I know this question has been asked a lot and I have already read these questions; Fourier Transform Dirac Delta, Fourier Transform of Dirac Delta Function, The inverse Fourier transform of $1$ is Dirac's Delta and Dirac Delta function inverse Fourier transform.

But in this particular proof shown below, I cannot understand the logic behind a certain step towards the end.

I'm using the following convention for the Fourier transform $$\hat f(k)=\frac{1}{2 \pi}\int_{-\infty}^{\infty}f(x)e^{-i k x}dx\tag{1}$$ and Fourier Integral (inverse transform) $$f(x)=\int_{-\infty}^{\infty}\hat f(k)e^{i k x}dk\tag{2}$$

Now suppose we take $f(x)=\delta(x)$, then by $(1)$ we get

$$\begin{align}\hat f(k)&=\frac{1}{2 \pi}\int_{-\infty}^{\infty}\delta(x)e^{-i k x}dx\\&=\frac{1}{2\pi}\end{align}$$

The last equality came from the sifting property $$\int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a)$$

If we now write $f(x)$ as a Fourier integral using $(2)$

$$f(x)=\delta(x)=\int_{-\infty}^{\infty}\hat f(k)e^{ikx}dk=\frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{i k x}dk$$


$$\delta(x)=\frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{i k x}dk\tag{3}$$

Now suppose we take another function, $g(x)=1$, and take the Fourier transform of it, then by $(1)$

$$\begin{align}\hat g(k)&=\frac{1}{2 \pi}\int_{-\infty}^{\infty}g(x)e^{-i k x}dx\\&=\frac{1}{2 \pi}\int_{-\infty}^{\infty}1 \cdot e^{-i k x}dx\\&=\frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{-i k x}dx\end{align}$$

This $$\hat g(k)=\frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{-i k x}dx\tag{4}$$ has a very similar form to $(3)$

Now it is this point for which I am confused:

My lecturer then states that 'if we replace $x$ with $k$ and $k$ with $x$' in $(3)$ then it follows that $\hat g(k)=\delta(-k)=\delta(k)$. Doing so, I obtain:

$$\delta(k)=\frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{i k x}dx\tag{5}$$

If I now replace $k$ with $-k$ and note that the delta function is even, then, the result follows.

I have 2 questions about this proof:

  1. It is like my lecturer thinks that the $k$ in $(3)$ is just some dummy variable for which you can just change to any another variable. I would understand this if we were writing $$\delta(x)=\frac{1}{2 \pi}\int_{k=-\infty}^{\infty}e^{i k x}dk=\frac{1}{2 \pi}\int_{k'=-\infty}^{\infty}e^{i k' x}dk'$$ But that is not what is going on here, the $x$ and $k$ are being exchanged. Since when can we just interchange $x$ and $k$? Don't we have to worry about the domain of the integral changing also?

  2. This may seem a little trivial, but, why can we simply replace $k$ with $-k$ in $(5)$? It's confusing me as this is an integral, not just some function where we make a change of variables using a substitution.


1 Answer 1


It seems like you give more meaning to variable names than is the case.

  1. $k$ in $(3)$ really is a dummy variable. It doesn't matter if we use $k$, $k'$, $x$, $x'$, $\theta$, $\theta'$, or whatever. If $F(x) = \int f(x,y) \, dy$ then we can equally well write $F(y) = \int f(y,x) \, dx$ or $F(k) = \int f(k,\theta) \, d\theta$.

  2. Again, if $F(x) = \int f(x,y) \, dy$ then we can replace $x$ with any expression (not containing the bound/dummy variable): $F(2) = \int f(2,y) \, dy$, $F(3z+5) = \int f(3z+5, y) \, dy$, $F(-x) = \int f(-x,y) \, dy$, $F(2x) = \int f(2x,y) \, dy$.

  • $\begingroup$ Thank you for your answer, nice explanation. Everything you said makes perfect sense. I'm not sure why I was overthinking it so much. Regards. $\endgroup$
    – BLAZE
    Nov 5, 2019 at 18:17

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