How to integrate $\int_0^1x^a(1-x)^bdx$ 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.
 A: 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:
$$
    B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}
$$
A: 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 $b \in \mathbb{N}.$
A: 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
$$B(a+1,b+1)=\frac{a!\,b!}{(a+b+1)!}$$
