# Question regarding a non-rigourous proof that the Fourier transform of $1$ is the Dirac-delta function

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 $$(2)$$

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

then

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

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

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): $$F(2) = \int f(2,y) \, dy$$, $$F(3z+5) = \int f(3z+5, y) \, dy$$, $$F(-x) = \int f(-x,y) \, dy$$, $$F(2x) = \int f(2x,y) \, dy$$.

• 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. – BLAZE Nov 5 '19 at 18:17