Define The Improper Integral in Fourier Transform Fourier Transform of $\mathcal{L_1}$-functions is defined as follows:
$\hat{f}(\omega)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)\cdot e^{i\omega x} dx $.
I can think of three ways to define an Improper Integral of the form $\int _{-\infty}^{\infty}$:


*

*Choose arbitrary $x\in \mathbb R$ and split $\int_{-\infty}^{x}  + \int_{x}^{\infty} $

*$\lim _{n\to \infty} \int_{-n}^{n}$

*$\lim _{n\to \infty, m\to -\infty } \int _{m}^{n}$   
Outside the context of fourier analysis, i'm using definition 3 for improper integrals of this form. 1 is freely used in calculations of fourier transforms, and 2 is used in the fourier inversion theorem (At least at some of it's formulations).
How is the integral in fourier transform defined? Perhaps the methods above are equivalent for $\mathcal L_1$-functions?
Thanks in advance!
 A: I would simply define the Fourier transform $\int_{-\infty}^\infty f(x) e^{i \omega x}$ to be the Lebesgue integral of $f(x) e^{i\omega x}$ over the whole of $\mathbb R$. Since $|f(x) e^{i\omega x}| = |f(x)|$, and since $f(x)$ is in $\mathcal L_1$ by assumption, we can be sure that this integral is well-defined.
My suggestion is equivalent to all of your suggestions (1), (2) and (3).


*

*Since $f(x) e^{i \omega x} = f(x) e^{i \omega x} \chi_{(-\infty, a]} + f(x) e^{i \omega x} \chi_{[a, \infty)}$ for any fixed $a \in \mathbb R$ (where $\chi_A$ always denotes the indicator function on $A \subset \mathbb R$), we immediately get
$$ \int_{-\infty}^\infty f(x) e^{i \omega x} = \int_{-\infty}^a f(x) e^{i \omega x} + \int_{a}^\infty f(x) e^{i \omega x}.$$
So my suggestion is the same as your suggestion number (1).

*Next, observe that $f(x)e^{i\omega x} \chi_{(-n,n)}$ is dominated by $|f(x)|$, which is integrable by assumption. By the dominated convergence theorem, we learn that 
$$\int_{-\infty}^\infty f(x) e^{i \omega x} = \lim_{n \to \infty} \int_{-n}^n f(x) e^{i \omega x}$$
So my suggestion is the same as your suggestion number (2).

*We can also apply the dominated convergence theorem to the sequence of functions $f(x) e^{i\omega x} \chi_{(-m,n)}$ with $n$ fixed and $m$ varying in $ \mathbb N$ to show that $$\lim_{m \to - \infty} \int_{m}^n f(x) e^{i\omega x} = \int_{-\infty}^n f(x) e^{i \omega x}.$$
Applying dominated convergence a second time, using the sequence of functions $f(x) e^{i\omega x} \chi_{(-\infty,n)}$, with $n \in \mathbb N$, we learn that $$\lim_{n \to \infty} \left( \lim_{m \to - \infty} \int_{m}^n f(x) e^{i\omega x} \right) = \lim_{n \to \infty}  \int_{-\infty}^n f(x) e^{i \omega x} = \int_{-\infty}^{\infty} f(x) e^{i \omega x}.$$
So my suggestion is the same as your suggestion number (3).
