Why is this function not ##L^1(\mathbb{R} \times \mathbb{R})##?

Hi everyone in the following expression
##f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega ##

the book says I can't swap integrals bacause the function

##f(u) e^{i \omega(t-u)}## is not ## L^1(\mathbb{R} \times \mathbb{R})##

why ? complex exponential is not always bounded?

What properties does ##f(u)## have?

FactChecker said:

What properties does ##f(u)## have?

it is Fuorier inverse theorem with ##f## in ##L^1(\mathbb{R})##

laurabon said:

Hi everyone in the following expression
##f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega ##

the book says I can't swap integrals bacause the function

##f(u) e^{i \omega(t-u)}## is not ## L^1(\mathbb{R} \times \mathbb{R})##

why ? complex exponential is not always bounded?

Note that ##\sin## and ##\cos## are not integrable over ##\mathbb R##.

PeroK said:

Note that ##\sin## and ##\cos## are not integrable over ##\mathbb R##.

I came up with this . Is correct? ##
\begin{aligned}
& \left|f(u) e^{i \omega(t-u)}\right|=|f(u)| \\
& \iint|f(u)|=\int_{-\infty}^{\infty} c = \infty
\end{aligned}##

how should I use the fact that sin and cosine are not integrable ?I mean to verify that is in L1 I always have to use the fact that ##|e^{i \omega(t-u)}|=1## I think

laurabon said:

I came up with this . Is correct? ##
\begin{aligned}
& \left|f(u) e^{i \omega(t-u)}\right|=|f(u)| \\
& \iint|f(u)|=\int_{-\infty}^{\infty} c = \infty
\end{aligned}##

how should I use the fact that sin and cosine are not integrable ?I mean to verify that is in L1 I always have to use the fact that ##|e^{i \omega(t-u)}|=1## I think

I don't think that works. First, let's take ##t = 0##, where the integral reduces to:
$$f(0) = \int_{-\infty}^{+\infty} \bigg ( \int_{-\infty}^{+\infty}f(u)e^{-iwu}\ du \bigg ) \ dw$$If we try to swap the integrals and look at:$$\int_{-\infty}^{+\infty} f(u) \bigg ( \int_{-\infty}^{+\infty}e^{-iwu}\ dw \bigg ) \ du$$We can see immediately that the inner integral does not converge and does not represent a well-defined function of ##u##. Because neither ##\cos(wu)## nor ##\sin(wu)## is integrable with respect to ##w## on all of ##\mathbb R##.

laurabon said:

Hi everyone in the following expression
##f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega ##

the book says I can't swap integrals bacause the function

##f(u) e^{i \omega(t-u)}## is not ## L^1(\mathbb{R} \times \mathbb{R})##

why ? complex exponential is not always bounded?

Notice f(x)==1 is also bounded. To be able to be in ##L^1(X)## , where ##X## itself is unbounded, your f has to decrease fast-enough.