Deriving the inverse Fourier transform without knowledge of the form it will take I've run by several proofs of the Fourier inversion theorem.  However, every proof I have come across starts by assuming the form that the inverse transform will take.  For example, Ron Gordon's answer to this question starts with the following assumption:

Now let's assume that the inverse FT may be written as
$$f(x) = A \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-i k x}$$

Suppose I am stupid and I do not know what the inverse Fourier transform should look like.  How do I derive it using my knowledge of the definition of the Fourier transform?
 A: I assume $-ikx$ on the right should be $ikx$. You can use what you have instead, but it is not standard.
\begin{align}
    \int_{-R}^{R}f(x)e^{-ixl}dx 
   & = A\int_{-\infty}^{\infty}\hat{f}(k)\int_{-R}^{R}e^{ix(k-l)}dxdk \\
   & = 2A\int_{-\infty}^{\infty}\hat{f}(k)\frac{\sin(R(k-l))}{k-l}dk
\end{align}
The sinc function
$$
           S_{R}(k-l)=\frac{\sin(R(k-l))}{k-l}
$$
is increasingly peaked as $R\rightarrow\infty$ with a peak value of $R$ at $k-l=0$ (i.e., at $l=k$.) The total integral in $k$ of $S_{R}$ remains constant with a value of $\pi$. And the oscillatory nature of $\sin$ causes cancelling so that, for every $\epsilon > 0$, one has
$$
           \lim_{R\rightarrow\infty}\left(\int_{-\infty}^{l-\epsilon}+\int_{l+\epsilon}^{\infty}\right)\hat{f}(k)\frac{\sin(R(k-l))}{k-l}dk = 0.
$$
Finally, if $\hat{f}$ is smooth at $l$,
\begin{align}
  \lim_{R\rightarrow\infty}\int_{-R}^{R}f(x)e^{-ixl}dx & = \lim_{R\rightarrow\infty}2A\int_{l-\epsilon}^{l+\epsilon}\hat{f}(k)\frac{\sin(R(k-l))}{k-l}dk \\
 & \approx 2A\hat{f}(l)\lim_{R\rightarrow\infty}\int_{l-\epsilon}^{l+\epsilon}\frac{\sin(R(k-l))}{k-l}dk \\
 & = 2A\pi\hat{f}(l)
\end{align}
