About a proof of fourier transformation of error function

Question

In a proof of fourier transformation of error function, the formula $$ \int_{-\infty}^{\infty} \frac{-i e^{-\pi^{2} t^{2}}}{\pi t} e^{i 2 \pi x t} d t=\frac{-i}{\pi} \int_{-\infty}^{\infty} i \pi \operatorname{sgn}(x-\tau) \frac{1}{\sqrt{\pi}} e^{-\tau^{2}} d \tau $$ is used. How can I proof this equation? I cannot understand why $\operatorname{sgn}$ appears.

Answer 1

Have a look at the Wikipedia page. The argument is that \begin{align} \int_{-\infty}^{\infty} \frac{-i}{\pi} \frac{1}{t} e^{-\pi^2 t^2} e^{2\pi i xt} ~ dt = \frac{-i}{\pi}\mathcal F\Big( \frac{1}{t} \cdot e^{-\pi^2 t^2}\Big)(-x) \end{align} where $\mathcal F$ denotes the Fourier transform. It then uses the convolution rule, $\mathcal F(f\cdot g) = \mathcal F(f) * \mathcal F (g)$, \begin{align} \int_{-\infty}^{\infty} \frac{-i}{\pi} \frac{1}{t} e^{-\pi^2 t^2} e^{2\pi i xt} ~ dt &= \frac{-i}{\pi}\Bigg( \mathcal F\Big( \frac{1}{t}\Big) * \mathcal F \Big( e^{-\pi^2 t^2}\Big) \Bigg) (-x) \\ &= \frac{-i}{\pi}\int_{-\infty}^{\infty} \mathcal F\Big( \frac{1}{t} \Big)(-x-s) \mathcal F\Big( e^{-\pi^2 s^2} \Big)(s) ~ds \end{align} and then use the standard results, $\mathcal F(1/t) = -i\pi \operatorname{sgn}(x)$ and $\mathcal F( e^{-\pi^2 t^2}) = \frac{1}{\sqrt{\pi}} e^{-x^2}$.

Put these together and recognise $\operatorname{sgn(y)}=-\operatorname{sgn}(-y)$, \begin{align} \int_{-\infty}^{\infty} \frac{-i}{\pi} \frac{1}{t} e^{-\pi^2 t^2} e^{2\pi i xt} ~ dt &= \frac{-i}{\pi}\int_{-\infty}^{\infty} -i\pi\operatorname{sgn}(-x-s) \cdot \frac{1}{\sqrt{\pi}}e^{-s^2} ~ds \\ &= \frac{-i}{\pi}\int_{-\infty}^{\infty} i\pi\operatorname{sgn}(x+s) \cdot \frac{1}{\sqrt{\pi}}e^{-s^2} ~ds \\ &= \frac{-i}{\pi}\int_{-\infty}^{\infty} i\pi\operatorname{sgn}(x-\tau) \cdot \frac{1}{\sqrt{\pi}}e^{-\tau^2} ~d\tau \end{align} where the last step resulted from a change of variable $\tau=-s$.