Distribution integral result? Does anyone know what the distribution defined by the following integral is?
$$D_{m}(y)=\int_0^\infty dx\,x^{m+iy}$$
where $m\in\mathbb{Z}$ and $y\in\mathbb{R}$. I know that for $m=-1$ it is the delta distribution $D_{-1}(y)=2\pi\delta(y)$. But what about other $m\in\mathbb{Z}$ cases?

Here is what I've come up with so far:
$$\begin{align}
\int_{-\infty}^\infty dz \phi(z)\int_0^\infty \frac{dx}{x} x^{n+iz} &=  \int_{-\infty}^\infty dz \phi(z)\int_{-\infty}^\infty dy\, e^{ny}e^{iyz}\\
&=  \int_{-\infty}^\infty dz \phi(z)\sum_{m=0}^\infty\frac{n^m}{m!}\int_{-\infty}^\infty dy\, y^m e^{iyz}\\
&=  \int_{-\infty}^\infty dz \phi(z)\sum_{m=0}^\infty\frac{n^m}{m!}\left(-i\frac{\partial}{\partial z}\right)^m\int_{-\infty}^\infty dy\,  e^{iyz}\\
&=  2\pi\int_{-\infty}^\infty dz \phi(z)\sum_{m=0}^\infty\frac{(-in)^m}{m!}\delta^{(m)}(z)\\
&=  2\pi\int_{-\infty}^\infty dz \phi(z)\left(e^{-in\frac{\partial}{\partial z}}\delta(z)\right)\\
&=  2\pi\int_{-\infty}^\infty dz \left(e^{in\frac{\partial}{\partial z}}\phi(z)\right)\delta(z)\\
&=2\pi \left(e^{in\frac{\partial}{\partial z}}\phi(z)\right)|_{z=0}
\end{align}$$
What do you think? Is this correct?
