# Find the Fourier Transform of a Generalized Function

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.

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*}
• 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?