Find the Fourier transform of generalized functions
${\rm exp}(ix^2)$
My solution: I take the integral for all $\mathbb{R}$
$F({\rm exp}(ix^2))=\frac{1}{\sqrt{2\pi}}\int {\rm exp}(iy^2-ixy)dy=\frac{1}{\sqrt{2\pi}}\int {\rm exp}[(\sqrt{i}y-\frac{ix}{2\sqrt{i}})^2-\frac{ix^2}{4}]dy=$
$\frac{1}{\sqrt{2\pi}}{\rm exp}(\frac{-ix^2}{4})\int {\rm exp}(\sqrt{i}y-\frac{\sqrt{i}x}{2})^2dy=\frac{1}{\sqrt{2\pi}}{\rm exp}(\frac{-ix^2}{4})\int {\rm exp}(-i(\sqrt{i}y-\frac{\sqrt{i}x}{2})i)^2dy$
Then I make a replacement
$u=-i(\sqrt{i}y-\frac{\sqrt{i}x}{2}), du=-i\sqrt{i}dy$
But I do not know in which area to integrate now. Maybe I counted incorrectly
${\rm exp}(x)$for ${\rm exp}(x)$. What you have now means $e\times x\times p$, etc. – Shaun Jun 03 '23 at 16:44\operatorname{exp}will produce better results in certain contexts than{\rm exp}, e.g., when not using parentheses around the argument. – Travis Willse Jun 03 '23 at 17:12