ormaybesaladqh

2022-10-23

How evaluate this $f(x)={x}^{2}+\frac{1}{1+2{x}^{4}}$ with fourier tranform

ehedem26

Beginner2022-10-24Added 13 answers

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 $\hat{f}(\omega )={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-i\omega t}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$. If you're using something else, the answer is a little different.)

First of all, Fourier transform of 1 is $2\pi \delta (\omega )$. Hence

$$\begin{array}{rl}t& \stackrel{\mathcal{F}}{\to}2\pi i{\delta}^{\prime}(\omega )\\ {t}^{2}& \stackrel{\mathcal{F}}{\to}2\pi {i}^{2}{\delta}^{\u2033}(\omega )=-2\pi {\delta}^{\u2033}(\omega ).\end{array}$$

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:

$$\hat{f}(\omega )=-2\pi {\delta}^{\u2033}(\omega )+\{\begin{array}{ll}\pi q{e}^{q\omega}(\mathrm{cos}q\omega -\mathrm{sin}q\omega ),& \omega <0\\ \pi q{e}^{-q\omega}(\mathrm{sin}q\omega -\mathrm{cos}q\omega ),& \omega \ge 0\end{array}$$

where $q={2}^{1/4}/2$

First of all, Fourier transform of 1 is $2\pi \delta (\omega )$. Hence

$$\begin{array}{rl}t& \stackrel{\mathcal{F}}{\to}2\pi i{\delta}^{\prime}(\omega )\\ {t}^{2}& \stackrel{\mathcal{F}}{\to}2\pi {i}^{2}{\delta}^{\u2033}(\omega )=-2\pi {\delta}^{\u2033}(\omega ).\end{array}$$

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:

$$\hat{f}(\omega )=-2\pi {\delta}^{\u2033}(\omega )+\{\begin{array}{ll}\pi q{e}^{q\omega}(\mathrm{cos}q\omega -\mathrm{sin}q\omega ),& \omega <0\\ \pi q{e}^{-q\omega}(\mathrm{sin}q\omega -\mathrm{cos}q\omega ),& \omega \ge 0\end{array}$$

where $q={2}^{1/4}/2$

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