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What is the easiest way to to derive the following equation:

$$\int_{-\infty}^{\infty}e^{ikx}dx = 2\pi\delta(k)$$

I understand the equation can be derived by assuming the Fourier integral theorem, but can anyone derive this without making reference to the Fourier integral theorem?

Kenshin
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    Please check this http://math.stackexchange.com/questions/558077/is-there-any-handwavy-argument-that-shows-that-int-infty-infty-e-ikx/560206#560206 – Mathlover May 21 '14 at 09:59
  • @Mathlover, thanks – Kenshin May 21 '14 at 10:18
  • @Mathlover, I struggle with the last part of that proof. I can't understand the jump from lim sin(ax)/x = 2pidelta(k). How can we prove the limit is 2pidelta(k)? – Kenshin May 21 '14 at 12:50
  • : It is just a defination : You could say $\lim\limits_{ a\to \infty } f_a(x) = \lim\limits_{ a\to \infty } 2\frac{\sin {ax}}{x}= g(x)$ and you could try to draw $g(x)$. You can see some $a$ values how affect the graphs. You will get $g(x)$ as impulse graph if $a--->\infty$ . Then If you use Priyatham result in other question, you could see that For $\epsilon>0$ $$ \int\limits_{-\epsilon}^\epsilon g(x) \mathrm{d}x = 2\pi$$

    Finally

    Somebody defined that $g(x)=2\pi \delta(x)$. Is it ok now proof?

     – Mathlover May 21 '14 at 13:51
  • @Mathlover, I'm asking how do you know that the integral of g(x) from -e to e is 2pi. You've stated it but haven't proven it, and this is the only step in the proof I don't understand. – Kenshin May 21 '14 at 14:21
  • Please see other answer .As i wrote, Priyatham's answer gave that proof – Mathlover May 21 '14 at 16:11
  • see the answer here. I pretty much asked the same question not so long ago.http://math.stackexchange.com/questions/796744/the-delta-function-written-as-the-integral-of-a-complex-number – mark leeds May 22 '14 at 00:12
  • @markleeds, unfortunately that answer makes use of the fourier theorem, but I am trying to use the above integral to prove the fourier theorem. – Kenshin May 22 '14 at 00:43

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