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I have a question about an equation I am trying to integrate, the integral is: $$\int_0^1 x^a (1 - x)^b ~dx,$$ where $a, b > 0$.

Any assistance with this problem would be appreciated.

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1  
Do you know the binomial theorem? – Kevin Carlson Aug 29 '12 at 2:32
2  
Or do you know the beta function? You need to provide us some info if you want to get useful help. – GEdgar Aug 29 '12 at 2:40

There is one more ingenious way that i want to share with you.

Define functions $F(t)=t^a$ and $G(t)=t^b$ and observe their convolution,

$$F \star G(t) = \int_{0}^{t} F(\lambda)G(t-\lambda) d \lambda$$ $$=> F \star G(t) = t^{a+b+1}\int_{0}^{1} y^a(1-y)^b dy, $$ (1) where we used the integral-transform determined by the relation $\lambda = ty$.

In a second step, we simply use the Laplace-transform, \begin{equation} \mathcal{L}(F)(s)= \frac{a!}{s^{a+1}} \end{equation} and \begin{equation} \mathcal{L}(G)(s)= \frac{b!}{s^{b+1}}. \end{equation}

Now we can use the property of the Laplace-transform which states that the corresponding time-domain function of a multiplication of two functions of the complex variable $s$ in the Laplace-domain, equals the convolution of the corresponding time-domain functions of each of the two respective Laplace-domain functions involved in the previous multiplication.

Applying this to the actual problem leads us to,

\begin{equation} \mathcal{L}(F)(s)\mathcal{L}(G)(s)= \frac{a!b!}{s^{a+b+2}} =\frac{a!b!}{(a+b+1)!}\frac{(a+b+1)!}{s^{a+b+2}} \end{equation}

and an easy calculation gives rise to a new expression for the above convolution of $F$ and $G$, which is,

$$F \star G(t) = \frac{a!b!}{(a+b+1)!}t^{a+b+1} $$ (2)

Expressing the equality of (1) and (2) leads directly to the desired result :)

\begin{equation} \int_{0}^{1} y^a(1-y)^b dy= \frac{a!b!}{(a+b+1)!}, \end{equation}

which holds for $a \in \mathbb{N}$ and $b \in \mathbb{N}.$

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The integral at hand is known as Euler's integral of the first kind. It's value, as function of $a$ and $b$ is know as beta function: $$ B(a+1,b+1) = \int_0^1 x^a (1-x)^b \mathrm{d} x $$ Integrating by parts, one can derive recurrence equations: $$ (a+1) B(a+1, b+1) = b B(a+2,b) $$ Change of variables $x \mapsto 1-x$ gives $B(a,b) = B(b,a)$. Recurrence equations can be solved to give expression of the beta function in terms of ratio of Euler's $\Gamma$-functions: $$ B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} $$

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Google "Beta function". What you have is $$B(b+1,a+1)=B(a+1,b+1)=\int_0^1x^a(1-x)^bdx=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$For example, if $\,a\,,\,b\in\Bbb N\,$ , then $$B(a+1,b+1)=\frac{a!\,b!}{(a+b+1)!}$$

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Rupert, here is the Wikipedia link for Beta function: en.wikipedia.org/wiki/Beta_function – user2468 Aug 29 '12 at 2:47

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