Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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. –  icurays1 Nov 18 '12 at 17:05
    
@icurays1 Where have I implicitly assumed k>0? –  Arthur Nov 18 '12 at 17:23
1  
Using Jordan's lemma in the upper half plane. –  icurays1 Nov 18 '12 at 17:25
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.