Distribution of $X+Y$ when $X$ and $Y$ are i.i.d with pdf $f(x)=\alpha\beta^{-\alpha}x^{\alpha-1}\mathbf1_{0I am working on the following problem:

Let $X$ and $Y$ be independent random variables with common density $f(x)=\alpha\beta^{-\alpha}x^{\alpha-1}\mathbf1_{0<x<\beta}$ where $\alpha\geqslant1$. Let $U=\min(X,Y)$ and $V=\max(X,Y)$. Find the joint density of $(U,V)$ and hence find the pdf of $U+V$.

As $U+V=X+Y$, I can simply find the pdf of $X+Y$ to see what the pdf of $U+V$ should be. 
I get the pdf of $T=X+Y$ to be $$f_T(t)=\int f(t-y)f(y)\,\mathrm{d}y=\alpha^2\beta^{-2\alpha}\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y\,\mathbf1_{0<t<2\beta}\tag{1}$$
Not sure if that integral can be simplified though. 
Coming back to the actual question, the joint pdf of $(U,V)$ is given by
$$f_{U,V}(u,v)=2f(u)f(v)\mathbf1_{0<u<v<\beta}=2\alpha^2\beta^{-2\alpha}(uv)^{\alpha-1}\mathbf1_{0<u<v<\beta}$$
I did a change of variables $(U,V)\to(W,Z)$ where $W=U+V$ and $Z=U$. Absolute value of jacobian is unity. Also, $0<u<v<\beta\implies 0<z<\frac{w}{2}<\beta$. So marginal pdf of $W$ is
$$f_W(w)=2\alpha^2\beta^{-2\alpha}\int_0^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{dz}\,\mathbf1_{0<w<2\beta}\tag{2}$$
It is possible that I have made some error in the proper supports of the random variables. It is also possible that the integral does not have a solution in terms of elementary functions. In any case, I could not proceed with the integral. So I couldn't even verify that $W=U+V$ has the same pdf as $T=X+Y$. It appears that I am getting different distributions of $W$ and $T$. And out of curiosity, does the distribution of $X$ have a name (in which case I would have searched for the convolution of two such random variables) ?
Edit.
Proceeding with the last integral I get by hand 
$$\int_0^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{dz}=w^{2\alpha-1}\int_0^{1/2}t^{\alpha-1}(1-t)^{\alpha-1}\,\mathrm{dt}=w^{2\alpha-1}I_{1/2}(\alpha,\alpha)B(\alpha,\alpha)$$ where $I_{x}$ is the regularized incomplete beta function. Using the property $I_x(a,b)=1-I_{1-x}(b,a)$, we get $I_{1/2}(\alpha,\alpha)=\frac{1}{2}$. So finally we have $$\int_0^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{dz}=\frac{1}{2}w^{2\alpha-1}B(\alpha,\alpha)$$
This implies that
$$f_W(w)=\alpha^2\beta^{-2\alpha}B(\alpha,\alpha)w^{2\alpha-1}\mathbf1_{0<w<2\beta}$$
That this is not a density in the given range of $w$ is easily seen. So I feel I have made a big mistake somewhere. I have checked my calculations with Mathematica and they seem to agree.
 A: Since
\begin{align}
\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y\,\mathbf1_{0<t<2\beta}&=\begin{cases}
\int_{0}^{t}(y(t-y))^{\alpha-1}\,\mathrm{d}y&\text{when }0\le t\le \beta\\
\int_{t-\beta}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y&\text{when }\beta\le t\le 2\beta\\
\end{cases}\\
\end{align}
we have ($t<\beta$)
$$\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y=
\int_{0}^{t/2}(y(t-y))^{\alpha-1}\,\mathrm{d}y+\int_{t/2}^{t}(y(t-y))^{\alpha-1}\,\mathrm{d}y$$
and by a change of variable $z=t-y$ in the second integral of the rhs
$$\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y=
2\int_{0}^{t/2}(y(t-y))^{\alpha-1}\,\mathrm{d}y$$
Similarly, when $t>\beta$,
\begin{align*}
\int_{t-\beta}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y&=
\int_{t-\beta}^{t/2}(y(t-y))^{\alpha-1}\,\mathrm{d}y+\int_{t/2}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y\\
&=2\int_{t/2}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y
\end{align*}
again by a change of variable $z=t-y$ in the second integral of the rhs. I am however unable to recover the same functional expression for the density in this second case, namely$$2\int_{0}^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{d}z$$
Now, as pointed out in the question,
$$2\int_{0}^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{d}z \propto w^{2(\alpha-1)+1}=w^{2\alpha-1}$$by a change of scale, which would imply that the distribution of interest has the density
$$f(w)\propto w^{2\alpha-1} \mathbf1_{0<w<2\beta}$$
which turns it into a Beta ${\cal B}(2\alpha,1)$ distribution rescaled on $(0,2\beta)$, hence with density
$$f(w) = \{2\beta\}^{-2\alpha}\dfrac{\Gamma(2\alpha+1)}{\Gamma(2\alpha)}w^{2\alpha-1} \mathbf1_{0<w<2\beta}=2\alpha\{2\beta\}^{-2\alpha}w^{2\alpha-1} \mathbf1_{0<w<2\beta}$$

This comes as a contradiction when considering the unbelievably
  detailed answer from W. Huber, since Uniforms are Beta ${\cal
  B}(1,1)$. And since the sum of two Uniforms is not a Beta ${\cal
  B}(2,1)$ random variable, but instead an rv with "tent" density.

Aside: More generally a sum of Beta variates is not another Beta variate, the "explanation" being straightforward when looking at Betas as two Gammas normalised by their sum. Adding two Betas sees different sums in the denominator.
The issue is thus with the derivation of the density of $W=U+V$: since
$$(U,V) \sim 2\alpha \beta^{-2}[uv]^{\alpha-1}\,\mathbb{I}_{0<u<v<\beta}$$
a change of variables $(Z,W)=(U,U+V)$ leads to
$$(Z,W) \sim 2\alpha \beta^{-2}[z(w-z)]^{\alpha-1}\,\mathbb{I}_{0<z<w-z<\beta}$$
and the indicator constraints are
$$0<z \quad 2z<w \quad z<\beta \quad z>w-\beta \quad 0<w \quad\text{and}\quad w<2\beta$$
Therefore, in conclusion,
$$W\sim 2\alpha^2 \beta^{-2\alpha}\int_{\max\{0,w-\beta\}}^{\min\{\beta,w/2\}}[z(w-z)]^{\alpha-1}\,\text{d}z\,\mathbb{I}_{0<w<2\beta}$$
namely (1) and not the proposed expression (2).
