Writing the Beta Function in terms of the Gamma Function I am studying the gamma and beta functions and I have seen an exercise which asks you to re-write the beta function in terms of the gamma function as follows:
$B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$
The only hint that the exercise gives is that you should start with a function of the form:
$f(\alpha,\beta,t)=\int_0^tx^{\alpha-1}(t-x)^{\beta-1}dx$
The first thing I notice is that the right-hand side can be considered as a convolution of two functions, so:
$L[f]=L[x^\alpha-1]L[(t-x)^{\beta-1}]$
However, this just ends up with the Laplace transform of f on the LHS and then a mess of transforms on the right.  The other thing is that if you set $t=1$, $f$ becomes the beta function:
$f(\alpha,\beta,1)=B(\alpha,\beta)=\int_0^1x^{\alpha-1}(1-x)^{\beta-1}dx$
Hence I am hoping that if I convert $f$ to this form we have the beta function and then take the product of the Laplace transforms of the two functions in the convolution, it somehow comes out as a set of integrals which are equivalent to the given identity once I use the integral formula for the gamma function.  However, I have tried this and got a mess, could someone assist (if this is the right way of doing it)?
Edit: I have looked on Wikipedia and seen that the identity can be proved just by taking the product of two gamma functions and then changing variables to show that this is $B(\alpha,\beta)\Gamma(\alpha+\beta)$: however, the exercise is in a section on Laplace transforms so I think it wants you to take the Laplace transform of the given function $f$ and work it out that way.
 A: $$f(\alpha, \beta, t) = \int_0^t x^{\alpha -1} (t-x)^{\beta -1}\;dx
= \big(t^{\alpha - 1} * t^{\beta -1}\big)(t)$$
so using the fact that convolution is multiplicative in the Laplace domain, 
\begin{align}
\mathcal{L}\{f(\alpha, \beta, t)\}(s)
&= \mathcal{L}\{t^{\alpha-1}\}(s) \cdot \mathcal{L}\{t^{\beta - 1}\}(s)\\
&= \frac{\Gamma(\alpha)}{s^{\alpha}}\cdot \frac{\Gamma(\beta)}{s^{\beta}}\\
&= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \cdot\underbrace{\frac{\Gamma(\alpha+\beta)}{s^{\alpha+\beta}}}_{\mathcal{L\{t^{\alpha+\beta - 1}\}(s)}}
\end{align}
Taking inverse Laplace we then find
$$f(\alpha,\beta,t) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}  t^{\alpha+\beta -1}$$
hence at $t=1$:
$$B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$
Added To address the extra question in the comments,
$$
\Gamma(\tfrac12)\cdot\Gamma(\tfrac12) = \Gamma(\tfrac12+\tfrac12)\cdot B(\tfrac12,\tfrac12)
= (1)\int_0^1 x^{-1/ 2} (1-x)^{-1/ 2}\;dx
= \int_0^1 \frac{dx}{\sqrt{x(1-x)}}$$
We can evaluate this integral by noting the symmetry about $x=\tfrac12$, so substitute $x =\frac{1+u}{2}$ to shift the symmetry about $u=0$
$$\Gamma(\tfrac12)^2 = \int_{-1}^1 \underbrace{\frac{du}{\sqrt{1-u^2}}}_{\text{even function}}
= 2\int_0^1 \frac{du}{\sqrt{1-u^2}}
= 2\cdot \left.\sin^{-1}(u)\right|_0^1
= 2 \cdot \left(\frac{\pi}{2} - 0\right) = \pi$$
Hence $\Gamma(\tfrac12) = \sqrt{\pi}$.
A: Here it is another approach, assuming we are just interested in real $\alpha,\beta>0$.
For a fixed value of $\beta$ the function $\int_{0}^{1}x^{\beta-1}(1-x)^{\alpha-1}\,dx $ is clearly positive, continuous and log-convex with respect to $\alpha$: the Cauchy-Schwarz inequality proves the midpoint-log-convexity and the continuity improves this to log-convexity. In particular
$$g(\alpha)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\beta)}\int_{0}^{1}x^{\beta-1}(1-x)^{\alpha-1}\,dx $$
is positive, continuous and log-convex. By integration by parts we have $g(\alpha+1)=\alpha\cdot g(\alpha)$ and $g(1)=1$ is trivial. By invoking the Bohr-Mollerup theorem we have $g(\alpha)=\Gamma(\alpha)$, hence
$$ \Gamma(\alpha+\beta)\,B(\alpha,\beta) = \Gamma(\alpha)\,\Gamma(\beta) $$
as wanted.
