# Where is the $1/2$ lost in Fourier transform?

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?

• Think about the domain of the Fourier transform. What kind of object does it act on? Commented Mar 10 at 19:48
• Sorry not get the point. Please... Commented Mar 10 at 19:50
• Do you have an answer to the question I asked? What kind of object does the FT act on? Commented Mar 10 at 19:54
• The time domain but with $u(t)$, right? Or I didn't get what object means? Commented Mar 10 at 20:13
• 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 $47$? Commented Mar 10 at 20:26

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 $$f(0) = \frac{0+1}{2} = 1/2$$.