I was supposed to evaluate the Fourier transform of the function $$f(x) = \frac{x^4}{1+x^4}$$ and apart from the standard way of integrating with complex contours, which I find pretty tedious, I thought that it could be a better way.
My first idea was to use the convolution theorem treating $f$ as a product of two functions $$g(x) = x^4\;\;\;\; h(x) = \frac{1}{1+x^4}$$ using the fact that $$\mathcal{F}\{x^n f(x)\} = i^n(\mathcal{F}\{f\})^{(n)}$$This seems to me the fastest way. Or maybe one could simply decompose the fraction in $$f(x) = 1+\frac{1}{1+x^4}$$ In either cases the problem now is on the evaluation of the Fourier transform of $h$ which, even in this case, I would prefer, if possible, to not use contour integration. My mind jumped on the idea that maybe there would be some connection with the Fourier transform of $$\frac{1}{1+x^2}$$ maybe by substitution of $y=x^2$. I get then $$h(y) = \frac{1}{1+y^2}$$ but there's not a property for the Fourier transform on such "scaling". Is there a property similar to the translation and scaling for Fourier transform that holds even for substitution like the latter?
If this is not the right way, is there some other properties or tricks that I could use to evaluate that transform?
Edit
What if I decompose the fraction in this manner
$$\frac{1}{1+x^4}=\frac{1}{(1+ix^2)(1-ix^2)} = \frac{1}{i^2(\frac{1}{i}+x^2)(\frac{1}{i}-x^2)} = \frac{i}{2(i+x^2)}-\frac{i}{2(-i+x^2)}$$
would this be a suitable way?
Using this "trick" I would get $${i\over 2}\mathcal{F}\left\{\frac{1}{i+x^2}\right\} = {i\over2}\sqrt{{\pi\over {2i}}}e^{\sqrt{i}|k|}$$ and a similar transform for the other factor
Just to be on the same page, I define the Fourier transform in the following way $$\hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-ikx}f(x)dx$$
Best Answer
For the sake of simplicity, let $t\in\mathbb{C}\setminus\mathbb{R}_{\geq 0}$ so that the polynomial $X^2-t\in\mathbb{C}[X]$ does not have a real root, but you can use the Cauchy principal values when $t\in\mathbb{R}_{\geq 0}$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=\dfrac{1}{z^2-t}\text{ for all }z\in\mathbb{C}\setminus\big\{-\sqrt{t},+\sqrt{t}\big\}\,,$$ where $\sqrt{t}$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots). That is, $$F_t(k)=\frac{1}{\sqrt{2\pi}}\,\int_{-\infty}^{+\infty}\,\exp(\text{i}kx)\,f_t(x)\,\text{d}x\text{ for }k\in\mathbb{R}\,.$$ For simplicity, I shall write $c:=\dfrac{1}{\sqrt{2\pi}}$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $k\geq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by $$[-R,+R]\cup\Big\{R\,\exp(\text{i}\theta)\,\Big|\,\theta\in[0,\pi]\Big\}\,.$$ Observe that $$2\pi\text{i}\,\text{Res}_{z=+\sqrt{t}}\big(f_t(z)\big)=\lim_{R\to\infty}\,\oint_{C_R}\,\frac{\exp(\text{i}kz)}{z^2-t}\,\text{d}z=\frac{1}{c}\,F_t(k)\,.$$ Hence, taking the sign of $k$ into account, we get $$F_t(k)=\frac{c\pi\text{i}}{\sqrt{t}}\,\exp\Big(\text{i}|k|\sqrt{t}\Big)\text{ for each }k\in\mathbb{R}\,.$$
For example, $\sqrt{+\text{i}}=\frac{+1+\text{i}}{\sqrt{2}}$ and $\sqrt{-\text{i}}=\frac{-1+\text{i}}{\sqrt{2}}$. This means $$F_{s\text{i}}(k)=\frac{c\pi\text{i}}{\left(\frac{s+\text{i}}{\sqrt{2}}\right)}\,\exp\Biggl(\text{i}\,\left(\frac{s+\text{i}}{\sqrt{2}}\right)\,|k|\Biggr)=c\pi\left(\frac{1+s\text{i}}{\sqrt{2}}\right)\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\exp\left(\frac{s\text{i}|k|}{\sqrt{2}}\right)\,,$$ where $s\in\{-1,+1\}$. Now, let $\varphi(z):=\dfrac{1}{z^4+1}$. Then, as you found out, $$\varphi(z)=-\frac{\text{i}}{2}\,\sum_{s\in\{-1,+1\}}\,\frac{s}{z^2-s\text{i}}\,.$$ Hence, the Fourier transform $\hat{\varphi}$ of $\varphi$ is given by $$\hat{\varphi}(k)=-\frac{\text{i}}{2}\,\sum_{s\in\{-1,+1\}}\,s\,F_{s\text{i}}(k)\,.$$ That is, $$\hat{\varphi}(k)=\frac{c\pi}{\sqrt{2}}\,\sum_{s\in\{-1,+1\}}\,\left(\frac{1-s\text{i}}{2}\right)\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\exp\left(\frac{s\text{i}|k|}{\sqrt{2}}\right)\,.$$ That is, $$\hat{\varphi}(k)=\frac{c\pi}{\sqrt{2}}\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\Biggl(\cos\left(\frac{|k|}{\sqrt{2}}\right)+\sin\left(\frac{|k|}{\sqrt{2}}\right)\Biggr)\,.$$
Now, if you want to determine the Fourier transform $\hat{\Phi}$ of $\Phi$, where $$\Phi(z):=\dfrac{z^4}{z^4+1}=1-\dfrac{1}{z^4+1}\,,$$ you will get a distribution. Note that the Fourier transformation $\hat{\kappa}_\epsilon$ of the constant function $\kappa_\epsilon\equiv \epsilon$ is $$\hat{\kappa}_{\epsilon}(k)=2c\pi\,\epsilon\,\delta(k)\,,$$ where $\delta$ is the Dirac delta distribution. Thus, using $\Phi=\kappa_1-\phi$, we conclude that $$\hat{\Phi}(k)=2c\pi\,\left(\delta(k)-\frac{1}{2\sqrt{2}}\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\Biggl(\cos\left(\frac{|k|}{\sqrt{2}}\right)+\sin\left(\frac{|k|}{\sqrt{2}}\right)\Bigg)\right)\,.$$ Nonetheless, be warned that $\hat{\Phi}$ is not a function.