what is the value of $e^{jx}$ as $x \to\infty?$

Asked Sep 29 '24 at 16:52

Active Sep 29 '24 at 17:00

Viewed 89 times

This came up when we started Fourier Transform in class but the teacher essentially handwaved it to be equal to zero.

if I were to treat the power as real then it obviously tends to zero

using the Euler expansion however, I am almost certain that it is undefined. but bounded (kind of like $\lim_{x\to \infty} \sin x$)

This exponential comes often enough that i have to know about this.

Btw, if anyone is interested, the specific question it was first used in was $$FT(u(t))=\int_{-\infty}^\infty u(t)e^{-jwt}$$ this resolves to $$\frac{e^{-jw\infty}-1}{-jw}=-\frac{1}{jw}$$

if any of you have a less... mathematically painful way to evaluate it, please do tell.

edited Sep 29 '24 at 17:00

Thomas Andrews

186,215

asked Sep 29 '24 at 16:52

Animesh Shukla

1

Is $j=\sqrt{-1},$ the imaginary unit? It definitely doesn't go to zero then. – Thomas Andrews Sep 29 '24 at 16:56
You really have to do some hand-wavy things with some Fourier transforms until you reach functional analysis and realize that the transform is really being applied to distributions, not functions. The Dirac $\delta$ function, for example, is not really a function, but a distribution. – Thomas Andrews Sep 29 '24 at 16:57
What is $u(t)$ defined to be for the "specific example?" – Thomas Andrews Sep 29 '24 at 17:02
@ThomasAndrews yes j^2=-1
and u(t)=0 fot t<0 and =1 for t=>1
– Animesh Shukla Sep 29 '24 at 17:04
@ThomasAndrews so, what i said in the third line is correct? $e^{-j \infty}$ is indeterminate but bounded? – Animesh Shukla Sep 29 '24 at 17:07
It is hand waving. The Fourier transform is defined for integrable functions, and the unit step is not integrable. From an engineering standpoint, one can apply a few rules to obtain useful results, but these are based on the mathematics of distributions. – copper.hat Sep 29 '24 at 17:08
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basically, you're asking what's the FT of the Heaviside function, right? – Sine of the Time Sep 29 '24 at 17:08
There is no limit of $e^{j \omega t}$ as $t \to \infty$ if $\omega \neq 0$. – copper.hat Sep 29 '24 at 17:09
@SineoftheTime yep – Animesh Shukla Sep 29 '24 at 17:12
@copper.hat i see. so, if it appears again, i cant just take the exponential to be zero? bummer. also, did what i said in third line correct? or is it that it the exponential $e^{-jw\infty}$ is literally undefined like 1/0? – Animesh Shukla Sep 29 '24 at 17:15
@AnimeshShukla It is undefined, it is like looking for a limit as a a point spins around in a circle. – copper.hat Sep 29 '24 at 17:16
Yes, the limit is undefined and the function is bounded. – Thomas Andrews Sep 29 '24 at 17:19
https://math.stackexchange.com/search?q=Fourier+transform+Heaviside – Anne Bauval Sep 29 '24 at 17:20
i guess it answers the question. and now i feel kinda dumb. should i delete the question? – Animesh Shukla Sep 29 '24 at 17:20
It is a reasonable question, and good that you are curious. – copper.hat Sep 29 '24 at 17:22
alright imma keep it up. – Animesh Shukla Sep 29 '24 at 17:26
Please put the definition of $u$ in your question. As written, the question is incomplete, and a reader should not have to dig through comments to know the question.@AnimeshShukla – Thomas Andrews Sep 29 '24 at 17:34
See THIS ANSWER – Mark Viola Sep 29 '24 at 20:33

what is the value of $e^{jx}$ as $x \to\infty?$

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