# Defined integral of $x^{\alpha} (1-x)^{\beta-1}$

How can one prove the following equality?

$$\int_{0}^{1} x^{\alpha} (1-x)^{\beta-1} \,dx = \frac{\Gamma(\alpha+1) \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)}$$

• Write definition of gamma function twice and do the change of variable t=xy and u=x(1-y). – zwim Feb 15 '17 at 22:05
• Write $\Gamma(1+\alpha) – Mark Viola Feb 15 '17 at 22:07 • $$\Gamma(a+1)\Gamma(b)=\int_0^{\infty}\int_0^{\infty}dxdyx^ay^{b-1}e^{-x-y}$$. Now sub$x\rightarrow x^2,y\rightarrow y^2$and then go to polar coordinates – tired Feb 15 '17 at 22:15 ## 3 Answers Function gamma :$\Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}dt$Function beta :$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt$Your expression is$B(\alpha+1,\beta)=\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+\beta+1)}$$$\Gamma(\alpha)\Gamma(\beta)=\int_0^{\infty}t^{\alpha-1}e^{-t}dt\int_0^{\infty}u^{\beta-1}e^{-u}du=\int_0^{\infty}\int_0^{\infty}t^{\alpha-1}u^{\beta-1}e^{-(t+u)}\,dt\,du$$ Change of variables$t=xy, u=x(1-y)$With$(x,y)\in\ ]0,\infty[\times]0,1[$for$(t,u)\in\ ]0,\infty[\times]0,\infty[$and jacobian$|J|=x$. (Rem:$x=t+u$and$0\le y=t/(t+u)\le t/t\le 1$since$u\ge 0$) $$\Gamma(\alpha)\Gamma(\beta)=\int_0^{\infty}\int_0^1x^{\alpha-1}y^{\alpha-1}x^{\beta-1}(1-y)^{\beta-1}e^{-x}\,x\,dx\,dy=$$ $$\int_0^{\infty}\int_0^1[x^{\alpha-1}x^{\beta-1}e^{-x}\,x][y^{\alpha-1}(1-y)^{\beta-1}]\,dx\,dy=\quad$$ $$\int_0^{\infty}x^{\alpha+\beta-1}e^{-x}\,dx\int_0^1y^{\alpha-1}(1-y)^{\beta-1}\,dy=\Gamma(\alpha+\beta)B(\alpha,\beta)$$ There are Fubini invocations here and there but since everything is in separated variables, it works everywhere$\Gamma$and$B$are defined. We show that using Convolution $$\beta(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma {(y+1)}}{\Gamma{(x+y+2)}}$$ $$proof$$ Let us choose some functions$f(t) = t^{x} \,\, , \, g(t) = t^y$Hence we get $$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds$$ So by definition we have $$\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y )$$ We can now use the laplace of the power $$\mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$$ Notice that we need to find the inverse of Laplace$\mathcal{L}^{-1}$$$\mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{- 1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!} {(x+y+1)!}$$ So we have the following $$(t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$ By definition we have $$t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds$$ This looks good , put$t=1$we get $$\frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$$ By using that$n! = \Gamma{(n+1)}$We arrive happily to our formula $$\int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma {(x+y+2)}}$$ which can be written as $$\int^{1}_0 s^{x-1}(1-s)^{y-1}\,ds= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma {(x+y+1)}}$$ • the identity$\mathcal{L}^{-1}\left(\frac{1}{s^{x+y+2}}\right)=\Gamma(x+y+1)^{-1}$should be elaborated – tired Feb 15 '17 at 22:29 • @tired, by definition of the gamma function. – Zaid Alyafeai Feb 15 '17 at 22:31 • You should absolutely define$f$in this way:$f(t):=t^x U(t)$where$U$=Heaviside function ; the same for$g$. Otherwise convolution$f*g$is not defined by$\int_0^t...$– Jean Marie Feb 15 '17 at 22:31 • is that so? remember you take a integral in the complex plane over a (possible) multivalued function....the result is correct but i think not true "by definition" – tired Feb 15 '17 at 22:33 • @tired, sorry that was wrong. I meant the Laplace transform is known for that function. So the inverse Laplace transform is easy to deduce. – Zaid Alyafeai Feb 15 '17 at 22:37 Let us prove the symmetrical form $$f(\alpha,\beta)=\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\tag{1}$$ for any positive$\alpha,\beta$.$f(\cdot,\beta)$and$f(\alpha,\cdot)$are continuous functions and moments, in particular they are log-convex functions by the Cauchy-Schwarz inequality. Clearly$f(\alpha,\beta)=f(\beta,\alpha)$(due to the substitution$x\mapsto 1-x$) and$f(\alpha,1)=\frac{1}{\alpha}$. By integration by parts $$f(\alpha+1,\beta)= \frac{\alpha}{\beta}\cdot f(\alpha,\beta+1) \tag{2}$$ hence, by induction,$(1)$holds for any$\alpha\in\mathbb{N}^*$or$\beta\in\mathbb{N}^*$. By log-convexity and the Bohr-Mollerup theorem it follows that$(1)$holds for any positive$\alpha,\beta$. With a little extra effort, it is not difficult to show that the same holds for any$\alpha,\beta\$ with positive real part.

• straight to the point as always (+1) – tired Feb 15 '17 at 22:36
• @tired: there was a small issue, now fixed. I found that the Bohr-Mollerup theorem gives a nice shortcut for proving this identity for the Beta function. It is not a "usual" proof, but I think it is rather efficient. – Jack D'Aurizio Feb 15 '17 at 22:40
• Nice Jack (+1)!. – Zaid Alyafeai Feb 15 '17 at 22:40
• @JackD'Aurizio it is the most efficent one i (now) know, thumps up! – tired Feb 15 '17 at 22:42