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From an old notebook, I found a question mark (not known existing how many years) about a contradition between the integral form and the original exponential decaying function $f(t)$.

$$ f(t)=\begin{cases} e^{-bt},& t\ge 0\\ 0 & t<0 \end{cases} $$ where $b>0$ and $f(0)=1$. The Fourier transform of $f(t)$ is $$ \mathcal{F}[f(t)](\omega)=\frac{1}{j\omega+b}, $$

But if $f(t)$ is recovered by inverse transform from $F(\omega)$ \begin{align} f(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{j\omega+b}e^{j\omega t}\mathrm d\omega\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{b\cos{\omega t}+\omega\sin{\omega t}}{\omega^2+b^2}\mathrm d\omega\\ &=\begin{cases} 0 & \text{ for } t<0 \text{ (this integral is a bit complexed to solve) }\\ \frac{1}{2\pi}\left[\arctan\frac{\omega}{b}\right]_{-\infty}^{\infty}=\frac12 & \text{ for } t=0\\ e^{-bt} & \text{ for } t>0 \end{cases} \end{align}

where in the second case $t=0$, $f(t)$ becomes only $\frac12\ne e^{-bt}|_{t=0}=1$.

Where in the steps lost the other half?

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  • $\begingroup$ Think about the domain of the Fourier transform. What kind of object does it act on? $\endgroup$
    – Xander Henderson
    Commented Mar 10 at 19:48
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    $\begingroup$ Sorry not get the point. Please... $\endgroup$
    – MathArt
    Commented Mar 10 at 19:50
  • $\begingroup$ Do you have an answer to the question I asked? What kind of object does the FT act on? $\endgroup$
    – Xander Henderson
    Commented Mar 10 at 19:54
  • $\begingroup$ The time domain but with $u(t)$, right? Or I didn't get what object means? $\endgroup$
    – MathArt
    Commented Mar 10 at 20:13
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    $\begingroup$ No. I am not talking about Stone-Weierstrass. I am talking about the definition of an $L^1$ "function". Two pointwise-defined functions which differ only on a set of measure zero will have the same FT, because the FT doesn't "see" the difference between those two functions. Instead, which if I take your function and redefine it so that $f(2) = 47$ (I am only changing the value at that one point). When you take the IVT of the FT, you are not going to get a function $\check{\hat{f}}$ with $\check{\hat{f}}(2) = 47$. Should I ask what happened to my $47$? $\endgroup$
    – Xander Henderson
    Commented Mar 10 at 20:26

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This is part of the Gibb's phenomenon. It is more thoroughly explained there, but the short version is:

  • Not all functions are recovered by the inverse Fourier transform of their Fourier transform.
  • The combination "inverse Fourier transform of the Fourier transform of $f(t)$" is a projection operator, onto some subset, $F_{\text{nice}}$, of the space of functions.
  • The more times a function is continuously differentiable, the "closer" it is to $F_{\text{nice}}$ and the smaller the discrepancy between finite initial sums of "inverse Fourier transform of the Fourier transform of $f(t)$" and $f$.
  • At jump discontinuities, the function isn't even zero times continuously differentiable, so we can expect larger changes in the projection.

These changes caused by a jump discontinuity, at $x = x_0$, say, are:

  • The projection function outputs the midpoint of the jump at $x_0$.
  • The projection function overshoots the height of the jump from both sides in an oscillatory manner.

For your function, \begin{align*} \lim_{t \rightarrow 0^-} f(t) &= 0 \text{ and } \\ \lim_{t \rightarrow 0^+} f(t) &= 1 \text{.} \end{align*} So we expect the inverse Fourier transform of the Fourier transform of $f$ to be a function having $f(0) = \frac{0+1}{2} = 1/2$.

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