Given beta distribution as: $ \mathcal{B}(x;a,b) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1} $
I am trying to show: $ \int_0^1 x^{a-1} (1-x)^{b-1}\,dx = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} $
which can be considered as the normalizing constant.
Here is what I have done so far (using the hints at Bishop's). Write $\Gamma(a)\Gamma(b)$ and change variable $t=y+x$ fixing $x$:
\begin{align*} \newcommand{\ints}{\int_0^\infty\!\!\int_0^\infty} \Gamma(a)\Gamma(b) &= \int_0^\infty \exp(-x)x^{a-1} \, dx \int_0^\infty \exp(-y)y^{b-1} \, dy \\ &= \ints \exp(-x)x^{a-1} \exp(-y)y^{b-1} \, dx\,dy \\ &= \ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx \end{align*}
Now if we make the variable change $x=\mu t$ \begin{align*} \ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx &= \ints e^{-t} (\mu t)^{a-1} (t-\mu t)^{b-1} t \, dt\, d\mu \\ &= \ints e^{-t} t^{a+b-1} \mu^{a-1} (1-\mu)^{b-1} \, d\mu \, dt \end{align*}
We can split the integrals: \begin{align*} &= \underbrace{\int_0^\infty e^{-t} t^{a+b-1} \,dt}_{\Gamma(a+b)} \underbrace{\int_0^\infty \mu^{a-1} (1-\mu)^{b-1} \, d\mu}_{\text{limits are not from 0 to 1}} \end{align*}
What is wrong with the limits? Is it my mistake? Or can it be substituted to from $0$ to $1$?
PS. sorry for the big space between double integrals, I don't know the right way to do them.