I keep getting the wrong answer for this problem!

Find the Fourier transform of $f(x) = \frac{a}{\pi} \frac{1}{a^2 + x^2}$ using the residue theorem.

Well, by definition:

$$\hat f(k) = \frac{1}{\sqrt{2\pi}} \frac{a}{\pi}\int_{-\infty}^{+\infty}\frac{e^{-ikx}}{a^2 + x^2}\mathrm{d}x$$

I define the complex function:

$$g(z) \doteqdot \frac{e^{-ikz}}{a^2 + z^2} = \frac{e^{-ikz}}{(z-ai)(z+ai)}$$

Let's pick the simple pole at $z=ai$; the residue is:

$$\text{Res}(g,ai) = \lim_{z \to ai}\frac{e^{-ikz}}{z+ai} = \frac{e^{ak}}{2ai}$$

Now for a contour, choose a line segment in the real axis from -R to +R and an arc of a circle of radius R centred at the origin connecting the two ends of the segment. This contour includes the pole ai. As R tends to infinity, the integral over the arc vanishes (Jordan's lemma) and the integral over the segment becomes an integral over the real line. So, by the residue theorem:

$$\int_{-\infty}^{+\infty}\frac{e^{-ikx}}{a^2 + x^2}\mathrm{d}x = 2\pi i\frac{e^{ak}}{2ai} = \pi\frac{e^{ak}}{a}$$

The Fourier transform then is: $$\hat f(k) = \frac{1}{\sqrt{2\pi}}\frac{a}{\pi} \pi\frac{e^{ak}}{a}= \frac{1}{\sqrt{2\pi}}e^{ak}$$

Which is wrong; the correct answer is $\frac{1}{\sqrt{2\pi}}e^{-a|k|}$ but I don't see how the absolute value can pop up.

  • 2
    $\begingroup$ You've assumed $k>0$ without really knowing it. You have to compute a separate integral for $k<0$, using the semi-circle in the lower half-plane. Not sure why your constant is off. $\endgroup$
    – icurays1
    Nov 18 '12 at 17:05
  • $\begingroup$ @icurays1 Where have I implicitly assumed k>0? $\endgroup$
    – Arthur
    Nov 18 '12 at 17:23
  • 2
    $\begingroup$ Using Jordan's lemma in the upper half plane. $\endgroup$
    – icurays1
    Nov 18 '12 at 17:25

You have assumed that $k>0$, so the integral is correct, but you must also consider negative values. To this, try taking a semicircular contour that is on the bottom half of the plane instead.

An example is here: http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28II.29_.E2.80.93_Cauchy_distribution


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