Difficult Fourier integral giving a distribution I would like to understand the distribution defined by
$$
b(x)=\int_{-\infty}^{\infty}\lvert y\rvert e^{-ixy} dy
$$
What I've understood so far is that
$$
b(x)=\lim_{\alpha\to0^+}\int_{-\infty}^{\infty}\lvert y\rvert e^{-ixy}e^{-\alpha y^2} dy
= \lim_{\alpha\to0^+} \frac{1}{\alpha}\left(1-\frac{x\sqrt{\pi}}{2\sqrt{\alpha}}\exp(-\frac{x^2}{4\alpha})\text{erfi}(\frac{x}{2\sqrt{\alpha}})\right)
$$
in terms of the imaginary error function $\text{erfi}(z)=\text{erf}(iz)/i$. With $u=\frac{x}{2\sqrt{\alpha}}$ we get $b(x)=\lim_{\alpha\to0^+}[1-\sqrt{\pi}u e^{-u^2}\text{erfi}(u)]/\alpha$, which can be plotted.
How can we deduce the limiting distribution $\alpha\to0^+$?
More generally, I'm looking for $b_n(x)=\int_{-\infty}^{\infty}\lvert y\rvert^n e^{-ixy} dy$, which is easy for even positive integers $n$ but hard for the odd ones. Any hints appreciated!
 A: Let $f(y)=|y|$ and $H(y)$ denote the Heaviside function.  Then, that we can write in distribution
$$\begin{align}
\mathscr{F}\{f\}(x)&=\int_{-\infty}^\infty |y|e^{-ixy}\,dy\\\\
&=2\text{Re}\left(\int_{-\infty}^\infty yH(y)e^{-ixy}\,dy\right)\\\\
&=2\text{Re}\left( i\frac{d}{dx}\mathscr{F}\{H\}(x)\right)\\\\
&=2\text{Re}\left( i\frac{d}{dx}\left(\pi\delta(x)-\frac ix \right)\right)\\\\
&=-\frac2{x^2}
\end{align}$$
More generally, we have for $g(y)=|y|^n$
$$\begin{align}
\mathscr{F}\{g\}(x)&=2\text{Re}\left( i^n\frac{d^n}{dx^n}\mathscr{F}\{H\}(x)\right)\\\\
&=2\text{Re}\left( i^n\frac{d^n}{dx^n}\left(\pi\delta(x)-\frac ix \right)\right)\\\\
&=2\text{Re}\left( i^n\pi\delta^n(x)+i^{n+1}\frac{n!}{x^{n+1}}\right)\\\\
\end{align}$$
A: Another way (although I like Mark's solution better as his result is more precise), we can make use of the fact that $|y|$, and later $|y|^n$, are even functions by manipulating the Fourier transform to the form of a Laplace transform.
\begin{align}
\int_{-\infty}^\infty |y| e^{-ixy} \, dy &= -\int_{-\infty}^0 y e^{-ixy} \, dy + \int_0^\infty y e^{-ixy} \, dy \\
&= 2 \int_0^\infty y e^{-ixy} \, dy \\
&= 2 \mathcal{L}\{y\} \\
&= \frac{2}{(ix)^2}.
\end{align}
In general, since $|y|^n$ is an even function, its Fourier transform must be real. Thus we have for odd $n$
\begin{align}
\int_{-\infty}^\infty |y|^n e^{-ixy} \, dy &= -\int_{-\infty}^0 y^n e^{-ixy} \, dy + \int_0^\infty y^n e^{-ixy} \, dy \\
&= 2 \mathcal{L}\{y^n\} \\
&= 2\frac{n!}{(ix)^{n+1}}
\end{align}
The same approach and result applies if $n$ is even; however, $n+1$ is then odd, making this transform complex valued with no real part which does not satisfy the requirement that the transform be real. Therefore we have
$$
\int_{-\infty}^\infty |y|^n e^{-ixy} \, dy =
\begin{align}
\begin{cases}
0 &\text{n even}\\
\frac{2n!}{(ix)^{n+1}} &\text{n odd}
\end{cases}
\end{align}.
$$
