Alleged trick for Fourier transform of $\frac{x^4}{1+x^4}$

Question

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 $$\hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-ikx}f(x)dx$$

Answer 1

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.

Answer 2

Note that $$ \frac{x^4}{x^4+1} = 1-\underbrace{\frac1{x^4+1}}_{f(x)}\ , $$ so there is a simpler way to reduce to the function $f$ isolated above. Now it is a matter of taste to go.

Personally, i would opt for the residue theorem, so that the integral is painless solved in some lines. We may assume first $t\ge 0$. (Else we substitute $x$ by $-x$ in the integral below, to reduce ourselves to $|t|$.) $$ \begin{aligned} \hat f(t) &= \frac1{\sqrt {2\pi}}\int_{\Bbb R} \frac{e^{itx}}{1+x^4}\; dx \\ &= \lim_{R\to \infty} \frac1{\sqrt {2\pi}}\int_{-R}^R \frac{e^{itx}}{1+x^4}\; dx \\ &= \lim_{R\to \infty} \frac1{\sqrt {2\pi}} \int_{C} \frac{e^{itz}}{1+z^4}\; dz - \lim_{R\to \infty} \frac1{\sqrt {2\pi}} \underbrace{ \int_S \frac{e^{itz}}{1+z^4}\; dz}_{\to 0} \\ &\qquad\qquad\text{where $C=J\sqcup S$ is the contour with $J=[-R,R]$, and $S$ is the semicircle with diameter $J$ in upper half plane,} \\ &= \lim_{R\to \infty} \frac1{\sqrt {2\pi}} \cdot 2\pi i \sum_{a\text{ pole inside }C} \operatorname{Res}_{z=a} \frac{e^{itz}}{1+z^4} \\ &= \frac1{\sqrt {2\pi}} \cdot 2\pi i \sum_{a\in\{z_1,z_2\}=\{(\pm1+i)/\sqrt 2\}} \operatorname{Res}_{z=a} \frac{e^{itz}}{z^4+1} \\ &\qquad\qquad\text{and we can isolate the residues in $z_1,z_2=\frac 1{\sqrt 2}(\pm1+i)$,} \\ &\qquad\qquad\text{by factoring $z^4+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$} \\ &= \frac1{\sqrt {2\pi}} \cdot 2\pi i \left( \exp(itz_1)\cdot\frac 1{(z_1-z_2)(z_1-z_3)(z_1-z_4)} + \exp(itz_2)\cdot\frac 1{(z_2-z_1)(z_2-z_3)(z_2-z_4)} \right) \\ &\qquad\qquad\text{now we draw the unit circle and the four roots $z_{1,2,3,4}$ on it to simply see the relations} \\ &\qquad\qquad\text{$z_1-z_2=\sqrt2$, $z_1-z_3=\sqrt2(1+i)$, $z_1-z_4=i\sqrt 2$, and} \\ &\qquad\qquad\text{$z_2-z_1=-\sqrt2$, $z_2-z_4=-\sqrt2(1-i)$, $z_2-z_3=i\sqrt 2$,} \\ &= \frac1{\sqrt {2\pi}} \cdot 2\pi i \cdot \sum_\pm \frac 1{\sqrt 2\cdot i\sqrt 2\cdot (1\pm i)\sqrt 2} \exp\left(\frac t{\sqrt 2}(i\pm 1)\right)\ . \end{aligned} $$ A Galois theorist would stop here. I also have to stop here, since there is an other answer along the same lines. What i wanted to say...

In the original post we have a separation, instead of working with $1/(z^4+1)$ there is a split and we have to work with the two fractions $1/(z^2\pm i)$ (up to a multiplicative constant). Each one has one root in the upper half plane, so this splitting does not sensibly simplify algebra (if we finally again pass to residues).