# Checking some work on a Fourier Transform

1. Aug 18, 2014

### Emspak

1. The problem statement, all variables and given/known data

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

$$f(x) = \int^{+\infty}_{-\infty} g(k) e^{ikx}dk$$

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

$f(x) = e^{-ax^2}$ and $f(x) = e^{-ax^2-bx}$

2. Relevant equations

3. The attempt at a solution

For the first one I tried the following:

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

To get the Fourier transform I used:

$$\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}e^{ikx}e^{-ax^2}dx$$

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

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

I can change the variable so that $y^2=(x+\frac{ik}{2a})^2$
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:

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

Which we know from before is $\sqrt{\frac{\pi}{a}}$.

And that leaves us with

$\frac{1}{\sqrt{2a}}e^{\frac{-k^2}{4a}}=g(k)$

Assuming all this is correct -- and I am not 100 percent sure it is -- then if I were to apply the same procedure to $e^{ax^2-bx}$ I should end up with the above multiplied by the integral of $e^{-bx}$. 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: Aug 18, 2014
2. Aug 18, 2014

### vanhees71

If you define your Fourier transform from $k$ to $x$ space as
$$f(x)=\int_{\mathbb{R}} \mathrm{d} k g(k) \exp(\mathrm{i} k x),$$
then the inverse is
$$g(k)=\frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} x f(x) \exp(-\mathrm{i} k x).$$
Concerning the factors $2 \pi$ 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. Aug 18, 2014

### Emspak

thanks, that helps. One other question: say I want to integrate

$$\int^{\infty}_{-\infty}e^{ikx}dx$$

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

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

$x = r\cos \theta$ and $dx = -r\sin \theta + \cos \theta d\theta$

which leads me to

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

where the first term would go to zero but I suspect I am approaching this the wrong way...

4. Aug 18, 2014

### pasmith

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

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