Difficult Fourier integral giving a distribution

Question

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!

Answer 1

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 $g(y)=|y|^n$

$$\begin{align} \mathscr{F}\{g\}(x)&=2\text{Re}\left( i^n\frac{d^n}{dx^n}\mathscr{F}\{H\}(x)\right)\\\\ &=2\text{Re}\left( i^n\frac{d^n}{dx^n}\left(\pi\delta(x)-\frac ix \right)\right)\\\\ &=2\text{Re}\left( i^n\pi\delta^n(x)+i^{n+1}\frac{n!}{x^{n+1}}\right)\\\\ \end{align}$$

Answer 2

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

$$ \int_{-\infty}^\infty |y|^n e^{-ixy} \, dy = \begin{align} \begin{cases} 0 &\text{n even}\\ \frac{2n!}{(ix)^{n+1}} &\text{n odd} \end{cases} \end{align}. $$