Let $G$ be a locally compact abelian group and $f:G \to \mathbb{C}$ a function. Its Fourier transform (when it exists) is defined to be
$$\widehat{f}(\chi) = \int\_G f(g) \bar{\chi}(g) \mathrm{d} g,$$
Here $\chi \in \widehat{G} := \mathrm{Hom}(G, S^1)$ and $\mathrm{d} g$ is a choice of Haar measure on $G$.


Why does $\bar{\chi}$ appear and not $\chi$? Is this for historical reasons or is there a mathematical reason?


In practice one considers all characters $\chi$ at once (for example when performing Poisson summation), so it doesn't really make a difference including the complex conjugate in the definition.



Why complex conjugate in definition of the Fourier transform?
This is an $L^2$ product: consider for instance the most classical Fourier expansion of $\mathbb Z$ periodic functions (or distributions) of one real variable.
You have for $f$ locally square integrable
$$
f(x)=\sum\_{k\in \mathbb Z}\langle f, e\_k\rangle\_{L^2} e\_k(x),
$$
that is
$
f(x)=\sum\_{k\in \mathbb Z}\left(\int\_0^1 f(y){e^{-2iπ ky}} dy\right) e^{2iπ kx},
$
simply since
$
\langle f, e\_k\rangle\_{L^2}=\int\_0^1 f(y)\overline{e\_k(y)}dy.
$


