I have a problem here that becomes quite difficult to manage. I have to find the fourier transform of: $$f(x)=\frac{x}{x^4+4}$$ I'm sure there will be many ways to do this and I'll post my method that I started but am having a hard time with.

First, this is a practice problem that was given with a solution but the steps are difficult to follow: Here is the solution i received.

$$\mathfrak{F}\left[\frac{x}{x^{4}+4}\right]=\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}\frac{x}{x^{4}+4}e^{-i\xi x}dx$$

"We evaluate this integral in the upper half plane, with two poles inside the contour, at $1+i$ and $-1+i$, when $x > 0$." (Note: This is Complex Variable Theory, specifically Contour Integrals. The chapter in the book is "Complex Integration and Residue Theory")

$$F = \frac{i\sqrt{2\pi}}{4}e^{-\xi}\sin(\xi) $$

"Odd extension gives"

$$F = \frac{i\sqrt{2\pi}}{4}e^{-|\xi|}\sin(\xi) $$

That's it for a solution he provides.

Generally to take the integral of something like this in the upper half plane requires one to first evaluate the poles.

I discover poles of $\dfrac{xe^{i\xi x}}{x^{4}+4}$ (i.e., look at the $x^{4} + 4$) at $$ \sqrt{2}e^{(\frac{i\pi}{4}(2k+1))}, k = 0,1,2,3$$

Once I have these I can draw a figure of the situation:

Contours with Poles

As you can see, given these chosen Contours, one from $-\infty$ to $\infty$ and the other contour where the values go into the positive imaginary axis. This will equal zero when considering now $x=z$ and $R \rightarrow \infty$:

$$\int_{0}^{\pi}\frac{z}{z^{4}+4}e^{-i\xi z}dz$$

and $z = Re^{i\theta}$ and $dz = RdRd\theta$ we get

$$\int_{0}^{\pi}\frac{R^{4}e^{4i\theta}}{R^{4}e^{4i\theta}+4}e^{-i\xi z}RdRd\theta$$

So it can be seen that this would go to zero because of the $e^{z}$ term as $ R \rightarrow \infty$ no matter what value of $\theta$ we integrate over.

Okay now that this part is over the difficulty I get is in finding the residues of my two poles that i have. I can do this by finding the general residue of $\sqrt{2}e^{\frac{i\pi}{n}(2k+1)}$:

First i rename $\sqrt{2}e^{\frac{i\pi}{n}(2k+1)}$ to $C_{k}$ which makes this more manageable. where we only care about k = 0,1.

$$ Res[C_{k}] = \lim_{z\rightarrow C_{k}} (z-C_{k})f(z)$$ $$ Res[C_{k}] = \lim_{z\rightarrow C_{k}} \frac{(z-C_{k})ze^{-i\xi z}}{(z^{4}-4)} $$

we need L'Hopital's rule here and then after evaluating this i get something odd that i dont like dealing with:

$$ Res[C_{k}] = \lim_{z\rightarrow C_{k}} \frac{e^{-i\xi z}}{(4C^{2})} $$

$$ Res[C_{k}] = \lim_{z\rightarrow C_{k}} \frac{e^{-i\xi z}}{(8e^{\frac{i2\pi}{n}(2k+1)})} $$

what bothers me is the $e^{i\xi z} $ term in the numerator. My approach may be odd too. I'm hoping for someone who has done similar problems to provide some insight. Using Complex integration techniques would be most helpful but if you have any other ideas it would be good to hear them too. Feel free to ask if you want to see any steps i skipped. I can type those out. I will get to doing that as soon as I have time.

  • 1
    $\begingroup$ You shouldn't have $dR$ by the way. $\endgroup$ Apr 19 '14 at 15:01

I don't know what Fourier Transform is and also this will help you or not. At least, I give some thought. Split the algebraic fraction form as \begin{align} \frac{x}{x^4+4}&=\frac{x}{(x^2-2x+2)(x^2+2x+2)}\\ &=\frac{1}{4(x^2-2x+2)}-\frac{1}{4(x^2+2x+2)}\\ &=\frac{1}{4((x-1)^2+1)}-\frac{1}{4((x+1)^2+1)}\\ \end{align} Then \begin{align} \frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}\frac{x}{x^{4}+4}e^{-i\xi x}\,dx=\frac{1}{4\sqrt{2}}\int_{-\infty}^{\infty}\left(\frac{e^{-i\xi x}}{(x-1)^2+1}-\frac{e^{-i\xi x}}{(x+1)^2+1}\right)\,dx \end{align} Let $u=x-1$ then $x=u+1$ and $dx=du$. Also, let $v=x+1$ then $x=v-1$ and $dx=dv$. Hence \begin{align} \frac{1}{4\sqrt{2}}\int_{-\infty}^{\infty}\left(\frac{e^{-i\xi x}}{(x-1)^2+1}-\frac{e^{-i\xi x}}{(x+1)^2+1}\right)\,dx&=\frac{1}{4\sqrt{2}}\left(\int_{-\infty}^{\infty}\frac{e^{-i\xi (u+1)}}{u^2+1}\,du-\int_{-\infty}^{\infty}\frac{e^{-i\xi (v-1)}}{v^2+1}\,dv\right)\\ &=\frac{1}{4\sqrt{2}}\left(e^{-i\xi}\int_{-\infty}^{\infty}\frac{e^{-i\xi u}}{u^2+1}\,du-e^{i\xi}\int_{-\infty}^{\infty}\frac{e^{-i\xi v}}{v^2+1}\,dv\right)\\ \end{align} I think the rest can be evaluated using this. I hope this really helps you. If not, please do not vote down my answer. Just ignore it. Also, thank you so much for whoever vote up my answer. I really appreciate it. ヾ(-^〇^-)ノ

  • $\begingroup$ Aha, I've just found this. You can use this technique to evaluate the last integral using residue theorem. $\endgroup$ Apr 19 '14 at 16:51
  • $\begingroup$ Your algebra at the beginning was enough to do this after knowing some key fourier transform pairs and properties. I used all your work up until the third line. From there you can use the shift property of fourier transforms: $\mathfrak{F}[f(x-a)]=e^{ia\xi}\mathfrak{\xi}[f(x)]$ and you can then use the hint that $\mathfrak{F}[e^{-ax^{2}}] = \frac{1}{\sqrt{2a}}e^{\frac{\xi^{2}}{4a}}$ in our case $a = 1$ each time. Thanks. $\endgroup$
    – Dr. Dan
    Apr 20 '14 at 22:02
  • $\begingroup$ Thank you @ZealotSveta for accepting my answer. I didn't know what Fourier transform is but I'm glad my answer could help you out. (>‿◠)✌ $\endgroup$ Apr 21 '14 at 6:27

The residue at $z_0$ of a function of the form $f(z)/g(z)$ when $f(z_0) = 0$ and $g(z)$ has a simple zero (a zero of order 1) at $z_0$ is just given by $f(z_0) /g'(z_0)$. Here $f(z) = ze^{-i\xi z}$ and $g(z) = z^4 + 1$, so the residue will be $e^{-i\xi z_0}/ 4z_0^2$.

Plug in the two poles you determined and add the terms from the residue theorem, this should give the answer. The numerators are not anything to be worried about, as they are bounded functions in the upper half-plane if $\xi \leq 0$. You can then get the $\xi > 0$ case just by taking complex conjugates of your original integral for $-\xi$.


$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\fermi\pars{x}\equiv{x \over x^{4} + 4} =\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}\expo{\ic k x} \,{\dd k \over 2\pi}\ \imp\ \tilde{\fermi}\pars{k}=\int_{-\infty}^{\infty} {x \over x^{4} + 4}\,\expo{-\ic k x}\,\dd x:\ {\large ?}}$.

\begin{align} \tilde{\fermi}\pars{k}& =-\ic\int_{-\infty}^{\infty}{x\sin\pars{k x} \over x^{4} + 4}\,\dd x =-\ic\sgn\pars{k}\int_{-\infty}^{\infty} {x\sin\pars{\verts{k}x} \over x^{4} + 4}\,\dd x \\[3mm]&=-\ic\sgn\pars{k}\,\Im\int_{-\infty}^{\infty} {x\expo{\ic\verts{k}x} \over x^{4} + 4}\,\dd x =-\ic\sgn\pars{k}\,\Im\bracks{2\pi\ic \sum_{n = 0}^{1}{x_{n}\expo{\ic\verts{k}x_{n}} \over 4x_{n}^{3}}} \end{align} where $\ds{x_{n} \equiv \root{2}\exp\pars{\bracks{{n \over 2} + {1 \over 4}}\pi\ic}}$ and $\ds{n = 0,1,2,3}$. $\ds{\braces{x_{n}\ \mid\ n = 0,1,2,3}}$ are the roots of $\ds{x^{4} + 4 = 0}$. In performing the integration, we 'close the contour' in the upper half of the complex plane.

\begin{align} \tilde{\fermi}\pars{k}& =-\ic\,{\pi \over 2}\sgn\pars{k}\,\Re\bracks{ \sum_{n = 0}^{1}{x_{n}^{2}\expo{\ic\verts{k}x_{n}} \over x_{n}^{4}}} =\ic\,{\pi \over 8}\sgn\pars{k}\,\Re\bracks{ \sum_{n = 0}^{1}x_{n}^{2}\expo{\ic\verts{k}x_{n}}} \\[3mm]&=\ic\,{\pi \over 8}\sgn\pars{k}\,\Re\bracks{% 2\expo{\pi\ic/2}\exp\pars{\ic\verts{k}\root{2}\expo{\ic\pi/4}} +2\expo{3\pi\ic/2}\exp\pars{\ic\verts{k}\root{2}\expo{3\ic\pi/4}}} \\[3mm]&=\ic\,{\pi \over 4}\sgn\pars{k}\,\Re\bracks{% \ic\exp\pars{\ic\verts{k}\,\pars{1 + \ic}} -\ic\exp\pars{\ic\verts{k}\,\pars{-1 + \ic}}} \\[3mm]&=-\ic\,{\pi \over 4}\sgn\pars{k}\,\Im\bracks{% \exp\pars{\verts{k}\,\pars{-1 + \ic}} -\exp\pars{\verts{k}\,\pars{-1 - \ic}}} \\[3mm]&=-\ic\,{\pi \over 2}\sgn\pars{k}\,\Im \exp\pars{\verts{k}\,\pars{-1 + \ic}} =-\ic\,{\pi \over 2}\,\sgn\pars{k}\expo{-\verts{k}}\sin\pars{\verts{k}} \end{align}

$$\color{#66f}{\large% \tilde{\fermi}\pars{k}=\int_{-\infty}^{\infty} {x \over x^{4} + 4}\,\expo{-\ic k x}\,\dd x =-\ic\,{\pi \over 2}\,\expo{-\verts{k}}\sin\pars{k}} $$


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