Imaginary exponent of Fourier transforms

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.

• The reason is that you lose unitarity when you do this. – Cameron Williams Apr 14 '16 at 2:32
• That's essentially a (bilateral) Laplace transform. One issue is convergence of the integral, because $e^{-tx}$ is growing exponentially when $t$ and $x$ have opposite signs. – carmichael561 Apr 14 '16 at 2:34
• if $t \in \mathbb{R}$ : it's called the real (axis) Laplace transform, it is not always invertible, and not orthogonal at all, in contrary to the usual Fourier/Laplace transform : hence it is much more complicated. – reuns Apr 14 '16 at 10:36
• @carmichael561 : I don't agree at all for separating the bilateral and the unilateral Laplace transform, why not only talk of the bilateral Laplace transform of $f(x)$ and of $f(x) 1_{x \ge 0}$ ? – reuns Apr 14 '16 at 10:42

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.
• This question has prompted me to think about the functions the bilateral Laplace Transform would converge for. It seems that it would converge for $L^1_{loc}$ super-exponentially decaying functions, but I can't quite convince myself of this. You are right that the Schwarz functions are an issue, but I am curious there is a relatively simple classification of functions for which this does converge. – Nicholas Stull Apr 14 '16 at 2:47
• @NicholasStull : the Laplace transform of $f(x) = e^{- |x|}$ converges in the strip $|Re(s)| < 1$ – reuns Apr 14 '16 at 10:43