Assuming that $a > 0,\ {\tt Mathematica}$ says that $$\!\!\! 2\int_{0}^{\infty}\!\!\!\int_{0}^{\infty} \frac{v^{a}\,\left(u + v\right)^{a - 1}\,\, \mathrm{e}^{-u -2 v}}{\Gamma^{2}\left(a\right)} \,\mathrm{d}u\,\mathrm{d}v = a - \frac{\Gamma\left(a + 1/2\right)}{\Gamma\left(1/2\right)\Gamma\left(a\right)}.$$
How can we arrive at the solution $?$. I don't see the proper change(s) of variables.
Note: $\displaystyle\Gamma\left(a\right) = \int_{0}^{\infty}t^{a - 1}\,\,\mathrm{e}^{-t} \,\mathrm{d}t$.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$
\begin{align} &\bbox[5px,#ffd]{2\int_{0}^{\infty}\int_{0}^{\infty} {v^{a}\pars{u + v}^{a - 1}\, \expo{-u - 2v} \over \Gamma^{2}\pars{a}}\dd u\,\dd v} \\[5mm] = &\ {2 \over \Gamma^{2}\pars{a}} \int_{0}^{\infty}\int_{v}^{\infty} v^{a}\expo{-v}u^{a - 1}\,\expo{-u}\,\,\dd u\,\dd v \\[5mm] = &\ {2 \over \Gamma^{2}\pars{a}} \int_{0}^{\infty}\int_{0}^{\infty}\ \underbrace{\bracks{\int_{-\infty}^{\infty} {\expo{\ic k\pars{u - v}} \over k - \ic 0^{+}}{\dd k \over 2\pi\ic}}} _{\ds{\Theta\pars{u - v}}}\ \times \\[2mm] &\ v^{a}\expo{-v}u^{a - 1}\,\expo{-u}\,\,\dd u\,\dd v \end{align} $\ds{\Theta}$ is the Heaviside Theta ( or Heaviside Step ) Function.
Then, \begin{align} &\bbox[5px,#ffd]{2\int_{0}^{\infty}\int_{0}^{\infty} {v^{a}\pars{u + v}^{a - 1}\, \expo{-u - 2v} \over \Gamma^{2}\pars{a}}\dd u\,\dd v} \\[5mm] = &\ {2 \over \Gamma^{2}\pars{a}} \int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}} \underbrace{\bracks{\int_{0}^{\infty}v^{a} \expo{-\pars{1 + \ic k}v}\,\,\dd v}} _{\ds{\pars{1 + \ic k}^{-a - 1}\,\,\Gamma\pars{a + 1}}} \ \times \\[2mm] &\ \underbrace{\bracks{\int_{0}^{\infty}u^{a - 1} \expo{-\pars{1 - \ic k}u}\,\,\dd u}} _{\ds{\pars{1 - \ic k}^{-a}\,\,\Gamma\pars{a}}} {\dd k \over 2\pi\ic} \\[5mm] = &\ {2 \over \Gamma^{2}\pars{a}}\,\Gamma\pars{a + 1} \Gamma\pars{a}\ \times \\[2mm] &\ \int_{-\infty}^{\infty} {\pars{1 + \ic k}^{-a - 1}\,\,\pars{1 - \ic k}^{-a} \over k - \ic 0^{+}}\,{\dd k \over 2\pi\ic} \\[5mm] = &\ 2a\left[% \mrm{P.V.}\int_{-\infty}^{\infty} {\pars{1 + k^{2}}^{-a}\,\pars{1 + \ic k}^{-1} \over k}\,{\dd k \over 2\pi\ic}\right. \\[2mm] &\ \left.+ \int_{-\infty}^{\infty} \pars{1 + k^{2}}^{-a}\,\pars{1 + \ic k}^{-1} \bracks{\ic\pi\,\delta\pars{k}}{\dd k \over 2\pi\ic}\right] \\[5mm] = &\ 2a\int_{0}^{\infty} {1 \over \pars{1 + k^{2}}^{a}\,k} \pars{{1 \over 1 + \ic k} - {1 \over 1 - \ic k}} \,{\dd k \over 2\pi\ic} \\[2mm] &\ + a \\[5mm] = &\ a - {2a \over \pi}\ \underbrace{\int_{0}^{\infty}{\dd k \over \pars{1 + k^{2}}^{a + 1}}} _{\ds{\root{\pi}\Gamma\pars{a + 1/2} \over 2\Gamma\pars{a + 1}}} \\[5mm] = &\ \bbx{a - {\Gamma\pars{a + 1/2} \over \Gamma\pars{1/2}\Gamma\pars{a}}}\quad \pars{\substack{\mbox{Note that}\\[1mm] \ds{\root{\pi} = \Gamma\pars{1 \over 2}}}}\\ & \end{align}
The last integral can be evaluated with Ramanujan's Master Theorem. Namely, \begin{align} &\int_{0}^{\infty} {\dd k \over \pars{1 + k^{2}}^{a + 1}} = {1 \over 2}\int_{0}^{\infty} {k^{\color{red}{1/2} - 1} \over \pars{1 + k}^{a + 1}}\,\dd k \end{align} and \begin{align} &{1 \over \pars{1 + k}^{a + 1}} = \sum_{n = 0}^{\infty}{-a - 1 \choose n}k^{n} \\[5mm] = &\ \sum_{n = 0}^{\infty}{a + n \choose n} \pars{-1}^{n}k^{n} = \sum_{n = 0}^{\infty} \color{red}{\Gamma\pars{a + 1 + n} \over \Gamma\pars{a + 1}}{\pars{-k}^{n} \over n!} \end{align} Then, \begin{align} &\int_{0}^{\infty} {\dd k \over \pars{1 + k^{2}}^{a + 1}} = {1 \over 2}\Gamma\pars{1 \over 2}\, {\Gamma\pars{a + 1 - 1/2} \over \Gamma\pars{a + 1}} \\[5mm] = & \root{\pi}\Gamma\pars{a + 1/2} \over 2\Gamma\pars{a + 1} \end{align}