Interchanging order of integration in the Mellin transform It is possible to understand (perhaps somewhat non-rigorously) the Fourier transform through interchanging order of integration and use of the delta function, like so:
$$\hat{f}(k)\equiv\int_{-\infty}^{\infty}dx \,f(x)e^{ikx},$$ $$\int_{-\infty}^{\infty}\frac{dk}{2\pi} \,\hat{f}(x)e^{-ikx}=\int_{-\infty}^{\infty}dy \,f(y)\int_{-\infty}^{\infty}\frac{dk}{2\pi}e^{-ik(x-y)}=\int_{-\infty}^{\infty}dy \,f(y)\delta(x-y)=f(x)$$
Now I'm curious to what extent such a formal manipulation is valid for a Mellin transform (with contour C chosen appropriately).
$$\hat{f}(z)\equiv\int_{0}^{\infty}dk \,k^{z-1}\,f(k),\qquad f(k)=\int_C\frac{dz}{2\pi i} k^{-z}\hat{f}(z)$$
$$\hat{f}(z)=^? \int_C\frac{dw}{2\pi i} \hat{f}(w)\int_{0}^{\infty}dk \,k^{-w+z-1}$$
The $k$ integral doesn't seem to make much sense since no matter what the exponent is it will diverge at one of the limits of integration. In dimensional regularization in physics, we take these integrals to vanish*, which clearly doesn't make any sense in this situation. So it seems that interchanging the order of integration is nonsense.
But in other situations in physics, interchanging the order of integration with the inverse Mellin transform does seem to make sense. Suppose we are calculating some integral $G(z,m)$
$$G(z,m)\equiv \int_{0}^{\infty}dk \frac{k^{z-1}}{k^2+m^2}f(k) =\int_C\frac{dw}{2\pi i} \hat{f}(w)\int_{0}^{\infty}dk \,\frac{k^{-w+z-1}}{k^2+m^2}$$
Now the $k$ integral can be given a meaning in dimensional regularization, which will involve some Euler gamma functions which introduce some new poles in addition to the poles of $\hat{f}(w)$ that we must take into account when we do the final integral over $C$.
So I am wondering how to make sense of the case involving $\hat{f}(z)$, which is equivalent (after a shift of $z$) to $G(z,0)$ which seems to work for non-zero $m$?

*However, note the paper Gorishnij and Isaev, Teor. Mat. Fiz.; 62.3 (1985), which argues it should be understood as a delta function $2\pi i \delta(w-z)$
 A: The Mellin and inverse Mellin transforms are defined as follows.

$$F(z)=\mathcal{M}_k[f(k)](z)=\int\limits_0^\infty f(k)\,k^{z-1}\,dk\tag{1}$$
$$f(k)=\mathcal{M}_z^{-1}[F(z)](k)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty} F(z)\,k^{-z}\,dz\tag{2}$$

Replacing $f(k)$ in formula (1) above using formula (2) above (where $z$ is also replaced by $s$) leads to the following.
$$F(z)=\int\limits_0^{\infty}\left(\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty} F(s)\,k^{-s}\,ds\right) k^{z-1}\,dk\tag{3}$$

Interchanging the order of integration and rearranging some terms in formula (3) above leads to the following.
$$F(z)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty} F(s)\left( \int_0^{\infty } k^{-s} k^{z-1} \, dk\right) \, ds\tag{4}$$

Noting the nested integral in formula (3) above is the Mellin transform $\mathcal{M}_k\left[k^{-s}\right](z)=2\,\pi\,\delta(i\,(z-s))$ (see here) leads to the following.
$$F(z)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty} F(s)\,2\,\pi\,\delta(i\,(z-s))\,ds\tag{5}$$

The integral in formula (5) above can be evaluated by setting $c=z$ and using the variable substitution $s=z+i\,t$ which leads to $ds=i\,dt$ and $t=-i\,(s-z)$. The lower integration limit $s=z-i\,\infty$ becomes $t=-i\,(z-i\,\infty-z)=-\infty$, the upper integration limit $z+i\,\infty$ becomes $t=-i\,(z+i\,\infty-z)=\infty$, $F(s)$ becomes $F(z+i\,t)$, and $\delta(i\,(z-s))$ becomes $\delta(i\,(z-(z+i\,t)))=\delta(t)$ which leads to the following.
$$F(z)=\frac{1}{2\,\pi\,i}\int\limits_{z-i\,\infty}^{z+i\,\infty} F(s)\,2\,\pi\,\delta(i\,(z-s))\,ds=\frac{1}{2\,\pi\,i}\int\limits_{-\infty}^{\infty} F(z+i\,t)\,2\,\pi\,\delta(t)\,i\,dt=F(z)\tag{6}$$
