# Prove that $\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, {\rm d}u =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)} {\Gamma(\alpha_1+\alpha_2)}$

I have the following equality in a textbook of mine

$$\frac{y^{\alpha_1+\alpha_2-1} e^{-y/\beta}}{\Gamma(\alpha_1+\alpha_2) \beta^{\alpha_1+\alpha_2}} \cdot \frac{\Gamma(\alpha_1+\alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)} \int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \frac{y^{\alpha_1+\alpha_2-1} e^{-y/\beta}}{\Gamma(\alpha_1+\alpha_2) \beta^{\alpha_1+\alpha_2}}$$

and I see that for the equality to be true we must have

$$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)} {\Gamma(\alpha_1+\alpha_2)}$$

However can someone give me an explanation why this is the case ?

• en.wikipedia.org/wiki/… – Meow Jan 4 '14 at 15:11
• I think I've posted answers to essentially this same question here before. I've up-voted the question, even though I may also (if I get around to finding one of those answers) vote to close this as a duplicate. – Michael Hardy Jan 4 '14 at 16:49
• Here's an answer I posted that proves this identity in case the exponents are integers: math.stackexchange.com/questions/86542/… – Michael Hardy Jan 4 '14 at 16:52
• Does the proof on wikipedia prove it in the case of integers only or is the proof valid for any real number ? – Shuzheng Jan 4 '14 at 18:22

## 1 Answer

For $\alpha_1, \alpha_2 \gt0$ We have

$$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \text{B}(\alpha_1,\alpha_2) \tag{1}$$

Where $\text{B}(a,b)$ is Beta Function \begin{align} \Gamma(\alpha_1) & =\int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \tag{2}\\ \Gamma(\alpha_2) & =\int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y \tag{3}\\ \end{align}

\begin{align} \Gamma(\alpha_1)\Gamma(\alpha_2) & = \int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y \tag{4}\\ & =\int\limits_0^\infty\int\limits_0^\infty\ e^{-x-y} x^{\alpha_1-1}y^{\alpha_2-1}\,\mathrm{d}x \,\mathrm{d}y \tag{5}\\ & =\int_{z=0}^\infty\int_{t=0}^1 e^{-z} \Big(zt\Big)^{\alpha_1-1}\Big(z(1-t)\Big)^{\alpha_2-1}z\,\mathrm{d}z \,\mathrm{d}t \tag{6}\\ & =\int_{z=0}^\infty\int_{t=0}^1 e^{-z} z^{\alpha_1+\alpha_2-1}t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}z \,\mathrm{d}t \tag{7}\\ & =\int_{0}^\infty e^{-z}z^{\alpha_1+\alpha_2-1} \,\mathrm{d}z\int_{0}^1t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}t \tag{8}\\ \Gamma(\alpha_1)\Gamma(\alpha_2) & =\Gamma(\alpha_1+\alpha_2){\rm B}(\alpha_1,\alpha_2) \tag{9}\\ \end{align}

Hence

$$\large{\rm B}(\alpha_1,\alpha_2)=\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, du =\frac{\Gamma(\alpha_1)\,\Gamma(\alpha_2)}{\Gamma(\alpha_1+\alpha_2)}$$

$\text{ Explanation: } (6)$ Substituting $x=zt$ and $y=z(1-t)$