Given the Fourier Tranform defined as $$\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi):=\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}e^{-2\pi i \xi t}dt=\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}$$ They say inverse Fourier Transform is given by $$\color{blue}{\mathcal{F}^{-1}\{}\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}\color{blue}{\}}(t)=\int_{-\infty}^{+\infty}\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}e^{2\pi i \xi t}d\xi=\color{green}{f(}t\color{green}{)}$$ I'm expecting that $$\color{blue}{\mathcal{F}^{-1}\{}\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi)\color{blue}{\}}(t)=\color{green}{f(}t\color{green}{)}$$ But actually $$\color{blue}{\mathcal{F}^{-1}\{}\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi)\color{blue}{\}}(t)=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}e^{-2\pi i \xi t}dt)e^{2\pi i \xi t}d\xi=$$$$=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}dtd\xi=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}d\xi)\color{green}{f(}t\color{green}{)}dt=\int_{-\infty}^{+\infty}0\cdot \color{green}{f(}t\color{green}{)}dt=0 \text{ ??}$$
-
1What the inverse should return is a function not necessary how its argument is denoted; that is, in the last integral you shouldn't write $$e^{2\pi\xi i t}$$ but $$e^{2\pi\xi i u}$$ or something else other than $t$, then you get some $\delta(\xi - u)$ when you integrate over $t$ – Physor Sep 20 '20 at 20:43
-
1How do you get zero from the inner integral? – md2perpe Sep 20 '20 at 20:45
-
1@md2perpe that is the magic part, just joking – Physor Sep 20 '20 at 20:46
-
@Physor. Yes, there are several errors in Amad's calculation. – md2perpe Sep 20 '20 at 21:06
-
@Physor Ah yes of course.. I will remember this important error I made that is easy to commit.. I'm trying to learn to invert integral transform operators, so I'm starting with the basic Fourier Transform, but derivation / proof of inversion formula seems very hard to grasp for me, but I'll keep trying.. – Amad Sep 20 '20 at 21:09
2 Answers
Let $f \in L^1(\mathbb{R})$ so that $\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} \, dx$ is defined. Then $$ |\hat{f}(\xi)| = \left|\int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} dx\right| \leq \int_{-\infty}^{\infty} \left|f(x) \, e^{-ix\xi}\right| \, dx = \int_{-\infty}^{\infty} \left|f(x) \right| \, dx < \infty $$ so $\hat{f}\in L^\infty(\mathbb{R})$ but we cannot be sure that $\hat{f} \in L^1(\mathbb{R}).$ Therefore, let us define $\hat{f_\epsilon}(\xi) := e^{-\epsilon \xi^2/2}\hat{f}(\xi)$ so that $\hat{f_\epsilon} \in L^1(\mathbb{R}),$ and at the end take the limit $\epsilon\to 0.$ We then have \begin{align} \int_{-\infty}^{\infty} \hat{f_\epsilon}(\xi) \, e^{iy\xi} \, d\xi &= \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2}\hat{f}(\xi) \, e^{iy\xi} \, d\xi \\ &= \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2} \left( \int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} \, dx \right) \, e^{iy\xi} \, d\xi \\ \\ &= \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2} \, e^{-ix\xi} \, e^{iy\xi} \, d\xi \right) \, dx \\ &= \int_{-\infty}^{\infty} f(x) \sqrt{\frac{2\pi}{\epsilon}} e^{-(x-y)^2/(2\epsilon)} \, dx \\ \\ &\overset{\{ x=y+z\sqrt{2\epsilon} \}}{=} \int_{-\infty}^{\infty} f(y+z\sqrt{2\epsilon}) \sqrt{\frac{2\pi}{\epsilon}} e^{-z^2} \, \sqrt{2\epsilon} \, dz \\ &= 2\sqrt{\pi} \int_{-\infty}^{\infty} f(y+z\sqrt{2\epsilon}) \, e^{-z^2} \, dz \\ &\to 2\sqrt{\pi} \int_{-\infty}^{\infty} f(y) \, e^{-z^2} \, dz = 2\sqrt{\pi} \int_{-\infty}^{\infty} \, e^{-z^2} \, dz \, f(y) = 2\pi \, f(y). \end{align}
-
but you interchanged the limit and the integral at the end, how is that justified. Of course I hope you have no problem with my edit – Physor Oct 11 '20 at 12:24
-
@Physor. Taking limits inside of the integral is justified by the dominated convergence theorem. – md2perpe Oct 11 '20 at 12:31
The first integral should be \begin{align} I(u) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t}e^{i2\pi\xi u}dtd\xi\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t+i2\pi\xi u}dtd\xi\\ &=\int_{-\infty}^{\infty}f(t)\left(\int_{-\infty}^{\infty}e^{i2\pi\xi (u-t)}d\xi\right)dt\\ \end{align} We have $$\delta(u-t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(u-t)x}dx = \int_{-\infty}^{\infty}e^{i2\pi(u-t)\xi}d\xi $$ Thereby it becomes \begin{align} I(u) &= \int_{-\infty}^{\infty}f(t)\delta(u-t)dt\\ &= f(u) \end{align} The function as mapping is returned, doesn't matter how its argument is denoted
- 4,586
-
this is similar to what I've done after fixing the mistake, but I didn't replace the inner Integral with Dirac delta. Do you have any reference to the proof that that inner integral is $\delta (u-t)$? – Amad Sep 20 '20 at 21:20
-
-
-
@Amad I said it's the magic part, that was about how you got the zero!, check this out https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_transform – Physor Sep 20 '20 at 21:37
-
yes, sorry, and it still seems like the magic part because when I plug $t=0$ in $\int_{-\infty}^{+\infty}e^{ixt}dx$ we get $\int_{-\infty}^{+\infty}e^{ix\cdot 0}dx=\int_{-\infty}^{+\infty}dx=0$ by the magic and no voodooich $2\pi\delta (0)=\infty$ – Amad Sep 20 '20 at 21:51
-
https://math.stackexchange.com/questions/3814073/fourier-representation-of-diracs-delta-function/3814112#3814112 – cmk Sep 20 '20 at 21:52
-
i didn't say it, I meant $\int_{-\infty}^{+\infty}e^{ix\cdot 0}dx=\int_{-\infty}^{+\infty}e^0 dx=\int_{-\infty}^{+\infty}1dx=0$ – Amad Sep 20 '20 at 21:55
-
but $\lim_{R\to\infty}\int_{-R}^{R}dx=\lim_{R\to\infty}\left[x\right]{-R}^{R}=\lim{R\to\infty}(R-R)=\lim_{R\to\infty}0=0$ – Amad Sep 20 '20 at 21:58
-
-
-
Dirac's delta shouldn't be used since it's not a function but a distribution. – md2perpe Sep 20 '20 at 23:30
-
2
-