Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments The Mellin transform/inverse transform pair are defined as follows:
(1) $\quad F(s)=\mathcal{M}_x[f(x)](s)=\int\limits_0^\infty f(x)\,x^{s-1}\,dx$
(2) $\quad f(x)=\mathcal{M}_s^{-1}[F(s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\infty}^{c+i\infty}F(s)\,x^{-s}\,ds$
I've been struggling to understand Mathmatica evaluations such as the following. Formula (6) below is the Mellin transform of the contribution of zeta zero $k$ (i.e. $\rho_k$) in von Mangoldt's explicit formula for the second Chebyshev function which was my initial interest, but formulas (3) and (4) below are simpler examples, and formula (5) below is a more general example.
(3) $\quad\mathcal{M}_x[x](s)=\delta (s+1)$
(4) $\quad \mathcal{M}_x\left[x^{1+i}\right](s)=\delta (s+(1+i))$
(5) $\quad\mathcal{M}_x\left[x^j\right](s)=\delta(s+j)$
(6) $\quad\mathcal{M}_x\left[-\frac{x^{\rho_k}}{\rho_k}\right](s)=-\frac{\delta\left(s+\rho_k\right)}{\rho_k}$
The results illustrated above seem inconsistent with the definition of the inverse Mellin transform illustrated in (2) above. The Dirac delta function makes sense to me in the context of integration along the real axis, whereas the direction of integration in (2) above is orthogonal to the real axis.
Consequently, inverse Mellin transforms such as the following don't seem to make sense to me.
(7) $\quad \mathcal{M}_s^{-1}[\delta(s+j)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\infty}^{c+i\infty}\delta(s+j)\,x^{-s}\,ds=x^j$
Question 1: Is the evaluation of the Mellin transform of $x^j$ in (5) above correct?
Question 2: Assuming the answer to Question 1 above is no, what is the correct Mellin transform of $x^j$?
Question 3: Assuming the answer to Question 1 above is yes, can anyone explain how the result of the Mellin transform of $x^j$ in (5) above makes sense in the context of the inverse Mellin transform illustrated in (7) above?
Assuming the answer to Question 1 above is yes and the answer to Question 3 above is no, it seems to me there's something wrong with the theory of Mellin transforms and/or Distributions which leads to the following two questions.
Question 4a: Can an alternate version of the Mellin transform/inverse transform pair be defined where the transform/inverse transform pair associated with $x^j$ makes more sense?
Question 4b: Can the distribution framework be extended to support a Dirac delta function evaluated along the imaginary axis in order to make more sense of the transform/inverse transform pair associated with $x^j$?
Question 4b above is explored in the answer I posted below.
I've also been trying to understand the evaluation of an integral associated with a Dirac delta function with a complex argument such as $\delta(s+(1+i))$ which was the result of evaluation (3) above. It seems to me integral (8) below should evaluate to 1, but I can't seem to get Mathematica to evaluate this integral. A simple substitution of variable $s=t-i$ in (8) below leads to (9) below which is obviously correct.
(8) $\quad\int\limits_{-\infty-i}^{\infty-i}\delta(s+(1+i))\,ds$
(9) $\quad\int\limits_{-\infty }^{\infty }\delta(t+1)\,dt=1$
Question (5): Shouldn't integral (8) above evaluate to 1? If not, why not and what is the proper interpretation of a Dirac delta function with a complex argument?
 A: $\mathcal{M}_s[1](x)=2\,\pi\,\delta(i\,s)$, is a nonsense.
$ \lim_{a \to \infty} \int_{-a}^a e^{-ik x }dx$ diverges for every $x$. 
What mathematica says is incorrect because it doesn't mention "convergence in the sense of distributions".
With $\tau(x)  =\int_{-\infty}^x \frac{2\sin(y)}{y}dy,\tau(-\infty) = 0,\tau(+\infty) = 2\pi$, we have a convergence in the sense of distributions
$$\int_{-\infty}^\infty e^{-ik x }dx = \lim_{a \to \infty} \int_{-a}^a e^{-ik x }dk = \lim_{a \to \infty} a\frac{2 \sin(ax)}{ax} \\ =\lim_{a \to \infty}\frac{d}{dx} \tau(ax) =\frac{d}{dx} 2\pi \ 1_{x > 0} = 2\pi \delta(x)$$ 
In the sense of distributions ? It means for any $\varphi \in C^\infty_c$ (more generally for any $\varphi \in L^1, \varphi' \in L^1$)
$$\lim_{a \to \infty}\int_{-\infty}^\infty \varphi(x) (\int_{-a}^a e^{-ik x }dk) dx = 2\pi \varphi(0)$$
This is equivalent to the Fourier inversion theorem because it means $$\int_{-\infty}^\infty \hat{\varphi}(k)e^{iky}dk =\lim_{a \to \infty}\int_{-a}^a\hat{\varphi}(k)e^{iky}dk= \lim_{a \to \infty}\int_{-a}^a e^{iky}\int_{-\infty}^\infty \varphi(x)e^{-ikx}dx dk\\= \lim_{a \to \infty}\int_{-a}^a e^{iky}\int_{-\infty}^\infty \varphi(x+y)e^{-ik(x+y)}dx dk
= \lim_{a \to \infty}\int_{-\infty}^\infty \varphi(x+y)\int_{-a}^a  e^{-ik x} dkdx=2\pi \varphi(y)$$
What about $\lim_{a \to \infty} \int_{-a}^a e^{ik (z-x) }dx$ ? It converges to an analytic functional. For any $\phi : \mathbb{C} \to \mathbb{C}$ complex analytic  such that for every $k,y$, $\int_{-\infty}^\infty |\phi(x+iy) x^k| dx < \infty$ then 
$$\lim_{a \to \infty}\int_{-\infty}^\infty \phi(x) \int_{-a}^a e^{ik(z- x) }dx = 2\pi \varphi(z)$$
For those $\phi$ complex analytic and Schwartz on every horizontal line (and only for those) it makes sense to talk for $z \in \mathbb{C}$ of the analytic functional $\delta_z$ such that $\langle  \delta_z, \phi \rangle = \phi(z)$.
