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| − | =Problem= | + | ==SOLUTION== |
| − | ''Fourier transform of the Gauss function''
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| − | Show that
| + | See [https://unapologetic.wordpress.com/2010/11/08/the-character-table-of-s4/] |
| − | $$\frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} dx = e^{-\frac{1}{2}tk^2} , \forall k \in \mathbb{R}$$
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| − | =Proof=
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| − | $$
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| − | \begin{align}
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| − | I \ &= \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} dx \\
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| − | &= \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{1}{2t}(x^2 + 2ikt)} dx \\
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| − | &= \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{1}{2t}((x + ikt)^2 + k^2t^2)} dx \\
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| − | &= \frac{1}{\sqrt{2\pi t}} e^{-\frac{k^2t}{2}} \int_{-\infty}^\infty e^{-\frac{1}{2t}(x + ikt)^2} dx .
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| − | \end{align}
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| − | $$
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| − | Now, we substitute \(u = x + ikx\) with \(du = dx\) and find
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| − | $$
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| − | \begin{align}
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| − | I \ &= \frac{1}{\sqrt{2\pi t}} e^{-\frac{k^2t}{2}} \int_{-\infty}^\infty e^{-\frac{1}{2t}u^2} du\\
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| − | &= \frac{1}{\sqrt{2\pi t}} e^{-\frac{k^2t}{2}} \sqrt{2\pi t} \\
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| − | &= e^{-\frac{k^2 t}{2}} .
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| − | \end{align}
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| − | $$
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| − | Where we have used that
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| − | $$\int_{-\infty}^\infty e^{-x^2 a} dx = \sqrt{\frac{\pi}{a}}$$
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| − | Thus, the proof follows.
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| − | <p style="text-align:right;">\(\square\)</p>
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| − | (No, I don't know how to right align a square. Change this if you do!) --[[User:Nik|Nik]] ([[User talk:Nik|talk]]) 11:16, 18 December 2014 (CET)
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