Laplace transform to prove relationship between the Gamma and Beta functions I am stuck in the following theoretical part:
I have the function:
$$f(t,\mu,\nu) = \int_{0}^{t}x^{\mu-1}(t-x)^{\nu-1}dx$$
Now I have to perform a Laplace transform to deduce that:
$$B(\mu,\nu) = \frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu)}$$
A "normal" Laplace transform I know how to use, but in this case: I don't know where to start. 
 A: 
In the ensuing analysis, we will make use of the Laplace Transform Pair
$$\begin{align}
f(t) &\leftrightarrow F(s)\\\\
t^{a-1} &\leftrightarrow \frac{\Gamma(a)}{s^a}\tag1
\end{align}$$
where $F(s)=\int_0^\infty f(t)e^{-st}\,dt$ and $f(t)=\frac1{2\pi i}\int_{\sigma -i\infty}^{\sigma +i\infty}F(s)e^{st}\,ds$.


Let $f(t,\mu,\nu)$ be represented by the convolution integral
$$f(t,\mu,\nu)=\int_0^t \tau^{\mu-1}(t-\tau)^{\nu-1}\,d\tau\tag2$$
Note that $f(1,\mu,\nu)=B(\mu,\nu)$, where $B(x,y)$ is the Beta function, which can be represented by the integral
$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$

Applying the Convolution Theorem of the Laplace Transform to $(2)$ and using $(1)$ reveals that
$$\begin{align}
\int_0^\infty f(t,\mu,\nu)e^{-st}\,dt&=\left(\int_0^\infty t^{\mu-1}e^{-st}\,dt\right)\left(\int_0^\infty t^{\mu-1}e^{-st}\,dt\right)\\\\
&=\left(\frac{\Gamma(\mu)}{s^\mu}\right)\left(\frac{\Gamma(\nu)}{s^\nu}\right)\\\\
&=\frac{\Gamma(\mu)\Gamma(\nu)}{s^{\mu+\nu}}\\\\
&=\frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu))}\int_0^\infty t^{\mu+\nu-1}e^{-st}\,dt\tag3
\end{align}$$

Taking the inverse Laplace Transform of both sides of $(3)$, we find that 
$$f(t,\mu,\nu)=\frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu)}\,t^{\mu+\nu-1}\tag4$$

Finally, setting $t=1$ in $(4)$ yields the coveted relationship

$$\bbox[5px,border:2px solid #C0A000]{B(\mu,\nu)=\frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu)}}$$

as was to be shown!
A: Although there is already a good answer, I needed to fill the missing steps for my understanding.  Perhaps will help others as well.
The convolution of $f(x) * g(x) = \int_0^x f(y)g(x-y)\,dy$
For functions $f(x)=x^{\alpha-1}$ and $g(x)=x^{\beta-1}$ we have 
$$ f*g = \int_0^x y^{\alpha-1} (x-y)^{\beta-1}\,dy 
    = x^{\alpha+\beta-1}\int_0^1 u^{\alpha-1} (1-u)^{\beta-1}\,du=x^{\alpha+\beta-1}B(\alpha,\beta)
$$
Now, the Laplace transform of $f(x)$
$$ \mathcal{L}\{x^{\alpha-1}\} = \int_0^\infty x^{\alpha-1} e^{-sx}\,dx = s^\alpha \int_0^\infty x^{\alpha-1} e^{-x} \, dx = s^\alpha \Gamma(\alpha)
$$
Using the fact that $\mathcal{L}\{f*g\} = \mathcal{L}\{f\} \mathcal{L}\{g\} $
$$
\mathcal{L}\{x^{\alpha+\beta-1}B(\alpha,\beta)\} = \mathcal{L}\{x^{\alpha-1}\} \mathcal{L}\{x^{\beta-1}\} 
$$
substituing left and right hand side with Laplace transforms
$$
s^{\alpha+\beta} \Gamma(\alpha+\beta) B(\alpha,\beta) = s^\alpha \Gamma(\alpha) s^\beta \Gamma(\beta)
$$
eliminating $s$ gives you the sought relationship
$$
\Gamma(\alpha+\beta) B(\alpha,\beta) = \Gamma(\alpha) \Gamma(\beta)
$$
