Dirac delta as a "mixture" of half-normal distributions

Question

Let's say I have a function $f$ which I want to represent as a "mixture" of half-normal distributions. It's not the mixture exactly because we allow negative "mixture density". To put it another way, can the function $g$ be found such that: $$\int_0^\infty \frac{1}{\sigma}\exp(-\frac{x^2}{2\sigma^2})g(\sigma)d\sigma=f(x)$$ For some function $f$? To solve this problem one has to obtain such a representation for all the Dirac deltas $f(x)=\delta_c(x)$ for all $c \ge 0$. But I don't know how to deal with solving this equation for any $c > 0$. I tried applying a Mellin transform to both sides of the equation. The Mellin transform of RHS is just $c^{s-1}$ while for the LHS we obtain: $$\int_0^{\infty}\int_0^{\infty} \frac{1}{\sigma}\exp(-\frac{x^2}{2\sigma^2})g(\sigma)x^{s-1}d\sigma dx=\int_0^{\infty}\int_0^{\infty} \frac{1}{\sigma}\exp(-\frac{x^2}{2\sigma^2})g(\sigma)x^{s-1}dxd\sigma $$ Which after making the substitution $u=\frac{x}{ \sigma}$ transformes to: $$\int_0^{\infty}\int_0^{\infty}\exp(-\frac{u^2}{2})g(\sigma)u^{s-1}\sigma^{s-1}dud\sigma=\left\{\mathscr{M}\exp(-\frac{x^2}{2})\right\}(s) \cdot \left\{\mathscr{M}g(x)\right\}(s)$$ Where $\mathscr{M}$ denotes the Mellin transform. Since the Mellin transform of the first factor is equal to: $$\left\{\mathscr{M}\exp(-\frac{x^2}{2})\right\}(s)=\Gamma(\frac{s}{2})2^{\frac{s}{2}-1}$$ We have that the Mellin transform of $g$, the function we are looking for is equal to: $$\left\{\mathscr{M}g(x)\right\}(s)=\frac{c^{s-1}}{\Gamma(\frac{s}{2})2^{\frac{s}{2}-1}}$$ But I don't know how to invert it. Is it possible to obtain closed form expression for $g$? Or maybe such function doesn't even exist?