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(a). In addition, it is given that the function g(a) is differentiable on (0,1).

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Linked Question 1

The value of $g (\frac{1}{2})$ is

$\frac{π}{4}$

2π

$\frac{π}{2}$

(i) g (a) = $\int_{0}^{1} t^{- a} {(1 - t)}^{a - 1} dt$ .......(1)

Using King g (a) = $\int_{0}^{1} {(1 - t)}^{- a} {(t)}^{a - 1} dt$ ......(2)

Now g (a) = g (1 – a)

Differentiating with respect to a

g ' (a) = – g ' (1 – a)

Put a = $\frac{1}{2}$ ; $∴ g' (\frac{1}{2})$ = 0 Ans.

Aliter: g (a) = $\int_{0}^{1} t^{- a} {(1 - t)}^{a - 1} dt$

differentiate under the sign of integral w.r.t. a keeping t constant

g ' (a) = $\int_{0}^{1} t^{- a} {(1 - t)}^{a - 1} \ln (1 - t) - {(1 - t)}^{a - 1} \ln t) dt$

Put a = $\frac{1}{2}$

$I = g' (\frac{1}{2}) = \int_{0}^{1} \frac{\ln (1 - t) - \ln t}{\sqrt{t - t^{2}}} dt$

Using King I = – I

$∴$ I = 0

Linked Question 2

The value of $g' (\frac{1}{2})$ is

$\frac{- π}{2}$