What is the reason behind stating inverse Fourier transform with a limit? The Inverse Fourier Transform theorem I have in my notes states:
Let $f \in L^1 (\mathbb{R}) $ be a piecewise continuous and piecewise differentiable function (the set of points of discontinuity is discrete and $f$ has left and right limit at every point, as well as left and right derivative at every point). Then, for every $t \in \mathbb{R}$ the following formula holds:
$$ \lim_{R \to \infty} \int_{-R}^R \hat{f}(\xi)e^{2 \pi i \xi t} d \xi = \frac{f(t+)+f(t-)}{2} $$
Then it says that it follows from this theorem that if $\hat{f} \in L^1(\mathbb{R})$, then $$\frac{f(t+)+f(t-)}{2} = \int_{-\infty}^{\infty} \hat{f}(\xi)e^{2 \pi i \xi t} d \xi$$
What I can't understand is why the theorem uses limit instead of just writing infinity in the bounds of the integral. Isn't the integral from $- \infty$ to $\infty$ defined as this limit? So, if this limit exists, then we should be able to write it as an integral? Where am I wrong? Does this limit always exist and if it does, why don't we write it as an integral with bounds at infinity?
 A: There is a slight differenced in the notation:
$$ \int_{-\infty}^{\infty} f(x) \,\mathrm{d}x
= \lim_{a\to-\infty} \int_{a}^{c} f(x)\,\mathrm{d}x + \lim_{b\to\infty} \int_{c}^{\infty} f(x)\,\mathrm{d}x, $$
where $c$ can be taken to be any real number.  In this case, both limits must exist in order for the integral to be defined.
On the other hand
$$\lim_{R \to \infty} \int_{-R}^{R} f(x)\,\mathrm{d}x $$
might exist when the previous integral does not.  For example,
$$ \lim_{R \to \infty} \int_{-R}^{R} x \,\mathrm{d}x
= \lim_{R \to \infty} \left[ \frac{1}{x} x^2 \right]_{x=-R}^{R}
= \lim_{R \to \infty} \left[ \frac{1}{2} R^2 - \frac{1}{2} R^2 \right]
= \lim_{R \to \infty} 0
= 0,
$$
while $\int_{-\infty}^{\infty} x \,\mathrm{d}x$ does not exist.
Of course, if $\int_{-\infty}^{\infty} f(x)\,\mathrm{x}$ exists, then it will coincide with the limit of the symmetric integral.  Since you are working in $L^1$, you know that the improper integral exists, and so the two notions reduce to the same thing.  But this is not necessarily an "obvious" result.
