Prove \(\int_{-\infty}^{\infty} \mathrm{e}^{ist^{2}} dt = \mathrm{e}^{is\pi /4}\sqrt{\pi}\) for \(s = \pm 1\)

First integral, for \(s=1\)
\begin{align}
2\int \mathrm{e}^{it^{2}} dt &= 2\frac{1}{\sqrt{i}} \int \mathrm{e}^{x^{2}} dx \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(x) \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(t\sqrt{i})
\end{align}

Thus
\begin{align}
2\int\limits_{0}^{\infty} \mathrm{e}^{it^{2}} dt &= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(t\sqrt{i}) \Big|_{0}^{\infty} \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} (i-0) = \sqrt{i}\sqrt{\pi} = \mathrm{e}^{i\pi /4}\sqrt{\pi}
\end{align}

Second integral, for \(s=-1\)
\begin{align}
2\int \mathrm{e}^{-it^{2}} dt &= 2\frac{1}{\sqrt{i}} \int \mathrm{e}^{-x^{2}} dx \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(x) \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(t\sqrt{i})
\end{align}

Thus
\begin{align}
2\int\limits_{0}^{\infty} \mathrm{e}^{-it^{2}} dt &= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(t\sqrt{i}) \Big|_{0}^{\infty} \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} (1-0) = \frac{1}{\sqrt{i}} \sqrt{\pi} = \mathrm{e}^{-i\pi /4}\sqrt{\pi}
\end{align}

And we have
\begin{equation}
\int\limits_{-\infty}^{\infty} \mathrm{e}^{ist^{2}} dt = \mathrm{e}^{is\pi /4}\sqrt{\pi}
\end{equation}
for \(s = \pm 1\)

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