# 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!

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}

• Brilliant, thanks @MarkViola! – Roman Jan 8 '19 at 22:09
• I think you can use $\lvert y \rvert = y[2H(y)-1]$ instead of $\lvert y \rvert = 2\text{Re}[y H(y)]$, so the result changes a tiny bit (the Dirac $\delta$ in the last result disappears, as does the Re); but the idea works great. – Roman Jan 8 '19 at 22:28
• @Roman The result should be unaffected. Yes, $|y|=y(2H(y)-1)$ and so, $\mathscr{F}\{f \}=\int_{-\infty}^\infty (2yH(y)-y)e^{-ixy}\,dy=i\frac{d}{dx}\mathscr{F}\{(2H-1)\}=i\frac{d}{dx}\left(2\pi \delta(x)-\frac{i2}{x}-2\pi \delta(x) \right)=-\frac2{x^2}$ – Mark Viola Jan 8 '19 at 23:50
• Thank you for the compliment. And you're welcome! – Mark Viola Jan 8 '19 at 23:51

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}.

• I don't see how you get to the second line of your first formula series. Also, your result is wrong for $n$ even, you should be getting $\frac{d^n\delta(x)}{dx^n}$. – Roman Jan 9 '19 at 7:38
• Hmm yes, I don't know much about distributions and I thought $\delta^{(n)}$ would evaluate to $0$ in this case without looking further into it - my apologies for that. I wanted to show that you can still utilize properties of other transforms, even if its not in the "right" context, if you can get it to that form. The second line comes from $|y|$ being even. The area under the curve on $\mathbb{R}^-$ is exactly the same as that on $\mathbb{R}^+.$ – AEngineer Jan 9 '19 at 8:16