Fourier transform of the error function, erf (x)

Claim: $$\mathcal{F}(\text{erf}(x))(t)=\frac{-i e^{-\pi^2t^2}}{\pi t}.$$

I will approach the problem by showing that the inverse transform of the above expression gives the error function. We have, by rules 105, 109, 206, and 309, of the WP Fourier transform page:

$$\int_{-\infty}^{\infty}\frac{-i e^{-\pi^2t^2}}{\pi t}e^{i2\pi xt}dt=\frac{-i}{\pi}\int_{-\infty}^\infty i\pi \text{sgn}(x-\tau)\frac{1}{\sqrt{\pi}}e^{-\tau^2}d\tau\\ =\frac{-1}{\sqrt{\pi}}\int_{-\infty}^\infty \text{sgn}(\tau-x)e^{-\tau^2}d\tau\\ =\frac{-1}{\sqrt{\pi}}\left ( -\int_{-\infty}^{-x} - \int_{-x}^{x} + \int_{x}^{\infty} e^{-\tau^2}d\tau \right )\\ =\frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{-\tau^2}d\tau \\ =\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-\tau^2}d\tau\\ =\text{erf}(x).$$

This result is reasonably well known; it is included in the Bateman MS (vol. I) and cited in Mehta's "Random Matrices" as A.31.1. A direct proof exploits the fact that erf is an odd function of its argument and a quick integration by parts.