For $a\in (0,1)$,$ \lim_{h\to 0^+} \int_h^{1-h} t^{-a}(1-t)^{a-1} dt$ exists. Let the limit be $g(a)$. Then evaluate $g’(\frac 12)$ $$g(a) = \int_0^1 t^{-a}(1-t)^{a-1}$$
$$g’(a) =0$$
The answer seems straightforward, but it doesn’t feel right, so I would like someone to reaffirm it so that i know this is the way to solve it
Thanks!
 A: I hope you're familiar with Leibniz integral rule (of differentiating under the integral sign).
$$\frac{d}{dx}\int_a^bf(x,t)dt = \int_a^b\frac{\partial}{\partial x} f(x,t)dt $$
Apply this rule on your integral. I believe you will easily find your answer then.
EDIT:
Your hypothesis that $g'(a) = 0$ is unfounded. Please perform the calculation again. If you face the problem again, just comment below. I or somebody will surely post the complete solution.
EDIT 2:
$g'(a) = 0$ does hold true at $a  = \frac{1}{2}$, so your answer might be correct too.
EDIT 3:
Differentiating under the integral sign, we get
$$g'(a) = \int_0^1 -t^{-a}\log{(t)}(1-t)^{a-1}+(1-t)^{a-1}t^{-a}\log{(1-t)} dt$$
Put $a = \frac{1}{2}$ here. You get
$$g'(\frac{1}{2}) = \int_0^1 \frac{\log{(1-t)}-\log{t}}{\sqrt{t}\sqrt{1-t}}dt$$
Let this eqn be e1.
Now, use the $a-x$ property in definite integration and replace $t$ by $1-t$ and you will get
$$g'(\frac{1}{2}) = \int_0^1 \frac{\log{(t)}-\log{(1-t)}}{\sqrt{t}\sqrt{1-t}} dt$$
Add them and I think you will get your answer. Please point out any mistake if you find.
