Gaussian integration with delta function An integration is given:
\begin{equation}
I = \int d^2 \vec{x}_1 \, d^2 \vec{x}_2  \, \delta^2 (\vec{x}_1  + \vec{x}_2 - \vec{x}) f_1(\vec{x}_1) f_2(\vec{x}_2)
\tag{1}
\end{equation}
Where we can assume that; the average of the $\langle x_1^2 \rangle = \langle x_2^2 \rangle$, where the functions are given:
\begin{align*}
f_1(\vec{x}_1) &= \frac{1}{\langle x_1^2 \rangle} e^{- x_1^2 / \langle x_1^2 \rangle} \\
f_2(\vec{x}_2) &= \frac{1}{\langle x_2^2 \rangle} e^{- x_2^2 / \langle x_2^2 \rangle}
\end{align*}
If we multiply the two functions, we get:
$$f_1(\vec{x}_1) f_2(\vec{x}_2) = \frac{1}{\pi^2 \langle x_1^2 \rangle\langle x_2^2 \rangle} e^{-(x_1^2 + x_2^2)/\langle x_1^2 \rangle} $$
Inserting this in the equation $\text{(1)}$ will give, with considering $(\vec{x}_1  + \vec{x}_2  = \vec x )$, which means:  $x_1^2 + x_2^2 = x^2+2x_2^2-2x\cdot x_2$ [just vector addition formula]
\begin{equation}
I = \int d^2 \vec{x}_1 \, d^2 \vec{x}_2 \, \delta^2 (\vec{x}_1 + \vec{x}_2 - \vec{x}) \frac{1}{\pi^2 \langle x_1^2 \rangle\langle x_2^2 \rangle} e^{-(x_1^2 + x_2^2)/\langle x_1^2 \rangle}
\tag{2}
\end{equation}
If we integrate with respect to the $d^2 \vec{x_1}$, we get setting $\vec x_1 = \vec x - \vec x_2 $ in equation $\text{(2)}$
\begin{equation}
I = \int d^2 \vec{x}_2 \, \frac{1}{\pi^2 \langle x_1^2 \rangle\langle x_2^2 \rangle} e^{-(x^2+2x_2^2-2x\cdot x_2+ x_2^2)}
\tag{2}
\end{equation}
Am I doing in the right way?
However the result needs to be
$$I = \frac{e^{-\frac{-x^2}{\langle x_1^2 \rangle + \langle x_2^2 \rangle}}}{{\pi(\langle x_1^2 \rangle + \langle x_2^2 \rangle)}}$$
The equation I'm trying to get is 6 in this article: which follows from equation 4.
 A: Given the form of the desired answer, I will drop the assumption $\langle x_1^2\rangle=\langle x_2^2\rangle$ for generality. It looks like you meant to include a $\frac{1}{\pi}$ factor in your definitions, so I'll take $f_j(x_j):=\frac{1}{\pi\langle x_j^2\rangle}\exp\frac{-x_j^2}{\langle x_j^2\rangle}$. Integrating out $x_1$,$$\begin{align}I&=\frac{1}{\pi^2\langle x_1^2\rangle\langle x_2^2\rangle}\int d^2x_2\exp\left(-\frac{(x-x_2)^2}{\langle x_1^2\rangle}-\frac{x_2^2}{\langle x_2^2\rangle}\right)\\&=\frac{1}{\pi^2\langle x_1^2\rangle\langle x_2^2\rangle}\int d^2x_2\exp\left[-\frac{\langle x_1^2\rangle+\langle x_2^2\rangle}{\langle x_1^2\rangle\langle x_2^2\rangle}\left(x_2^2-2\frac{\langle x_2^2\rangle}{\langle x_1^2\rangle+\langle x_2^2\rangle}x\cdot x_2+\frac{\langle x_2^2\rangle}{\langle x_1^2\rangle+\langle x_2^2\rangle}x^2\right)
\right].\end{align}$$Next we complete the square with $y:=x_2-\frac{\langle x_2^2\rangle}{\langle x_1^2\rangle+\langle x_2^2\rangle}x$, so$$\begin{align}I&=\frac{1}{\pi^2\langle x_1^2\rangle\langle x_2^2\rangle}\int d^2y\exp\left[-\frac{\langle x_1^2\rangle+\langle x_2^2\rangle}{\langle x_1^2\rangle\langle x_2^2\rangle}\left(y^2+\frac{\langle x_1^2\rangle\langle x_2^2\rangle}{(\langle x_1^2\rangle+\langle x_2^2\rangle)^2}x^2\right)\right]\\&=\frac{1}{\pi(\langle x_1^2\rangle+\langle x_2^2\rangle)}\exp\frac{-x^2}{\langle x_1^2\rangle+\langle x_2^2\rangle},\end{align}$$as required.
