2

How is the Fourier transform of $g(x)=\frac{1}{x^2-a^2}, a\in \mathbb{R}$ calculated?

I know that the Fourier transfom of $f(x)=\frac{1}{x^2+a^2}, a\in \mathbb{R}$ can be calculated by just closing the counter up and down depending on the sign of the transformed variable $p$ and we get something like $\tilde f(p)=\frac{\pi e^{a p} \theta (-p)}{a}+\frac{\pi e^{-a p} \theta (p)}{a}$

But I do not know to compute an integral fo the form \begin{equation} \int e^{-ipx}\frac{1}{x^2-a^2} \end{equation} calculated.

A direct calculation on Mathematica gave answer to be undefined but using Mathematica command to compute this Fourier translation gave \begin{equation} \tilde g(p)=-\frac{\pi \text{sgn}(p) \sin (a p)}{a} \end{equation}

I wonder how is this expression obtained.

Moreover:

Mathematica integration of $\tilde g(p)$ to find $g(x)$ gives back the correct result.

user824530
  • 713

2 Answers2

4

As @Calvin Khor and @herb steinberg mentioned, the integral should be considered in the principal value sense.

$$I(p)=P.V.\int_{-\infty}^{\infty}e^{-ipx}\frac{1}{x^2-a^2}dx$$

Let's take $p>0$ and consider the following contour

enter image description here

The big half-circle of radius $R$ in the lower half-plane (will go clockwise) and two small half circles of radius $r$ (will go counter-clockwise) are added - to make a closed contour.

$$\oint e^{-ipx}\frac{1}{x^2-a^2}dx=2\pi i \sum Res=0\, \,\text{ - there are no poles inside the contour}$$

On the other hand $$\oint=I(p)+\int_{C_1}+\int_{C_2}+\int_R=0$$ Integral along big half-circle $\int _R\to0$ as $R\to\infty$

$$\int_{C_2}=\lim_{r\to0}\int_\pi^{2\pi}\frac{e^{-ip(a+re^{i\phi})}}{2are^{i\phi}}ire^{i\phi}d\phi=\lim_{r\to0}\int_\pi^{2\pi}\frac{e^{-ipa}(1-ipre^{i\phi}+...)}{2are^{i\phi}}ire^{i\phi}d\phi=\frac{\pi i}{2a}e^{-ipa}$$ $$\int_{C_1}=-\frac{\pi i}{2a}e^{ipa}$$ $$I(p>0)=-\frac{\pi}{a}\sin(ap)$$

For $p<0$ we close the contour in the upper half-plane and go around $x=a$ and $x=-a$ along small half-circles clockwise in the upper half-plane. Negative direction gives additional minus. $$I(p)=-\frac{\pi \text{ sgn}(p) \sin (a p)}{a}$$

Svyatoslav
  • 15,657
  • +1, For $\alpha=0$ one can define $1/x^2$ as the derivative of $-PV\frac1x$ (which apparently gives the finite part of $-1/x^2$) whose Fourier transform is $-\pi|p|$, matching the limit $a\to 0$ of your expression – Calvin Khor Apr 17 '21 at 13:23
  • @Calvin Khor, thank you. Interesting point of $a\to0$ : we really can get a finite part, as you wrote; thank you for the link. This makes sense, I think, if we plan further $I(p)$ manipulations - taking derivative, for example, where only the part depending on $p$ plays, and the constant (even though infinite) does not contribute. – Svyatoslav Apr 17 '21 at 13:47
1

I will consider a simpler version of this example, but the same technique can be used. Try splitting your integral up into multiple regions, that is to say: $$\int_{-\infty}^\infty=\int_{-\infty}^{-a-\epsilon}+\int_{-a+\epsilon}^{a-\epsilon}+\int_{a+\epsilon}^\infty$$ now let $\epsilon\to0$.

$$\frac{1}{x^2-a^2}=\frac1{2a}\frac{1}{x-a}-\frac1{2a}\frac{1}{x+a}$$ so for ease I will just look at: $$\int_{-\infty}^\infty\frac{1}{x-a}-\frac{1}{x+a}\,dx$$

$$\int_{-\infty}^{-a-\epsilon}\frac{1}{x-a}-\frac{1}{x+a}\,dx=\left[\ln\left(\frac{x-a}{x+a}\right)\right]_{-\infty}^{-a-\epsilon}$$ $$\int_{-a+\epsilon}^{a-\epsilon}\frac{1}{x-a}-\frac{1}{x+a}\,dx=\left[\ln\left(\frac{x-a}{x+a}\right)\right]_{-a+\epsilon}^{a-\epsilon}$$ $$\int_{a+\epsilon}^{\infty}\frac{1}{x-a}-\frac{1}{x+a}\,dx=\left[\ln\left(\frac{x-a}{x+a}\right)\right]_{a+\epsilon}^{\infty}$$ combining it all together gives (I have now included absolutes): $$\left(\ln|-2a-\epsilon|-\ln|-\epsilon|-0\right)+\left(\ln|-\epsilon|-\ln|2a-\epsilon|-\ln|-2a+\epsilon|+\ln|\epsilon|\right)+\left(0-\ln|\epsilon|+\ln|2a+\epsilon|\right)$$ Now you can see that a lot of stuff cancels and it won't be too hard to take the limit wrt. $\epsilon$

Henry Lee
  • 12,215
  • when there's a complex exponential isnt it a lot more work? – Calvin Khor Apr 17 '21 at 13:39
  • 1
    @Henry Lee I think this is good and detailed example of how the principal value is defined. $PV\int_{-\infty}^{\infty}\frac{1}{x-a}=PV\int_{-\infty}^{\infty}\frac{1}{x}=0$ (it is also clear due to the fact that the integrand is odd), so we get some non-zero result if we have the integrand $\frac{1}{x}f(x)$, where the even part is non-zero. Making changes $t=x-a$ and $t=x+a, , I(p)\sim\int_{-\infty}^\infty(\frac{1}{x-a}-\frac{1}{x+a})e^{-ipx}dx$ $=(e^{-ipa}-e^{ipa})\int_{-\infty}^\infty\frac{e^{-ipt}}{t}dt$, and - due to symmetry, only $-i\sin pt$ survives from ${e^{-ipt}}$ in the integrand – Svyatoslav Apr 17 '21 at 14:57
  • 1
    @Svyatoslav yes exactly, when you have odd functions (which $e^{ix}$ is) things tend to clear up nicely but I thought putting all that in one answer would get very complicated. I might add a link to the Cauchy Principle Value though as that can be useful – Henry Lee Apr 17 '21 at 16:59
  • 1
    https://en.wikipedia.org/wiki/Cauchy_principal_value – Henry Lee Apr 17 '21 at 16:59
  • 1
    @Henry Lee Thank you! – Svyatoslav Apr 17 '21 at 18:01