a formula in Gamma function

Question

I want to prove the following formula
$$ \int\limits_{1}^{s}t^{-\frac{\alpha}{\beta}-1}(s-t)^{\frac{\alpha}{\beta}-1}dt=\frac{\Gamma(\frac{\alpha}{\beta})}{\Gamma(1+\frac{\alpha}{\beta})} \times \frac{1}{s}\times (s-1)^{\frac{\alpha}{\beta}} $$ Here is what I did $$ \int\limits_{1}^{s}t^{-\frac{\alpha}{\beta}-1}(s-t)^{\frac{\alpha}{\beta}-1}dt= \int\limits_{1}^{s}t^{-\frac{\alpha}{\beta}-1}s^{\frac{\alpha}{\beta}-1} (1-\frac{t}{s})^{\frac{\alpha}{\beta}-1}dt=$$ $$= \frac{1}{s^2}\int\limits_{1}^{s}\frac{t^{-\frac{\alpha}{\beta}-1}}{s^{\frac{-\alpha}{\beta}-1}} (1-\frac{t}{s})^{\frac{\alpha}{\beta}-1}dt = \frac{1}{s^2}\int\limits_{1}^{s}(\frac{t}{s})^{-\frac{\alpha}{\beta}-1} (1-\frac{t}{s})^{\frac{\alpha}{\beta}-1}dt$$ $$=\frac{1}{s}\int\limits_{\frac{1}{s}}^{1}t^{-\frac{\alpha}{\beta}-1}(s-t)^{\frac{\alpha}{\beta}-1}dt $$ I thought I can use the Beta function $$ Beta(x,y)=\int\limits_{0}^{1}t^{x-1}(1-t)^{y-1}dt\;\; for\;\;\; x>0,\, y>0. $$ But it seems that it's not OK. For this formula, I need a hint not a whole solution. A small hint tending me to the answer can be acceptable as acomplete answer to me.

Answer 1

I think you complicate the matter.

By replacing your $\alpha/\beta$ by $a$, your formula is equivalent to $$\int_1^s t^{-a-1}(s-t)^{a-1} dt = \frac{\Gamma(a)}{\Gamma(1+a)}\frac{(s-1)^a}{s}$$

The left hand side equals $$\int_1^s \frac{1}{t^2}\left(\frac{s-t}{t}\right)^{a-1} dt = \int_1^s \left(\frac{s}{t}-1\right)^{a-1} d(-\frac{1}{t}) = \int_{1/s}^1 (st-1)^{a-1} dt = \frac{(s-1)^a}{sa} $$ note that the last integral has elementary antiderivative.

While for the right hand side, using the identity $\Gamma(x+1)=x\Gamma(x)$, it is equal to $\frac{(s-1)^a}{sa}$, so both sides are indeed equal.

Answer 2

Here is a hint.

I thought at first it was a simple application of the definition of $Beta$ integral, which is not the case.

A first direction of calculation would be the incomplete Beta integral (https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function), but what I propose you is to consider your integral :

• either as a fractional derivative expression (https://en.wikipedia.org/wiki/Fractional_calculus)

• or plainly under the form of a convolution

$$\tag{1}\dfrac{t^{a-1}}{\Gamma(a)} * \dfrac{t^{b-1}}{\Gamma(b)} $$

for which you have to prove that it is equal to a function belonging to the same family of power functions.

The easiest way to achieve this goal is to take the Laplace Transform (LT) of (1), by applying rules $LT(f(t)*g(t))=LT(f(t)).LT(g(t))$ (a convolution gives a product) and $LT(\dfrac{t^{a-1}}{\Gamma(a)})=\dfrac{1}{s^{a}}$.

Remark : (1) should be written under the form:

$$\tag{1}U(x)\dfrac{t^{a-1}}{\Gamma(a)} * U(x)\dfrac{t^{b-1}}{\Gamma(b)} $$

with $U$ the Heaviside function ($1$ for $x>0$, $0$ otherwise) in order to enhance the fact that we deal with causal functions.