Complex conjugate of Fourier coefficient

Given the (inverse) Fourier transform \(F(x) = \frac{1}{\sqrt{2\pi}} \int dk \space \tilde{F}(k) e^{ikx}\), I will show that the complex conjugate of Fourier coefficient can be expressed as \(\tilde{F}^\ast(k) = \tilde{F}(-k).\)

Motivation

Why do we want this result? This is because you can convert any complex function \(f^\ast(x)\) you obtained from Fourier transform into a real function with a negative parameter, \(f(-x)\). So you don’t need to panic when you get complex quantities in your results.

Original wrong derivation

We take the complex conjugate of Eq. 1 to obtain \(F^\ast(x) = \frac{1}{\sqrt{2\pi}} \int dk \space \tilde{F}^\ast(k) e^{-ikx}.\) Next, treating \(F^\ast(x)\) as just another function of \(x\), and directly do Fourier transform on it to obtain \(\begin{align} F^\ast(x) &= \frac{1}{\sqrt{2\pi}} \int dk' \tilde{F}(k') e^{ik'x}\\ &= \int d(-k) \tilde{F}(-k) e^{-ikx}. \end{align}\) In the second equality sign, I did a change of variable \(k'\rightarrow -k\), motivated by the fact that after this change, the exponential factor looks the same as that in Eq. 2, and we can directly equate Eq. 2 with Eq. 4 to obtain

Why it is wrong

We used a sleight-of-hand: the Fourier transform of \(F^\ast(x)\), since we are treating it as an independnet function, is not necessarily \(\tilde{F}(k')\) itself, but instead some other function \(G(k')\) that is not necessarily related to \(\tilde{F}(k')\).

Updated correct derivation

We take the complex conjugate of Eq. 1 to obtain \(F^\ast(x) = \frac{1}{\sqrt{2\pi}} \int dk\space \tilde{F}^\ast(k) e^{-ikx}\) By the reality condition of $F(x)$, we have \(F(x) = F^\ast(x)\) We change the dummy integration variable for Eq. 1 to get

where for the 2nd equality sign, we replace \(d(-k)\rightarrow dk\) since the integration domain is symmetric (from \(-\infty\) to \(\infty\)).

Combining the above 3 equations, we finally obtain the desired result