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How can one prove the following equality?

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

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  • $\begingroup$ Write definition of gamma function twice and do the change of variable t=xy and u=x(1-y). $\endgroup$ – zwim Feb 15 '17 at 22:05
  • $\begingroup$ Write $\Gamma(1+\alpha) $\endgroup$ – Mark Viola Feb 15 '17 at 22:07
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    $\begingroup$ $$\Gamma(a+1)\Gamma(b)=\int_0^{\infty}\int_0^{\infty}dxdyx^ay^{b-1}e^{-x-y}$$. Now sub $x\rightarrow x^2,y\rightarrow y^2$ and then go to polar coordinates $\endgroup$ – tired Feb 15 '17 at 22:15
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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.

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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

$$ \int^{1}_0 s^{x-1}(1-s)^{y-1}\,ds= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma {(x+y+1)}}$$

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  • $\begingroup$ the identity $\mathcal{L}^{-1}\left(\frac{1}{s^{x+y+2}}\right)=\Gamma(x+y+1)^{-1}$ should be elaborated $\endgroup$ – tired Feb 15 '17 at 22:29
  • $\begingroup$ @tired, by definition of the gamma function. $\endgroup$ – Zaid Alyafeai Feb 15 '17 at 22:31
  • $\begingroup$ 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 $\int_0^t...$ $\endgroup$ – Jean Marie Feb 15 '17 at 22:31
  • $\begingroup$ 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" $\endgroup$ – tired Feb 15 '17 at 22:33
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    $\begingroup$ @tired, sorry that was wrong. I meant the Laplace transform is known for that function. So the inverse Laplace transform is easy to deduce. $\endgroup$ – Zaid Alyafeai Feb 15 '17 at 22:37
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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.

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  • $\begingroup$ straight to the point as always (+1) $\endgroup$ – tired Feb 15 '17 at 22:36
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Feb 15 '17 at 22:40
  • $\begingroup$ Nice Jack (+1)!. $\endgroup$ – Zaid Alyafeai Feb 15 '17 at 22:40
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    $\begingroup$ @JackD'Aurizio it is the most efficent one i (now) know, thumps up! $\endgroup$ – tired Feb 15 '17 at 22:42

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