I'm reading about Fourier transforms. I'm curious why the imaginary unit is needed in exponent. Why not instead define it as:
$$ \hat f(t)=\int_xe^{-tx}f(x) \, dx $$
I'm looking at the proofs of some of the basic properties and I don't see why the above definition wouldn't suffice.
There's a general framework that the Fourier transform fits into using Pontryagin duality and studying the characters of a locally compact abelian group, such as $\mathbb{R}$. The characters of $\mathbb{R}$ are exactly the maps $x \mapsto e^{itx}$, which is where the complex factor comes from. This has all sorts of wonderful consequences, like the fact that the Fourier transform is unitary and that we have inversion and Plancherel.
Alternatively, for a somewhat silly reason: The integral $\int e^{-tx} f(x) \, dx$ will not converge for too many functions because it is very poorly behaved when $t$ and $x$ have opposite signs. Hence you lose Plancherel (in any sense), together with the fact that $\langle f, g \rangle = \langle \hat{f}, \hat{g} \rangle$, definition on all of $L^2$, and so on. It doesn't even converge for all Schwarz functions, so this is an issue. If you restrict to $x \ge 0$, then you've defined the Laplace transform.