Limit of sin(x) as x approaches infinity
$begingroup$
This question comes from Fourier Transforms, specifically the evaluation of $mathcal{F}(e^{2pi iat})$. Normally I see this derived by first finding the Inverse FT of a delta function, i.e.
begin{align}
mathcal{F}^{-1}(delta(omega-a)) &= int_{-infty}^infty delta(omega-a) e^{2pi i omega t},domega = e^{2pi i at}
end{align}
But suppose we wanted to be stubborn and directly evaluate
begin{align}
mathcal{F}(e^{2pi i a t}) &= int_{-infty}^infty e^{2pi i a t} e^{-2pi i omega t} ,dt
end{align}
It should work, right? Why wouldn't it? If we continue, we get to
begin{align}
mathcal{F}(e^{2pi i at}) &= lim_{x to infty}int_{-x}^x e^{2pi i (a - omega) t} ,dt \ &= frac{1}{pi(a-omega)}lim_{x to infty}sin(2pi(a-omega) x) \
&= frac{1}{pi (a - omega)}lim_{x to infty} sin(x)
end{align}
Now, how do we convincingly evaluate $displaystyle lim_{x to infty} sin(x)$? If I didn't have the context of evaluating a Fourier Transform, I would simply answer that the limit doesn't exist.
However, in this case I know what the answer should be, but I don't at all understand why it works or how to make it rigorous. It seems to contradict my previous understanding of limits.
limits fourier-transform
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
$endgroup$
add a comment |
$begingroup$
This question comes from Fourier Transforms, specifically the evaluation of $mathcal{F}(e^{2pi iat})$. Normally I see this derived by first finding the Inverse FT of a delta function, i.e.
begin{align}
mathcal{F}^{-1}(delta(omega-a)) &= int_{-infty}^infty delta(omega-a) e^{2pi i omega t},domega = e^{2pi i at}
end{align}
But suppose we wanted to be stubborn and directly evaluate
begin{align}
mathcal{F}(e^{2pi i a t}) &= int_{-infty}^infty e^{2pi i a t} e^{-2pi i omega t} ,dt
end{align}
It should work, right? Why wouldn't it? If we continue, we get to
begin{align}
mathcal{F}(e^{2pi i at}) &= lim_{x to infty}int_{-x}^x e^{2pi i (a - omega) t} ,dt \ &= frac{1}{pi(a-omega)}lim_{x to infty}sin(2pi(a-omega) x) \
&= frac{1}{pi (a - omega)}lim_{x to infty} sin(x)
end{align}
Now, how do we convincingly evaluate $displaystyle lim_{x to infty} sin(x)$? If I didn't have the context of evaluating a Fourier Transform, I would simply answer that the limit doesn't exist.
However, in this case I know what the answer should be, but I don't at all understand why it works or how to make it rigorous. It seems to contradict my previous understanding of limits.
limits fourier-transform
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
$endgroup$
1
$begingroup$
The limit does not exist; you've proved that $e^{2pi i a t}$ is not an integrable function. The "Fourier transform" you discuss is actually a generalized Fourier transform; it's very useful, but one shouldn't expect procedures related to integrable functions and regular Fourier transforms to apply.
$endgroup$
– Greg Martin
Nov 30 '18 at 6:13
$begingroup$
Is it possible to generalize the notion of integration to match the generalized Fourier Transform? Or a way to directly evaluate $mathcal{F}(e^{2pi i at})$ that does not make use of the inverse fourier transform?
$endgroup$
– user2520385
Nov 30 '18 at 6:26
add a comment |
$begingroup$
This question comes from Fourier Transforms, specifically the evaluation of $mathcal{F}(e^{2pi iat})$. Normally I see this derived by first finding the Inverse FT of a delta function, i.e.
begin{align}
mathcal{F}^{-1}(delta(omega-a)) &= int_{-infty}^infty delta(omega-a) e^{2pi i omega t},domega = e^{2pi i at}
end{align}
But suppose we wanted to be stubborn and directly evaluate
begin{align}
mathcal{F}(e^{2pi i a t}) &= int_{-infty}^infty e^{2pi i a t} e^{-2pi i omega t} ,dt
end{align}
It should work, right? Why wouldn't it? If we continue, we get to
begin{align}
mathcal{F}(e^{2pi i at}) &= lim_{x to infty}int_{-x}^x e^{2pi i (a - omega) t} ,dt \ &= frac{1}{pi(a-omega)}lim_{x to infty}sin(2pi(a-omega) x) \
&= frac{1}{pi (a - omega)}lim_{x to infty} sin(x)
end{align}
Now, how do we convincingly evaluate $displaystyle lim_{x to infty} sin(x)$? If I didn't have the context of evaluating a Fourier Transform, I would simply answer that the limit doesn't exist.
However, in this case I know what the answer should be, but I don't at all understand why it works or how to make it rigorous. It seems to contradict my previous understanding of limits.
limits fourier-transform
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
$endgroup$
This question comes from Fourier Transforms, specifically the evaluation of $mathcal{F}(e^{2pi iat})$. Normally I see this derived by first finding the Inverse FT of a delta function, i.e.
begin{align}
mathcal{F}^{-1}(delta(omega-a)) &= int_{-infty}^infty delta(omega-a) e^{2pi i omega t},domega = e^{2pi i at}
end{align}
But suppose we wanted to be stubborn and directly evaluate
begin{align}
mathcal{F}(e^{2pi i a t}) &= int_{-infty}^infty e^{2pi i a t} e^{-2pi i omega t} ,dt
end{align}
It should work, right? Why wouldn't it? If we continue, we get to
begin{align}
mathcal{F}(e^{2pi i at}) &= lim_{x to infty}int_{-x}^x e^{2pi i (a - omega) t} ,dt \ &= frac{1}{pi(a-omega)}lim_{x to infty}sin(2pi(a-omega) x) \
&= frac{1}{pi (a - omega)}lim_{x to infty} sin(x)
end{align}
Now, how do we convincingly evaluate $displaystyle lim_{x to infty} sin(x)$? If I didn't have the context of evaluating a Fourier Transform, I would simply answer that the limit doesn't exist.
However, in this case I know what the answer should be, but I don't at all understand why it works or how to make it rigorous. It seems to contradict my previous understanding of limits.
limits fourier-transform
limits fourier-transform
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
asked Nov 30 '18 at 5:45
user2520385user2520385
27616
27616
1
$begingroup$
The limit does not exist; you've proved that $e^{2pi i a t}$ is not an integrable function. The "Fourier transform" you discuss is actually a generalized Fourier transform; it's very useful, but one shouldn't expect procedures related to integrable functions and regular Fourier transforms to apply.
$endgroup$
– Greg Martin
Nov 30 '18 at 6:13
$begingroup$
Is it possible to generalize the notion of integration to match the generalized Fourier Transform? Or a way to directly evaluate $mathcal{F}(e^{2pi i at})$ that does not make use of the inverse fourier transform?
$endgroup$
– user2520385
Nov 30 '18 at 6:26
add a comment |
1
$begingroup$
The limit does not exist; you've proved that $e^{2pi i a t}$ is not an integrable function. The "Fourier transform" you discuss is actually a generalized Fourier transform; it's very useful, but one shouldn't expect procedures related to integrable functions and regular Fourier transforms to apply.
$endgroup$
– Greg Martin
Nov 30 '18 at 6:13
$begingroup$
Is it possible to generalize the notion of integration to match the generalized Fourier Transform? Or a way to directly evaluate $mathcal{F}(e^{2pi i at})$ that does not make use of the inverse fourier transform?
$endgroup$
– user2520385
Nov 30 '18 at 6:26
1
1
$begingroup$
The limit does not exist; you've proved that $e^{2pi i a t}$ is not an integrable function. The "Fourier transform" you discuss is actually a generalized Fourier transform; it's very useful, but one shouldn't expect procedures related to integrable functions and regular Fourier transforms to apply.
$endgroup$
– Greg Martin
Nov 30 '18 at 6:13
$begingroup$
The limit does not exist; you've proved that $e^{2pi i a t}$ is not an integrable function. The "Fourier transform" you discuss is actually a generalized Fourier transform; it's very useful, but one shouldn't expect procedures related to integrable functions and regular Fourier transforms to apply.
$endgroup$
– Greg Martin
Nov 30 '18 at 6:13
$begingroup$
Is it possible to generalize the notion of integration to match the generalized Fourier Transform? Or a way to directly evaluate $mathcal{F}(e^{2pi i at})$ that does not make use of the inverse fourier transform?
$endgroup$
– user2520385
Nov 30 '18 at 6:26
$begingroup$
Is it possible to generalize the notion of integration to match the generalized Fourier Transform? Or a way to directly evaluate $mathcal{F}(e^{2pi i at})$ that does not make use of the inverse fourier transform?
$endgroup$
– user2520385
Nov 30 '18 at 6:26
add a comment |
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$begingroup$
The limit does not exist; you've proved that $e^{2pi i a t}$ is not an integrable function. The "Fourier transform" you discuss is actually a generalized Fourier transform; it's very useful, but one shouldn't expect procedures related to integrable functions and regular Fourier transforms to apply.
$endgroup$
– Greg Martin
Nov 30 '18 at 6:13
$begingroup$
Is it possible to generalize the notion of integration to match the generalized Fourier Transform? Or a way to directly evaluate $mathcal{F}(e^{2pi i at})$ that does not make use of the inverse fourier transform?
$endgroup$
– user2520385
Nov 30 '18 at 6:26