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?
 A: Formally,
$$\begin{align}
\langle D_m, \phi \rangle
& = \int_{-\infty}^{\infty} D_m(y) \phi(y) \, dy \\
& = \int_{-\infty}^{\infty} \left( \int_0^\infty x^{m+iy} \, dx \right) \phi(y) \, dy \\
& = \int_0^\infty \left( \int_{-\infty}^{\infty} x^{m+iy} \phi(y) \, dy \right) \, dx \\
& = \int_0^\infty x^m \left( \int_{-\infty}^{\infty} x^{iy} \phi(y) \, dy \right) \, dx \\
& = \int_0^\infty x^m \left( \int_{-\infty}^{\infty} e^{iy \ln(x)} \phi(y) \, dy \right) \, dx \\
& = \int_0^\infty x^m \hat\phi(-\ln(x)) \, dx \\
& = \{ \text{substitution } x = e^t \} \\
& = \int_{-\infty}^{\infty} e^{mt} \hat\phi(-t) \, e^t \, dt \\
& = \int_{-\infty}^{\infty} e^{(m+1)t} \hat\phi(-t) \, dt
\end{align}$$
The case $m = -1$:
$$
\langle D_{-1}, \phi \rangle
= \int_{-\infty}^{\infty} e^{(-1+1)t} \hat\phi(-t) \, dt
= \int_{-\infty}^{\infty} e^{i0t} \hat\phi(-t) \, dt
= 2\pi \phi(0) 
= \langle 2\pi \delta, \phi \rangle
$$
by the Fourier inversion theorem.
When $m \neq -1$ the factor $e^{(m+1)t}$ grows too, either at $+\infty$ (when $m>-1$) or at $-\infty$ (when $m<-1$) fast to be a tempered distribution and we can't use Fourier transform. Therefore $D_m$ for $m \neq -1$ is not defined on the Schwartz space.
However, it is defined when $\hat\phi \in \mathscr E'$, which I think is the case when $\phi \in C^\omega$, i.e. when $\phi$ is analytic. Then there exists $R>0$ such that $\hat\phi(-t) = 0$ for $|t|>R$, so
$$\begin{align}
\int_{-\infty}^{\infty} e^{(m+1)t} \hat\phi(-t) \, dt
& = \int_{-R}^{R} e^{(m+1)t} \hat\phi(-t) \, dt \\
& = \int_{-\infty}^{\infty} \chi_{(-R, R)}(t) \, e^{(m+1)t} \hat\phi(-t) \, dt \\
& = \int_{-\infty}^{\infty} \chi_{(-R, R)}(t) \, e^{-(m+1)t} \hat\phi(t) \, dt \\
& = \int_{-\infty}^{\infty} \mathscr F \left\{ \chi_{(-R, R)}(t) \, e^{-(m+1)t} \right\}(\omega) \, \phi(\omega) \, d\omega
\end{align}$$
Thus,
$$\begin{align}
D_m(\omega) 
& = \mathscr F \left\{ \chi_{(-R, R)}(t) \, e^{-(m+1)t} \right\}(\omega) \\
& = \int_{-R}^{R} e^{-(m+1)t} \, e^{-i\omega t} \, dt \\
& = \frac{e^{((m+1)-i\omega)R}-e^{-((m+1)-i\omega)R}}{(m+1)-i\omega}
\end{align}$$
There's a problem here, though, that I don't know how to handle. We chose $R$ rather arbitrary. Nothing prevents us from taking a larger $R$. Hopefully someone else comes by and can sort it out.
A: 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?
