Fourier transform of some tempered distributions Problem: Find the Fourier transform of the tempered distributions $F_1(f) = \int_0^{\infty} f(x)xdx$, and $F_2(f) = (\frac{d^n}{dx^n} \delta)(f)$, where $f$ is assumed to be a Schwartz function.
Attempt:
$$(1)\,\begin{align*}\mathcal{F}F_1(f) 
&= \int_0^{\infty} \widehat f(x)x\,dx\\
&= \int_0^{\infty} \int_{-\infty}^{\infty}f(y)e^{-2 \pi i yx} x\,dy\,dx\\
&= \int_0^{\infty} 0 + \frac{1}{2\pi i}\int_{-\infty}^{\infty} f'(y)e^{-2 \pi i xy}\,dy\,dx
\end{align*}$$
where the last step follows from integration by parts and the fact the $f$ is a Schwartz function. Now, I am not sure how to proceed from here to reach a significant new tempered distribution. Moreover, the integrand doesn't seem to belong to $L^1(\mathbb{R} \times \mathbb{R_{\gt 0}})$, so I can't even use Fubini...
$$(2)\begin{align*}
\mathcal{F}F_2(f) 
&= \int_{-\infty}^{\infty} \widehat{(f^{(n)})}(x)\,dx\\
&= \widehat {(f^{(n)})}(0)\\
&= \int_{-\infty}^{\infty} f(y) \frac{e^{-2 \pi i xy}}{-2 \pi i y}\,dy
\end{align*}$$
At this point, this seems like a complex analysis problem... I am not sure how to proceed with the integral though as my background in complex analysis is weak.
Any help is appreciated!
 A: The definition of Fourier transform that we are using:
$$\hat{f}(y) = \int f(x) \, e^{-i \, 2\pi \, xy} \, dx$$
The inverse is
$$f(x) = \int \hat{f}(y) \, e^{i \, 2\pi \, xy} \, dy$$
and
$$\hat{\hat{f}}(t) = f(-t).$$

First we note that $\hat{\delta}(y) = 1,$ that $\widehat{f'} = i \, 2\pi \, y \, \hat{f}(y)$ and that $\widehat{xf}(y) = \frac{i}{2\pi} \hat{f}'(y).$ Also note that the property of being an even (odd) function is preserved under the Fourier transform.
We can use this directly for $F_2$:
$$\widehat{F_2}(y) = \widehat{\delta^{(n)}}(y) = (i \, 2\pi \, y)^n \hat{\delta}(y) = (i \, 2\pi \, y)^n.$$
For $F_1$ we note that $F_1(x) = x H(x),$ where $H$ is the Heaviside step function, having derivative $H' = \delta.$ Therefore, $\widehat{F_1}(y) = \frac{i}{2\pi} \hat{H}(y).$ But what is $\hat{H}$?
We have $H' = \delta$ so $i \, 2\pi \, y \, \hat{H}(y) = 1,$ i.e. $\hat{H} = \frac{1}{i \, 2\pi \, y} + C \, \delta(y)$ for some constant $C.$ But we also have $H(x) = \frac12 \theta(x) + \frac12,$ where $\theta$ is the sign function. Thus we have
$$\frac{1}{i \, 2\pi \, y} + C \, \delta(y) = \frac12 \hat{\theta}(y) + \frac12 \hat{1}(y).$$
Now it's important to note that the first term on both sides is an odd function while the second term is an even function. We must therefore have
$$
\frac{1}{i \, 2\pi \, y} = \frac12 \hat{\theta}(y)
\quad \text{and} \quad
C \, \delta(y) = \frac12 \hat{1}(y).
$$
But $\hat{1}(y) = \hat{\hat{\delta}}(y) = \delta(-y) = \delta(y).$ Therefore we must have $C = \frac12$ and thus
$$\hat{H} = \frac{1}{i \, 2\pi \, y} + \frac12 \, \delta(y)$$
and therefore
$$\widehat{x H}(y) 
= \frac{i}{2\pi} \frac{d}{dy} \left( \frac{1}{i \, 2\pi \, y} + \frac12 \, \delta(y) \right)
= \frac{i}{2\pi} \left( -\frac{1}{i \, 2\pi \, y^2} + \frac12 \, \delta'(y) \right) \\
= -\frac{1}{(2\pi)^2 y^2} + \frac{i}{4\pi} \delta'(y)
$$
