# How evaluate this f(x)=x^2+(1)/(1+2x^4) with fourier tranform

How evaluate this $f\left(x\right)={x}^{2}+\frac{1}{1+2{x}^{4}}$ with fourier tranform
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ehedem26
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 $\stackrel{^}{f}\left(\omega \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-i\omega t}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$. If you're using something else, the answer is a little different.)
First of all, Fourier transform of 1 is $2\pi \delta \left(\omega \right)$. Hence
$\begin{array}{rl}t& \stackrel{\mathcal{F}}{\to }2\pi i{\delta }^{\prime }\left(\omega \right)\\ {t}^{2}& \stackrel{\mathcal{F}}{\to }2\pi {i}^{2}{\delta }^{″}\left(\omega \right)=-2\pi {\delta }^{″}\left(\omega \right).\end{array}$
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 $\frac{1}{1+2{x}^{4}}$. The result (and especially all the intermediate steps) are very messy though. I get:
$\stackrel{^}{f}\left(\omega \right)=-2\pi {\delta }^{″}\left(\omega \right)+\left\{\begin{array}{ll}\pi q{e}^{q\omega }\left(\mathrm{cos}q\omega -\mathrm{sin}q\omega \right),& \omega <0\\ \pi q{e}^{-q\omega }\left(\mathrm{sin}q\omega -\mathrm{cos}q\omega \right),& \omega \ge 0\end{array}$
where $q={2}^{1/4}/2$