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Difficult integral involving exponential

  • Thread starter zandria
  • Start date
  • #1
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Homework Statement


I'm trying to verify the Fourier transform but am getting stuck on the integration. Here is the pair:
[tex]f(x) = e^{-ax^2}[/tex]
[tex]\hat{f}(k) = \frac{1}{\sqrt{2a}}e^{-k^2/4a}[/tex]
[tex]a>0[/tex]

Homework Equations



I know that
[tex]\hat{f}(k)=\int_{-\infty}^{\infty}f(x)e^{ikx}dx[/tex]

The Attempt at a Solution



So I have
[tex]\hat{f}(k)=\int_{-\infty}^{\infty}e^{-ax^2}e^{ikx}dx[/tex]
[tex]\hat{f}(k)=\int_{-\infty}^{\infty}e^{-ax^2+ikx}dx[/tex]

I tried using integration by parts and I'm not sure that's the right way to go. If it is I'm not sure how to go about it without getting a more complicated integral.
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
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You need to complete the square, which means:

[tex]-(ax^2+ikx)=-\left[(\alpha x+\beta)^2+\gamma \right][/tex].

Find [itex]\alpha,\beta[/itex] and [itex]\gamma[/itex].

Edit1: it seems you have either listed [itex]\hat{f}(k)[/itex] wrong or the book where you got [itex]\hat{f}(k)[/itex] from is wrong, because the answer should be:

[tex]
\hat{f}(k) = \sqrt{\frac{\pi}{a}}e^{-k^2/4a}
[/tex]

edit2: While making no difference to the final answer in this case, shouldn't it be [tex]\hat{f}(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/tex], note the minus sign.
 
Last edited:

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