Difference between Fourier integral and Fourier transform 
What is the difference between Fourier integral and Fourier transform?

I know that for Fourier integral, the function must satisfy
$\int_{-\infty}^\infty  |f(t)| dt < \infty$, but what if I have a function that satisfies this condition: what does it mean to calculate Fourier transform and Fourier integral?


*

*Fourier Integral: 
$$f(t) =  \int_0^ \infty  A(\omega)\cos( \omega t) + B( \omega )\sin( \omega t)  d \omega$$
where
$A(\omega ) =  \frac{1}{\pi}   \int_{- \infty }^\infty f(t)\cos( \omega t) dt$ and $B(\omega ) = \frac{1}{\pi}  \int_{- \infty }^ \infty  f(t) \sin( \omega t) dt$

*Fourier Transform: 
$$F(\omega ) =  \frac{1}{2 \pi }   \int_{- \infty }^\infty  f(t) e^{-i \omega t}  dt$$

Fourier transform of a function $f$ is the function $\mathscr{F}f$ defined by
\begin{eqnarray}
\mathscr{F}f(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt \ .
\end{eqnarray}
Fourier integral is any integral of the form
\begin{eqnarray}
\int_{-\infty}^\infty y(\omega) e^{i\omega t} d\omega \ .
\end{eqnarray}
Fourier integral of a function $f$ is any Fourier integral, that satisfies
\begin{eqnarray}
x(t) = \int_{-\infty}^\infty y(\omega) e^{i \omega t} d\omega \ .
\end{eqnarray}
You can choose $y=\mathscr{F}x$ to find a suitable $y$.
The Fourier transform is usually defined with an expression such that it has to exist everywhere. Also the Fourier integral have to exist everywhere if we want the Fourier inversion theorem to be true. For simplicity this is usually shown using the assumption $\mathscr{F}f \in L^1$.
The Fourier integral
\begin{equation}
x(t) = \lim_{M \rightarrow \infty} \int_{-M}^M y(\omega) e^{i\omega t} d\omega
\end{equation}
has the advantage that it converges everywhere if $y=\mathscr{F}x$ and $x(t) = \mu(t) e^{-t}$, where $\mu(t)$ is the step function. However, the integral converges to the value $\frac{1}{2}(f(x+)+f(x-))$ at $0$, where $f(x+) = \lim_{t\rightarrow x^+} f(t)$ and $f(x-) = \lim_{t \rightarrow x-} f(t)$. The form has also another advantage: the residue theorem can be easily applied. However the integral is a Cauchy principal value integral and not necessarily for example an improper Riemann-integral of the first kind.
If you want to be able to transform the function $f(x) = \bigg\{ \begin{eqnarray} & \frac{\sin(x)}{x} & \ , x \neq 0 \\
& 1 & , \ x = 0 \end{eqnarray}$, you may want to write also the Fourier transform in the Cauchy principal value form. This however is not equivalent to the original definition, but an extension of it. The same result as in the extension can be also achieved using cosine and sine transforms and the definition
\begin{eqnarray}
\mathscr{F} = \mathscr{F}_c + i \mathscr{F}_s \ ,
\end{eqnarray}
where
\begin{eqnarray}
\mathscr{F}_c f(\omega) & = & \frac{2}{\pi} \int_0^\infty P_e f(t) \cos(\omega t) d\omega \\
\mathscr{F}_s f(\omega) & = & \frac{2}{\pi} \int_0^\infty P_o f(t) \sin(\omega t) d\omega \ ,
\end{eqnarray}
where
\begin{eqnarray}
P_e f(t) & = & \frac{1}{2}(f(t)+f(-t)) \\
P_o f(t) & = & \frac{1}{2}(f(t)-f(-t)) \ .
\end{eqnarray}
