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Checking some work on a Fourier Transform

  • Thread starter Emspak
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  • #1
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Homework Statement



OK, we're given to practice Fourier transforms. We are given

[tex]f(x) = \int^{+\infty}_{-\infty} g(k) e^{ikx}dk[/tex]

and told to get a Fourier transform of the following, and find g(k):

[itex]f(x) = e^{-ax^2}[/itex] and [itex]f(x) = e^{-ax^2-bx}[/itex]

Homework Equations





The Attempt at a Solution



For the first one I tried the following:

The integral of [itex]e^{-ax^2}[/itex] is [itex]\sqrt{\frac{\pi}{a}}[/itex]. I saved that information for later.

To get the Fourier transform I used:

[tex]\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}e^{ikx}e^{-ax^2}dx[/tex]

"completing the square" inside the integral I get [itex]e^{-a(x+\frac{ik}{2a})^2-\frac{k^2}{4a}}[/itex] and that leaves me with

[tex]\frac{1}{\sqrt{2\pi}}e^{\frac{-k^2}{4a}}\int^{+\infty}_{-\infty}e^{-a(x+\frac{ik}{2a})^2}dx[/tex]

I can change the variable so that [itex]y^2=(x+\frac{ik}{2a})^2[/itex]
and dx=dy in this case.

After all that I can change the limits (but they are still infinity, right?) So that I get this:

[tex]\frac{1}{\sqrt{2\pi}}e^{\frac{-k^2}{4a}}\int^{+\infty+\frac{ik}{2a}}_{-\infty+\frac{ik}{2a}}e^{ay^2}dy[/tex]

Which we know from before is [itex]\sqrt{\frac{\pi}{a}}[/itex].

And that leaves us with

[itex]\frac{1}{\sqrt{2a}}e^{\frac{-k^2}{4a}}=g(k)[/itex]

Assuming all this is correct -- and I am not 100 percent sure it is -- then if I were to apply the same procedure to [itex]e^{ax^2-bx}[/itex] I should end up with the above multiplied by the integral of [itex]e^{-bx}[/itex]. But if I try to integrate that I get infinities.

But I wasn't sure if that was a completely wrong approach, or if I even did the above bit correctly. The thing I have trouble getting my head around is whether e is periodic or not and what period to use when applying Fourier equations. I might be approaching this entirely wrongly.

Thanks in advance for whatever assistance/ criticism/ telling me I am foolish.
 
Last edited:

Answers and Replies

  • #2
vanhees71
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If you define your Fourier transform from [itex]k[/itex] to [itex]x[/itex] space as
[tex]f(x)=\int_{\mathbb{R}} \mathrm{d} k g(k) \exp(\mathrm{i} k x),[/tex]
then the inverse is
[tex]g(k)=\frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} x f(x) \exp(-\mathrm{i} k x).[/tex]
Concerning the factors [itex]2 \pi[/itex] this is the opposite of what physicists (at least in the high-energy particle/nuclear physics community) use.

In principle the way you do the integral is correct.
 
  • #3
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thanks, that helps. One other question: say I want to integrate

[tex]\int^{\infty}_{-\infty}e^{ikx}dx[/tex]

would I be better off using the Euler formula, which would get me [itex]\int^{\infty}_{-\infty}\cos x + i \sin x dx[/itex]? But then the cosine integral diverges...

I could convert it to polar coordinates, but then I end up with

[itex]x = r\cos \theta[/itex] and [itex]dx = -r\sin \theta + \cos \theta d\theta[/itex]

which leads me to

[tex]\int^{\infty}_{-\infty}e^{ikx}dx = \int^{\infty}_{0}e^{ikr\cos \theta}(-r \sin \theta d\theta) + \cos \theta dr= \int^{\infty}_{0}-re^{ikr\cos \theta}(\sin \theta) d\theta + \int^{\infty}_{0}e^{ikr\cos \theta}(\cos \theta) dr[/tex]

where the first term would go to zero but I suspect I am approaching this the wrong way...
 
  • #4
pasmith
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thanks, that helps. One other question: say I want to integrate

[tex]\int^{\infty}_{-\infty}e^{ikx}dx[/tex]
That integral diverges. There is a sense in which [tex]
\int_{-\infty}^\infty e^{ikx}\,dx = 2\pi\delta(k)
[/tex] where [itex]\delta[/itex] is the Dirac delta distribution, since then the inverse formula gives [tex]
\frac{1}{2\pi}\int_{-\infty}^\infty 2\pi \delta(k)e^{-ikx}\,dk = \frac{1}{2\pi} 2\pi e^0 = 1
[/tex]
 

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