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Find the Fourier Transform ($F$) of a Generalized Function

$f(x) = e^{ix^2}$

Here is my solution:

$(F[e^{ix^2}])(y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{iy^2}e^{-ixy}dy=$

$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}exp[i(y^2-xy+\frac{x^2}{4}-\frac{x^2}{4})]dy=$

$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}exp[(\sqrt{i}y-\sqrt{i}\frac{x}{2})^2]exp[-\frac{ix^2}{4}]dy=$

$=\frac{1}{\sqrt{2\pi}}e^{\frac{-ix^2}{4}}\int_{-\infty}^{\infty}exp([-i(\sqrt{i}y-\sqrt{i}\frac{x}{2})i]^2)dy=$

I tried to make this substitution $u=-i(\sqrt(i)y-\sqrt{i}\frac{x}{2}), du = -i\sqrt{i}dy$

But there were problems with the boundaries of integration.

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1 Answer 1

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It is known that all Gaussian functions that can be written in the form $be^{-\pi x^2}$ are equal to their Fourier transforms. Specifically, $${F(e^{-\pi x^2})(y)=e^{-\pi y^2}}.$$ Let $g(x)=e^{-\pi x^2}$. You can then try to write $f$ in terms of $g$, to see if you can use any properties of the FT. Indeed, if you look closely, $$ f(x)=g(ax) \text{, where $\displaystyle{a=\frac{i^{3/2}}{\sqrt{\pi}}}$}$$ Therefore, \begin{align*} F(f(x))(y) &= F(g(ax))(y) \\ & = \frac{1}{a}F(g(x))\left(\frac{y}{a}\right) \\ &\text{and since $F(g)=g$} \\ &= \frac{\sqrt{\pi}}{i^{3/2}} \exp\left(-\pi\left(\frac{\sqrt{\pi}}{i^{3/2}}x\right)^2\right) \\ &= \frac{\sqrt{\pi}}{i^{3/2}} \exp\left(-\pi^2\frac{1}{i^{3}}x^2\right) \\ &= \frac{\sqrt{\pi}}{i^{3/2}} \exp\left(-\pi^2ix^2\right) \\ \end{align*}

There has been a similar question at https://dsp.stackexchange.com/questions/59468/complex-numbers-and-fourier-transform

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  • $\begingroup$ But the function $f(x) = e^{ix^2}$ is generalized. And the Fourier transform for generalized functions is defined as $(F[f])(y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ixy}dx$. Does this not affect the decision? $\endgroup$
    – Ben
    Aug 23, 2023 at 7:02

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