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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?

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Ryan
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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}

Disintegrating By Parts
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    Forgive me if this is a stupid question. I think I understand every jump in logic, except for at the very first step. How is the LHS equal to the RHS? – Ryan Sep 02 '15 at 00:42
  • @Ryan I started with your assumption in the yellow box, multiplied both sides by $e^{-ixl}$, and integrated over $[-R,R]$ with respect to $x$. The order of integration was then swapped on the right side. – Disintegrating By Parts Sep 02 '15 at 01:12