If $G=\mathbb {R}$, how $\hat{f}(\chi)=\int_{G}f(x)\overline{\chi (x)}dx$ becomes $\hat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-ix\xi}dx$?

Question

The dual group of a locally compact Abelian group is used as the underlying space for an abstract version of the Fourier transform. If a function $f$ is in $L^{1}(G)$, then the Fourier transform is the function $\hat{f}$ on $\hat{G}$ defined by

$\hat{f}(\chi)=\int_{G}f(x)\overline{\chi (x)}dx$

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if $G=\mathbb{R}$, then how can I prove that the Fourier transform is of the form below?

$\hat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-ix\xi}dx$

Answer 1

As it was said in the comments, $$\Psi : G=\Bbb R \to \hat{G} = \{\chi_{\xi} \,\colon x \mapsto e^{ix\xi} \mid \xi \in \mathbb{R}\}, \xi \mapsto \chi_{_{\xi}}$$ is a group isomorphism.

Then define $$\hat f(\xi) := \hat f(\Psi(\xi)) = \hat f(\chi_{\xi}) = \int_{G}f(x)\overline{\chi_{\xi} (x)} dx$$

We get $$\hat f(\xi) = \int_{G}f(x)\overline{\chi_{\xi} (x)}dx = \int_{G}f(x)\overline{e^{ix\xi}}dx = \int_{G}f(x)e^{-ix\xi}dx$$

as desired.