Given that for each $a \in(0,1)$, $\lim\limits_{h \rightarrow 0^{+}} \int\limits_h^{1-h} t^{-a}(1-t)^{a-1} d t$ exists. If this limit be $g(a)$, then the value of $g\left(\frac{1}{2}\right)$, is
Answer & explanation
Correct answer: option 1
We have,
$g(a)=\lim\limits_{h \rightarrow 0^{+}} \int\limits_h^{1-h} t^{-a}(1-t)^{a-1} d t$
∴ $g\left(\frac{1}{2}\right)=\lim\limits_{h \rightarrow 0^{+}} \int\limits_h^{1-h} t^{-1 / 2}(1-t)^{-1 / 2} d t$
$\Rightarrow g\left(\frac{1}{2}\right)=\lim\limits_{h \rightarrow 0^{+}} \int\limits_h^{1-h} \frac{1}{\sqrt{t-t^2}} d t$
$\Rightarrow g\left(\frac{1}{2}\right)=\lim\limits_{h \rightarrow 0^{+}} \int\limits_h^{1-h} \frac{1}{\sqrt{\left(\frac{1}{2}\right)^2-\left(t-\frac{1}{2}\right)^2}} d t$
$\Rightarrow g\left(\frac{1}{2}\right)=\int\limits_0^1 \frac{1}{\sqrt{\left(\frac{1}{2}\right)^2-\left(t-\frac{1}{2}\right)^2}}$
$\Rightarrow g\left(\frac{1}{2}\right)=\left[\sin ^{-1}\left(\frac{t-\frac{1}{2}}{1 / 2}\right)\right]_0^1=\sin ^{-1} 1-\sin ^{-1}(-1)=\pi$