Integral $\int_{\mathbb{R^2}}\exp(-\gamma\vert x\vert^p)\exp(-\gamma\vert y-x\vert^p)dxdy$ Let $2\ge p>1$ and $\gamma>0.$

Is it possible to compute $$\int_{\mathbb{R^2}}\exp(-\gamma\vert x\vert^p)\exp(-\gamma\vert y-x\vert^p)dxdy\;?$$

for $p=2$ it's not difficult because we can expand the square but for general $p$ it cannot work.
Also note that is the $L^1$ norm of the convolution for $f(t)=\exp(-\gamma\vert t\vert^p).$ 
Note that $\int_{\mathbb{R^2}}\exp(-\vert y-x\vert^p)dxdy\;$ diverges so that the exposant $\exp(-\vert x\vert^p)$ is very important.
We can give a explicit bound using Young inequality but I would prefer the value if possible of course.
 A: Indeed, as the other answer points out, you should use $u = x, v = y-x$ and evenness of the integrand to arrive at $$I = 4\left( \int^\infty_0 e^{-\gamma z^p} dz\right)^2.$$ Now use $t = \gamma z^p$, so $dt = \gamma p z^{p-1} dz$ or $\frac{t^{1/p - 1}dt}{\gamma^{1/p} p} = dz$ to see $$I = \frac{4}{\gamma^{2/p}p^2} \left(\int^\infty_{0} t^{\frac{1}{p}-1}e^{-t}dt \right)^2 = \frac{4}{\gamma^{2/p} p^2} \Gamma\left(\frac{1}{p}\right)^2,$$ where $\Gamma$ is the usual Gamma function. (You'll definitely want to write down the transformation yourself and make sure I got the exponent right, but I think it is correct.) At this point, if you want, you can also use that $z\Gamma(z) = \Gamma(z+1)$ to re-write this as $$I = \frac{4}{\gamma^{2/p}} \Gamma\left(1+\frac 1 p \right)^2.$$ Since the Gamma function is non-elementary, I believe this is the simplest expression you can hope for. 
A: For first step, do a change of variables: $u=x$ and $v=y-x$. The Jacobian of the transformation is $1$, so you need to calculate $$I=\int_{\mathbb R^2}\exp(-\gamma |u|^p)\exp(-\gamma |v|^p) du dv$$
The two integrals are independent, and also you can integrate only from $0$:
$$I=\left(2\int_0^\infty e^{-\gamma z^p}dz\right)^2$$
The integral is solved by changing the variable $t=\gamma z^p$ and you obtain the integral for the Gamma function
