fourier transform of $f(x) = x^2+\frac{1}{1+2x^4}$

Question

I really have no thought on this. I can't seem to use residue thm., nor could I find a inverse transform for it. by some Fourier Calculator I know it's solvable but how?

Answer 1

The Fourier transform of $f(x)$ doesn't exist in the usual sense, but since $f$ can be viewed as a tempered distribution, we can interpret the Fourier transform in that setting. (I'm using the normalization $\hat f(\omega) = \int_{-\infty}^\infty e^{-i\omega t}f(t)\,dt$. If you're using something else, the answer is a little different.)

First of all, Fourier transform of $1$ is $2\pi\delta(\omega)$. Hence \begin{align*} t &\xrightarrow{\mathscr{F}} 2\pi i\delta'(\omega) \\ t^2 &\xrightarrow{\mathscr{F}} 2\pi i^2\delta''(\omega) = -2\pi\delta''(\omega). \end{align*}

The second term is less problematic, and exists in the usual sense. It is a standard exercise in residue calculus to compute the Fourier transform of $\frac{1}{1+2x^4}$. The result (and especially all the intermediate steps) are very messy though. I get: $$ \hat f(\omega) = -2\pi \delta''(\omega) + \begin{cases} \pi q e^{q\omega} (\cos q\omega - \sin q\omega), & \omega < 0 \\ \pi q e^{-q\omega} (\sin q\omega - \cos q\omega), & \omega \ge 0 \end{cases} $$ where $q = 2^{1/4}/2$.