Question

Given that for each $a\in (0,1),\underset{h\to 0^{+}}{lim}{\int }_h^{1-h}{t}^{-a}(1-t{)}^{a-1}dt$ exists. Let this limit be $g\left(a\right)$
In addition, it is given that the function $g\left(a\right)$ is differentiable on $(0,1)$, then The value of $g^{\prime }\left(\frac{1}{2}\right)$ is?

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $-\frac{\pi }{2}$
• D. 0

Answer 1

Answer: (D)

0

$g\left(a\right){=}_h^{l-h}\int t^{-a}(1-a{)}^{a-1}dt$ if $h⟶\infty$

${}_0^1\int t^{-a}(1-a{)}^{a-1}dt$

$g^{\prime }\left(a\right){=}_0^1\int \left[t^{-a}lnt\right(1-t{)}^{a-1}+t^{-a}(1-t{)}^{a-1}ln(1-t\left)\right]dt{g}^{\prime }\left(\frac{1}{2}\right){=}_0^1\int \left[t^{-\frac{1}{2}}lnt\right(1-t{)}^{\frac{1}{2}-1}+t^{-\frac{1}{2}}(1-t{)}^{\frac{1}{2}-1}ln(1-t\left)\right]dt{g}^{\prime }\left(\frac{1}{2}\right){=}_0^1\int \left[t^{-\frac{1}{2}}\right(1-t{)}^{-\frac{1}{2}}](-lnt+(1-t\left)dt\right)...........\left(1\right)g^{\prime }\left(\frac{1}{2}\right){=}_0^1\int \left[t^{\frac{1}{2}}\right(1-t{)}^{-\frac{1}{2}}](lnt-(1-t\left)dt\right)...........\left(2\right)$

Adding (1) and (2) we get

$g^{\prime }\left(\frac{1}{2}\right)=0$