lbrack \sin^{- 1}(1 - 2h) - \sin^{- 1}(2h - 1) \right\rbrack$$
$$= \frac{\pi}{2} - \left( - \frac{\pi}{2} \right) = \pi$$