If $f(x) = e^{-a x^2}$ and $g(x) = e^{-b x^2}$ with $a, b > 0$, does there exist a pair of positive numbers $(A, c)$ such that the equation
$$A e^{-cx^2} = f(x) + g(x) $$ is satisfied? I tried Gaussian integration but it does not seem to work. I am actually looking for the answer to the more general question:
Let $\vec \xi = \left[x_1 \ x_2 \ ... \ x_n \right]^T$ be an $n$ dimensional vector. Suppose we have two Gaussian functions: $$f(\vec \xi) = \frac{1}{\sqrt{\mathrm{det}A}} \exp(\vec \xi^T A^{-1}\vec \xi),$$ and $$g(\vec \xi) = \frac{1}{\sqrt{\mathrm{det}B}} \exp(\vec \xi^T B^{-1}\vec \xi),$$ where $A$ and $B$ are two $n \times n$ symmetric and positive definite matrices.
If $k (\vec \xi) = f (\vec \xi) + g (\vec \xi)$, can we find a matrix $C$ (with the same properties) such that $$k(\vec \xi) = \frac{1}{\sqrt{\mathrm{det}C}} \exp(\vec \xi^T C^{-1}\vec \xi)$$
Impossible unless $a=b$. Suppose you can find such $A$ and $c$. From the behavior at infinity, if $a<b$, you need $c=a$. But then if you write $$ Ae^{-cx^2}=e^{-ax^2}+e^{-bx^2} $$ and you divide both sides by $e^{-ax^2}=e^{-cx^2}$ and take the limit at $\infty$, you conclude $A=1$ necessarily. After cancellation, you get $e^{-ax^2}=0$ identically, which is impossible.