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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.

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