Negative exponent Fourier Transform of Sequences Why the exponent must be a negative in the Fourier transform of any sequence?
What happens with expressions 
$$x(m)=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}X(w)\exp(jmw)dw$$
if we define the Fourier transform of sequences as:
$$X(w)=\sum_{-\infty}^{\infty}x(m)\exp(jmw)$$ tal que $-\pi\leq w \leq \pi.$
recall that Fourier Transfrom of sequence $x(n)$ is $X(w)=\sum_{-\infty}^{\infty}x(n)\exp(-jnw)$
I am found that $x(m) = x(-m)$ this is true?
 A: The reason comes from what we want to end product to be: an approximation to the given function $x$ in the form of a trigonometric polynomial: $$x(t)\sim \sum c_k \exp(ikt) \tag1$$ 
The formula for coefficients $c_k$ is derived from (1), and necessarily involves the minus sign. In other words, we have the minus sign in the transform formula because we don't want to have it in  (1).
That said, there is no real  difference between $i$ and $-i$ -- the difference is  purely imaginary :). If you go through all Fourier analysis formulas changing the sign of $i$ everywhere, all statements will remain true.
As for your other question: the Fourier coefficients of a real-variable function $f$ satisfy $c_{-k}=\overline{c_k}$, as you can find by conjugating the integral that gives $c_k$.
A: Because Fourier wanted to make a convergent transform. If a function is convergent then it has a lot of nice proprieties very useful in math. And if you make the exponent positive, then the integral would not be convergent and you can't find those proprieties that Fourier found.
