Sorry for the lack of details in the title, the character limit was difficult to deal with. My aim here is to prove that for $t \in \Bbb R$ we have : $$f=F,f(t)=\int_{-\infty}^\infty \frac{{e}^{2i\pi tx}}{1+{x}^{2}}dx, F(t)=\iint_{\Bbb R^2} {e}^{\frac{2i\pi ty}{x}}{e}^{-(x^2+y^2)}dxdy$$
We know that $f$ and $F$ are continuous in $\Bbb R$, and that $f \notin {C}^{1}(\Bbb R)$. I tried switching from exponential to trigonometric notation for the complex expression inside $F(t)$ then proving that $$\int_{-\infty}^{\infty} \sin\left( \frac{2\pi ty}{x}\right){e}^{-(x^2+y^2)}dx =0$$ to simplify the expression but I end up with long unpalatable expressions which I'm not able to manipulate. My guess is there might be a simpler way to do this using Fourier transform but I couldn't find it. Any tips on how to approach this exercise would be appreciated!
Thanks in advance,
Arthur.
Surprisingly this problem doesn't have a whole lot of work to it. Consider the 2D integral $F(t)$ and convert to polar coordinates:
$$ = \int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_0^\infty e^{i2\pi t \tan\theta} re^{-r^2}drd\theta = \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{i2\pi t \tan\theta} d\theta$$
Because of the tangent singularities at odd multiples of $\frac{\pi}{2}$, we will split the integral up into two pieces and then use the substitution $z=\tan\theta$:
$$=\frac{1}{2}\Biggr( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{i2\pi t \tan \theta} d\theta + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{i2\pi t \tan \theta} d\theta \Biggr) = \frac{1}{2}\Biggr(\int_{-\infty}^\infty \frac{e^{i2\pi t z}}{1+z^2}dz + \int_{-\infty}^\infty \frac{e^{i2\pi t z}}{1+z^2}dz \Biggr)$$
$$= \int_{-\infty}^\infty \frac{e^{i2\pi t z}}{1+z^2}dz $$
which is exactly the integral that defines $f(t)$.