I'm going to try to derive it here and illustrate any questions I have about it in the process. I tried to understand it from here, and it's where my confusions are based off of. If you wonder why I write something down and then proceed to ask about what I wrote it down, it's because I'm trying to read off from the link above and I don't know why it is the way it is.

$$f(x)= (1/2\pi) \int_{-\infty}^{\infty}\bar f(k) \ e^{ikx} \ dk \tag{1}\label{eq1}$$

Where $f(x)$ and $\bar f(k)$ are Fourier Transform pairs. Define a function, $f(x)^*$, which is the complex conjugate of $f(x)$.

$$\implies f(x)^* = (1/2\pi) \int_{-\infty}^{\infty}\bar f(k')^* \ e^{-ik'x} \ dk' $$

The reason for this, as far as my understanding is concerned, is we're essentially taking the complex conjugate of the RHS of $\eqref{eq1}$. However, I don't see why $k$ is now $k'$. If we're saying $f(x)^*$ is the complex conjugate of $f(x)$, I don't see why we're now using a different dummy variable $k'$, I don't see why this is strictly necessary (Question $1$).

The next confusing part for me is a bit further along. Hopefully you don't mind if I skip a few steps:

$$\int_{-\infty}^{\infty} \lvert f(x) \rvert^2 \ dx = 1/2\pi)\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\bar f(k)e^{-ikx}\ dk \ \int_{-\infty}^{\infty}\bar f(k)^* e^{ik'x} dk'\right) dx \tag{2} \label{2}$$

$$\int_{-\infty}^{\infty} \lvert f(x) \rvert^2 \ dx = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \bar {f(k}) \bar f(k')^* e^{ix(k'-k)}\ dx \ dk' \ dk \tag{3} \label{3}$$

I know how it works from here, but I have a question on going from $\eqref{2}$ to $\eqref{3}$. I'm confused as to how the liberal use of moving around integrals and differentials can be done so liberally. Perhaps I'm rusty from my multivariable calculus class, but it looks like integrands had to have been picked up and moved around, yet I thought anything in the integral that depends on the differential cannot be moved outside it. I know that order of integration is a thing for triple integrals, and that we treat none differential-dependent variables per integral as constant, but still getting to $\eqref{3}$ seems very unnatural to me (Question 2). I suppose in multivariable calculus I'm more used to integrals looking like $\eqref{2}$ since the order is more apparent to me.


I answer your second question first since this is more general:

Your second question: When the author writes, for example, $g(w)^*$ what he really means is $(g(w))^*$. Likewise at other occasions of the $^*$. I.e. we take whatever value $g(w)$ has (for a given $w$) and take the complex conjugate of that value. This also means there is no variable $w^*$ or $(w)^*$ introduced.

Your first question: in line (3), the author missed to say that he will introduce the Inverse Fourier Transformations of $f(t)$ AND $g(t)$. Since the Inverse Fourier Transformations require a different integration variable each, the author cannot use $w$ twice but has to introduce for $g(t)$ a second variable which he calls $w'$. He then has to complex conjugate the full expression for $g(t)$, ie.

$$ (g(t))^* = ( \int \bar g(w') e^{i w' t} dw' )^* = \int (\bar g(w'))^* e^{-i w' t} dw' $$


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