I have to integrate the following: $$\int_0^\infty \frac{1}{(\sqrt{(\pi)}l)^3} \exp({\frac{- (\textbf{r} + \textbf{r}_0)^2}{l^2}}) \exp({-i\textbf{q}. \textbf{r}}) d\textbf{r}$$.
The result is supposedly $$\exp({\frac{-l^2 q^2}{4}}) \exp{i\textbf{q}.\textbf{r}_0}$$
but I am unsure how to evaluate this with the vectors
EDIT: It's from a physics paper and I'll post the full expressions leading to that integral: $$D(q) = \int_0^\infty d\textbf{r} [ (D_v - D_c) |\psi(\textbf{r})|^2 ] \exp^{-i\textbf{q}. \textbf{r}}$$ it then uses Gaussian ground-state wavefunctions given by: $$\psi(\textbf{r}) = \frac{1}{(\sqrt{\pi} l)^{3/2} }\exp(-{\frac{(\textbf{r} + \textbf{r}_0)^2}{2l^2}}).$$ You can ignore the $D_v - D_c$ and $(\sqrt{\pi} l)^{3/2}$ factors if you wish as they're just constants. The integral $D(q)$ is then given by the first equation of the post.
The essential step is to put the exponentials together, and then "complete the square" in $\mathbf{r}$, writing $$-\frac{(\mathbf{r}-\mathbf{r_0})\cdot(\mathbf{r}-\mathbf{r_0})}{l^2}-i\mathbf{q}\cdot\mathbf{r}$$ as $$-\frac{(\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q})\cdot(\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q})}{l^2}+i\mathbf{q}\cdot\mathbf{r_0}-\tfrac{1}{4}l^2\mathbf{q} \cdot \mathbf{q}.$$ Now, when you integrate over space (which I think is what is required here) the Gaussian just gives 1, even though it is shifted to centre about $\mathbf{r_0}+\tfrac{1}{2}il^2\mathbf{q}$, and you are left with the factors you want.
In detail, and using clearer notation for what I think is meant by the problem $$\int_{\text{all of space}} \frac{1}{(\sqrt{(\pi)}l)^3} \exp({\frac{- (\textbf{r} + \textbf{r}_0)^2}{l^2}}) \exp({-i\textbf{q}. \textbf{r}}) \, dV $$ $$=\int \frac{1}{(\sqrt{(\pi)}l)^3} \exp(-\frac{(\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q})\cdot(\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q})}{l^2}+i\mathbf{q}\cdot\mathbf{r_0}-\tfrac{1}{4}l^2\mathbf{q} \cdot \mathbf{q}) \, dV $$ taking out the factors independent of $\mathbf{r}$ $$=\exp{(i\mathbf{q}\cdot\mathbf{r_0}-\tfrac{1}{4}l^2\mathbf{q}} \cdot \mathbf{q})\int \frac{1}{(\sqrt{(\pi)}l)^3} \exp(-\frac{(\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q})\cdot(\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q})}{l^2}) \, dV $$ and substituting $\mathbf{r'}=\mathbf{r}-\mathbf{r_0}-\tfrac{1}{2}il^2\mathbf{q}$, noting that translating the whole of space just gives the whole of space, $$=\exp{(i\mathbf{q}\cdot\mathbf{r_0}-\tfrac{1}{4}l^2\mathbf{q}} \cdot \mathbf{q})\int_{\text{all of space}} \frac{1}{(\sqrt{(\pi)}l)^3} \exp(-\frac{\mathbf{r'}\cdot\mathbf{r'}}{l^2}) \, dV' $$ and the Gaussian integral is equal to 1.