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One question that I've had for a week or so now is how the following integral can equal a Dirac delta function:

[tex] \frac{1}{2\pi} \int_{-\infty}^{\infty}{dt} \:e^{i(\omega - \omega^{'})t}\: = \: \delta(\omega - \omega^{'})[/tex]

A text that I was reading discusses Fourier transforms and eventually arrives at the above equation through the use of definitions. Since the book was taking an inverse Fourier transform, the solution is already known just to be f(t) (the function first operated on by the regular Fourier transform.)

We know the definition of the Dirac delta function:

[tex]f(t)\: = \: \int_{-\infty}^{\infty}{d\tau f(\tau) \delta(t - \tau)[/tex]

And here we have a function f(t) being transformed and inverse transformed (with a dummy variable tau):

[tex]f(t)\: = \: \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{d\omega\:e^{i\omega t}} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} d\tau \: e^{-i\omega \tau} f(\tau)[/tex]

which can be manipulated to become

[tex]f(t)\: = \: \int_{-\infty}^{\infty}{d\tau f(\tau) [ \frac{1}{2\pi} \int_{-\infty}^{\infty}{d\omega} \:e^{i(t - \tau)\omega}][/tex]

Together these equations produce:

[tex] \frac{1}{2\pi} \int_{-\infty}^{\infty}{dt} \:e^{i(t - \tau)\omega}\: = \: \delta(t - \tau)[/tex]

(which is essentially the first equation)

I see how a delta function operating within an integral applies to the above case to show how the first equation works out, however, I wasn't sure if there was a direct mathematical formulation that would get from the original indefinite integral of e^(w-w') to the dirac delta function.

Obviously if I try and integrate this function and evaluate the solution at infinity and negative infinity the function diverges.

Let me know if there are any extra ways to look at this integral (or evaluate it for that matter) that would lead to the solution of a delta function.

Thanks,

pcalhoun

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# Getting a delta function from an indefinite integral

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