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!

• What makes you think that $g'(a) = 0$? Nov 18, 2020 at 15:06
• @MartinR the limits are constants Nov 18, 2020 at 15:21
• But $\int_0^1 t^{-a}(1-t)^{a-1}dt$ does depend on $a$. Nov 18, 2020 at 15:22
• @MartinR I know, that’s why i asked it over here. Nov 18, 2020 at 15:35
• surprisingly this question was asked today math.stackexchange.com/questions/3912236/… Nov 18, 2020 at 15:48

I hope you're familiar with Leibniz integral rule (of differentiating under the integral sign).

$$\frac{d}{dx}\int_a^bf(x,t)dt = \int_a^b\frac{\partial}{\partial x} f(x,t)dt$$

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

$$g'(\frac{1}{2}) = \int_0^1 \frac{\log{(t)}-\log{(1-t)}}{\sqrt{t}\sqrt{1-t}} dt$$

• Well @Aditya $g'(\frac{1}{2})$ is indeed 0. But this does not hold true for $g'(a)$. As you must have seen, a special case occurs at a = $\frac{1}{2}$. (-a = a-1). If your answer is wrong, perhaps the correct question is whose link Albus gave above. In that question $g(\frac{1}{2})$ is asked instead of $g'(\frac{1}{2})$. Please do see to that.... Nov 18, 2020 at 16:10
• Can you please write how $g’(1/2)=0$ if my method is wrong Nov 18, 2020 at 16:19