Discussion Forum

Given that for each   $a \epsilon (0,1), \lim_{h \rightarrow 0}$ $\int_{h}^{1-h}  t^{-a}(1-t)^{a-1}dt$  exists. Let this limit be g(a). In addition , it is given that the function g(a) differentiable (0,1)

The value of g'(1/2) is 

Answer:

Explanation:

 Plan    $\int_{0}^{a}f(x) dx=\int_{0}^{a} f(a-x) dx$

 As g(a) is given, use the numerical value of "a" given in the question and then proved .

 Given that ,

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

 Clearly  , g(a)=g(1-a)

  [Using   $\int_{0}^{a}f(x) dx=\int_{0}^{a}  f(a-x)dx]$

 Now, differentiate w.r.t 'a' , we get

   g'(a) =g'(1-a)(-1)

 Now, for    $a=-\frac{1}{2}$ , we have

   $-g'(\frac{1}{2})=g'(\frac{1}{2})$

  So,  $g'(\frac{1}{2})=0$