How is the exponential in the Fourier transform pulled out of the integrand? I'm looking at Fourier Transforms in a Quantum Physics sense, and it's useful to associate the Fourier Series with the Dirac Delta. The book I'm using follows this argument (Shankar, Quantum Mechanics):
The Dirac Delta has the following property:
$\int \delta(x-x^{\prime})f(x^{\prime}) \,\,\mathrm{d}x^{\prime} = f(x)$
We can represent a function by it's transform in the frequency domain:
$\hat{f}(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x)\,\, \mathrm dx$
and can also perform an inverse transform:
$f(x^{\prime})=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \hat{f}(k)\,\,\mathrm{d} k$
Substituting the first transform in the second:
$f(x^{\prime})=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x) \,\,\mathrm{d} x  \right) \,\,\mathrm dk$
Now, in the discussion that I'm reading, this is rearranged as:
$f(x^{\prime})=\int_{-\infty}^{\infty} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathrm{d} k\,\, e^{jk(x^{\prime}-x)}\right)f(x) \,\,\mathrm dx$
Which, by comparison with the first function allows us to associate the parenthesized part of this equation with the dirac delta. 
My Question:
Why can we remove the integrand from the inside of the dk integral, since it clearly depends on k, and $x-x^{\prime}$ is not necessarily zero?
 A: The steps missing:
$$\begin{align}f(x^{\prime})&=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x) \,\,\mathrm{d} x  \right) \,\,\mathrm dk\tag{0}\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dx\,dk\tag{1}\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dk\,dx\tag{2}\\
&=\int_{-\infty}^{\infty} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathrm{d} k\,\, e^{jk(x^{\prime}-x)}\right)f(x) \,\,\mathrm dx\tag{3}
\end{align}$$
The original proof jumped from (0) to (3).
(1) is bringing the $e^{jkx'}$into the integral, which you can do by distributive law.
(2) is a switching in the order of integration. 
(3) is pulling out the $f(x)$ and $\frac{1}{2\pi}$ out, because it is a constant in the inner integral:
$$\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dk$$
None of this is valid, mathematically, but it all can be made rigorous by doing the work in "distribution theory." Most physicists don't give a damn about that part, though, because they are mostly dealing with wave functions that are "close enough to" $\delta(x)$, not actually $\delta(x)$.
