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I googled this, but I didn't really find anything good.

So I just want a proof that for positive integers $x$ and $y$:

$$\int_{0}^{1} t^{x-1} \cdot (1-t)^{y-1} dt = \frac{(x-1)! \cdot (y-1)!}{(x+y-1)!}$$

Any help is appreciated.

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One can do the following:

  1. Note that for integer $m$ one has $$\int_0^{\infty} p^{m-1}e^{-p}dp=(m-1)! \tag{0}$$ This can be proved by iterative integration by parts.

  2. Now consider the product of two such integrals \begin{align} P(m,n)&=(m-1)!(n-1)!=\tag{1}\\ &=\int_0^{\infty}p^{m-1}e^{-p}dp\times \int_0^{\infty}q^{n-1}e^{-q}dq=\\ &=2\int_0^{\infty}x^{2m-1}e^{-x^2}dx\times 2\int_0^{\infty}y^{2n-1}e^{-y^2}dy, \end{align} and rewrite the last expression as a double integral over the region $x,y\geq0$ in the $(x,y)$-plane (these $x,y$ are not your $x,y$).

  3. In this double integral, pass to the polar coordinates: $$x=r\cos\theta,\qquad y=r\sin\theta,\qquad dx\,dy\rightarrow rdr\,d\theta.$$ This transforms it into \begin{align} P(m,n)&=4\int_0^{\infty}r^{2(m+n-1)}e^{-r^2}rdr\times \int_0^{\pi/2}\cos^{2m-1}\theta\sin^{2n-1}\theta \,d\theta=\\ &=2(m+n-1)!\int_0^{\pi/2}\cos^{2m-1}\theta\sin^{2n-1}\theta \,d\theta \end{align} Comparing this with (1), we find that $$2\int_0^{\pi/2}\cos^{2m-1}\theta\sin^{2n-1}\theta \,d\theta=\frac{(m-1)!(n-1)!}{(m+n-1)!}.\tag{2}$$

  4. Now it remains to show that the integral on the left of (2) is equal to your beta integral. For that, make the change of variables $t=\sin^2\theta$ so that $dt=2\sin\theta\cos\theta\,d\theta$ and $\cos^2\theta=1-t$. This gives \begin{align} 2\int_0^{\pi/2}\cos^{2m-1}\theta\sin^{2n-1}\theta \,d\theta&=\int_0^{\pi/2} \underbrace{\cos^{2m-2}\theta}_{(1-t)^{m-1}}\times\underbrace{\sin^{2n-2}\theta}_{t^{n-1}}\times\underbrace{2\sin\theta\cos\theta\,d\theta}_{dt}= \\&=\int_0^1 t^{n-1}(1-t)^{m-1}dt. \end{align}


P.S. Actually, we have just proved a more general formula. The only place where it was used that $m,n$ are integers, was the formula (0). If we define the gamma function by $$\displaystyle \int_0^{\infty}p^{m-1}e^{-p}dp=\Gamma(m),$$ then what we have shown above is the formula $$B(m,n)=\int_0^{1}t^{m-1}(1-t)^{n-1}dt=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}.$$

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  • $\begingroup$ Why can you just sub that stuff in, in the last step? $\endgroup$ – Prankster Aug 24 '13 at 12:11
  • $\begingroup$ @Prankster Sorry, I don't understand your question. Are you asking about the change from $\theta$ to $t$? $\endgroup$ – Start wearing purple Aug 24 '13 at 12:14
  • $\begingroup$ just the very last step where you just said that 2 sin $\theta$ cos $\theta$ $d\theta$ = dt And how you then changed the terminals of the integral $\endgroup$ – Prankster Aug 24 '13 at 16:02
  • $\begingroup$ @Prankster I have added a small clarification. For $dt$ the explanation is $$d(\sin^2\theta)=(\sin^2\theta)'_{\theta} d\theta=2\sin\theta \cos\theta\,d\theta$$ Also, as $\theta$ increases from $0$ to $\pi/2$, $\sin\theta$ (and $\sin^2\theta$) increases from $0$ to $1$. $\endgroup$ – Start wearing purple Aug 24 '13 at 16:06
  • $\begingroup$ thanks, I understand it now. Just not why you call it a double integral. Isn't a double integral 2 integral signs after each other? $\endgroup$ – Prankster Aug 25 '13 at 20:32
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For an alternative proof I will use the Laplace Transform, and the Convolution theorem.

The Convolution Theorem says that given two functions then $$ \mathcal{L}(f*g) = \mathcal{L}(g)\cdot \mathcal{L}(f)\,, \tag{1} $$ where the convolution for two functions is given as $$ f*g = \int_0^t f(s)g(t-s)\,\mathrm{d}t\,. \tag{2} $$

We want to show that $$ B(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}\,. $$ Start by defining the following functions $ f(t) = t^x$ and $g(t) = t^y $. By using $(2)$ we have $$ f * g = \int_0^t s^x (t-s)^y\,\mathrm{d}s\,, \tag{3} $$ and by the convolution theorem $(1)$ we also have that $$ \mathcal{L}(t^x * t^y) = \mathcal{L}(t^x)\mathcal{L}(t^y) = \frac{x!}{s^{x+1}} \cdot \frac{y!}{s^{y+1}} = \frac{x! \cdot y!}{s^{x+y+2}}\,. $$ Now taking the inverse Laplace of the above equation yield $$ t^x * t^y = \mathcal{L}^{-1}\left( \frac{x! \cdot y!}{s^{x+y+2}} \right) = t^{x+y+2} \frac{x! \cdot y!}{(x+y+1)!}\,. \tag{4} $$ Where we have to remember that $x!$ and $y!$ are simply constants. By comparing $(3)$ and $(4)$ we arrive at the equation $$ \int_0^t s^x(t-s)^y \mathrm{d}s = t^{x+y+2} \frac{x! \cdot y!}{(x+y+1)!} $$ Setting $t=1$, we now simply obtain $$ \int_0^1 s^x(1-s)^y \mathrm{d}s = \frac{x! \cdot y!}{(x+y+1)!} $$ as wanted.


For a proof of $(4)$ one can proceed as follows: The Laplace Transform of a function $f$ is defined as $$ \mathcal{L}\bigl(f(t)\bigr) := \int_0^\infty e^{-st}f(t)\mathrm{d}t\,. $$ Setting $f(t) = t^n$ and using the substitution $y = st$ gives $$ \mathcal{L}(t^n) = \int_0^\infty e^{-st} t^n\mathrm{d}t = \int_0^\infty e^{-y} \left(\frac{y}{s}\right)^n \frac{\mathrm{d}y}{s} = \frac{1}{s^{n+1}}\int_0^\infty e^{-y} y^n \mathrm{d}y = \frac{n!}{s^{n+1}} $$ Where the last integral is known as Eulers second integral, or The Gamma function. Hence $$ \mathcal{L}^{-1}\left( \frac{1}{s^{n+1}}\right) = \frac{t^n}{n!} $$ the proof is now complete by setting $n = x+y+1$, and multiplying by the constants $x!\cdot y!$.

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Following Rudin, one can use the Bohr Mollerup theorem.

Claim (Bohr Mollerup) The Gamma function is the unique function $f:\Bbb R^+\to\Bbb R$ such that

$(\rm i)$ $f(1)=1$

$(\rm ii)$ $f(x+1)=xf(x)$

$(\rm iii)$ $\log f(x)$ is convex.

Claim For $y>0$, $f(x):=B(x,y)\Gamma(x+y)/\Gamma(y)$ satisfies the conditions of the Borh Mollerup theorem, so $f(x)=\Gamma(x)$.

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