Integrals involving complex exponentials

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September 28th 2011, 09:01 AM
rebghb

Integrals involving complex exponentials

Hello Everyone!

I'm being reluctant when finding integrals involving complex exponentials, of the form:

1. $\int^{+\infty}_{-\infty}e^{-iat}dt$
2. $\int^{+\infty}_{0}e^{-iat}dt$

This is because I do not know what to do with infinities multiplied by a complex number, but I know as $t \rightarrow \infty$

Is there a way to solve this without getting into Fourier transform? I know that generalized functions are involved in those integrals, but can we got an answer without going through that?

Thanks!
September 28th 2011, 10:35 AM
Ackbeet

Re: Integrals involving complex exponentials

Those integrals do not converge. Period.
September 29th 2011, 09:58 AM
rebghb

Re: Integrals involving complex exponentials

Quote:

Originally Posted by Ackbeet

Those integrals do not converge. Period.

Okay. So let's put $2\pi f$ in place of $a$ , we get $\int ^{+\infty}_{-\infty}e^{-i2\pi ft}dt$ kinda reminds me of something (Happy)

Isn't that the Fourier transform of $f(t) = 1$ ?
September 29th 2011, 07:14 PM
CaptainBlack

Re: Integrals involving complex exponentials

Quote:

Originally Posted by rebghb

Okay. So let's put $2\pi f$ in place of $a$ , we get $\int ^{+\infty}_{-\infty}e^{-i2\pi ft}dt$ kinda reminds me of something (Happy)

Isn't that the Fourier transform of $f(t) = 1$ ?

Whatever some people think that is not how the FT of f(x)=1 is defined (at least without specifying what convergence method you want to use)

CB
September 30th 2011, 11:26 AM
rebghb

Re: Integrals involving complex exponentials

I know this integral, eventually gives $\delta (f)$ . But that's like giving infity a function in other variable
September 30th 2011, 12:12 PM
CaptainBlack

Re: Integrals involving complex exponentials

Quote:

Originally Posted by rebghb

I know this integral, eventually gives $\delta (f)$ . But that's like giving infity a function in other variable

Not under a conventional definition of integration. You need to resort to the theory of distributions to get the delta functional.

CB
October 1st 2011, 01:44 AM
rebghb

Re: Integrals involving complex exponentials

Alright, before closing the thread, I would like to say, In a course on basic signal proecssing, they establish the following:

$\mathcal{F} {\delta (t)} =\int ^{+\infty}_{-\infty}\delta (t) e^{-i2\pi ft dt} = 1$

Now, by inverse transform, and exchanging variables, we get the that the Fourier transform of f(t)=1 is $\delta (f)$ . Is this mathematically accepted?

NB The delta here is the Dirac delta
October 1st 2011, 06:24 AM
CaptainBlack

Re: Integrals involving complex exponentials

Quote:

Originally Posted by rebghb

Alright, before closing the thread, I would like to say, In a course on basic signal proecssing, they establish the following:

$\mathcal{F} {\delta (t)} =\int ^{+\infty}_{-\infty}\delta (t) e^{-i2\pi ft dt} = 1$

Now, by inverse transform, and exchanging variables, we get the that the Fourier transform of f(t)=1 is $\delta (f)$ . Is this mathematically accepted?

NB The delta here is the Dirac delta

Yes, in a slightly hand-waving way, but acceptable in signal processing and physics

CB