Why does Fourier Transform use a negative exponent in its formula? I realize this question has been asked before but I don't understand the explanations. For example I have read this (https://en.wikipedia.org/wiki/Negative_frequency) as well as numerous other answers. 
I understand e^-jwt is simply a point going clockwise in the complex domain. What I don't understand is why you multiply a signal x(t) with a clockwise complex number. 
What would be the result of let's say Fourier Transform using the positive exponent?  
I guess one theory I had is they define shifted impulses using delta[t-to] and the Fourier transfform of that is going to be e^-jwto. We think of things shifted after time 0 rather than before, so is this why we have a negative exponent? If I'm wrong, please provide clear sequential steps. 
 A: As far as I know, there is no special meaning to the negative sign. It is just a convention. In fact, we could have defined the Fourier transform with the positive sign. All interesting properties would be kept, or transformed with symmetries.
There is a bit of niceness in using the negative sign because then the Inversion Formula has a positive sign:
$$f(t)=\int_{\mathbb R} \hat f(\xi)e^{2i\pi  \xi t} d\xi$$
which shows the signal $f$ is a sum of complex exponentials. It is more natural (again, just a more natural convention) to look at exponentials with a positive sign in it.
A: I don't know how much you know about function analysis, but the set of all complex-valued squared integrable functions on $[-\pi,\pi]$, $L^2(-\pi,\pi)$, behaves like a vector space. In fact, it is an infinite dimensional vector space. Given $\mathbb{C}^k$ and an orthonormal basis $\{u_n\}_{n=1}^k$, we can write any vector $v \in \mathbb{R}^k$  in terms of that basis like so: $v=\sum_{n=1}^k \langle v,u_n\rangle u_n$, where $\langle a,b\rangle =\sum_{n=1}^k a_n\overline{b_n}$ for $a,b \in \mathbb{C}^k$. I'm sorry if I'm talking about stuff you haven't learned, but this is really just vectors, orthonormal bases, and the dot product but for $\mathbb{C}^k$ instead of $\mathbb{R}^k$. Similarly, $L^2(-\pi,\pi)$ undergoes the same rule. Given an orthonormal "basis" $\{\phi_n\}_{n=1}^\infty $, we can write any function $f \in L^2(-\pi,\pi)$ as $f=\sum_{n=1}^\infty \langle f, \phi_n\rangle \phi_n$, where $\langle f, g\rangle =\int_{-\pi}^\pi f(x)\overline{g(x)}dx$. As it turns out, $\left\{\frac{1}{\sqrt{2\pi}}e^{in\pi}\right\}_{n\in \mathbb{Z}}$ is indeed an orthonormal basis. Then given $f \in L^2(-\pi,\pi)$ , we can write $f(x)=\sum_{n \in \mathbb{Z}}\langle f, \frac{1}{\sqrt{2\pi}}e^{in\pi}\rangle \frac{1}{\sqrt{2\pi}}e^{in\pi}=\sum_{n \in \mathbb{Z}}\left(\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-in\pi}dx\right)e^{in\pi} =\sum_{n \in \mathbb{Z}}c_ne^{in\pi}$. Note the negative in the exponential in the integrand comes from conjugating $\frac{1}{\sqrt{2\pi}}e^{in\pi}$. Also note $\left\{\frac{1}{\sqrt{2\pi}}e^{-in\pi}\right\}_{n\in \mathbb{Z}}$ is an orthonormal basis. We can then write $f(x)=\sum_{n \in \mathbb{Z}}\langle f, \frac{1}{\sqrt{2\pi}}e^{-in\pi}\rangle \frac{1}{\sqrt{2\pi}}e^{-in\pi}=\sum_{n \in \mathbb{Z}}\left(\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{in\pi}dx\right)e^{-in\pi}$. Notice how I didn't write a negative in the exponent in the integrand, since the conjugation removes it. 
You mentioned the negative simplifies the math. Honestly, I can't say. It doesn't make a whole lotta difference. What matters at the end of the day is you're writing a function in $L^2(-\pi,\pi)$ in terms of an orthonormal basis.
A: Here's an intuitive viewpoint that I suspect could be made rigorous by someone who knows enough functional analysis:
The Fourier transform expresses a function $f$ as a linear combination of Fourier basis functions $e^{i \omega x}$, as follows: 
$$
\tag{1} f(x) = \int c(\omega) e^{i \omega x} \, dx.
$$ 
This is the main idea and there is no minus sign yet.
To solve for the coefficient $c(\omega_0)$, use Fourier's trick: take the inner product of both sides with $e^{i \omega_0 x}$ and observe the wonderful cancellation that occurs (thanks to orthogonality of the Fourier basis functions). The minus sign appears only because that's how the inner product is defined. 
What would actually be weird is if we included a minus sign in the exponent in equation (1). Given that equation (1) is our goal, the minus sign in the formula for $c(\omega)$ is just a result of using Fourier's trick.
