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?

  • 1
    $\begingroup$ Are they totally consistent with standard $\int (\mbox{integrand})\ dx$ through the rest of the book? I have seen physicists write $\int dx\ (\mbox{integrand})$ before. If that's the case, then they just changed the order of integration. $\endgroup$
    – Neal
    Jan 23, 2015 at 1:41
  • $\begingroup$ Should you have i in the exponential instead of j? Also, if you want to put a hat symbol on top of the f, to denote the Fourier transform of f, use \hat{f} : $\hat{f}$ $\endgroup$
    – Clyde
    Jan 23, 2015 at 1:42
  • $\begingroup$ I also suspect exactly what Neal has implied. Since this is a math forum, FYI, Fubini's theorem, en.wikipedia.org/wiki/Fubini%27s_theorem, allows you to switch the order of integration $\endgroup$
    – Clyde
    Jan 23, 2015 at 1:44
  • $\begingroup$ That should be $dx'$ in the first integral. $\endgroup$ Jan 23, 2015 at 1:49
  • $\begingroup$ There's a lot of notation error in the above. It is not true that $f(k)=\dots$ in the second integral, either, but rather $\hat{f}(k)=\dots$, and the next integral also should be $\hat{f}(k)$ inside the integral. $\endgroup$ Jan 23, 2015 at 1:52

1 Answer 1


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:


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 $\delta(x)$.

  • $\begingroup$ In Eq. 3, you have the integrand transposed with the differential, is this just a matter of style as Neal suggested in the comments of my question? $\endgroup$
    – kypalmer
    Jan 23, 2015 at 2:20
  • $\begingroup$ Yeah, that's irrelevant. I just copied to the expression directly from your proof, so I didn't even notice, but that order is irrelevant. Was just watching MIT QM lectures the other day where the prof kept writing $\int dx f(x)$. Makes the variable of integration clearer. $\endgroup$ Jan 23, 2015 at 2:22

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