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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?

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You seem to have confused Fourier transforms with Fourier series. –  Thomas Andrews Nov 13 '12 at 15:12
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Why? I'am talking about sequences ... –  juaninf Nov 13 '12 at 15:23
    
So $X$ is a sequence, and $\int_{-\infty}^{\infty}$ is really $\sum_{-\infty}^{\infty}$? –  Thomas Andrews Nov 13 '12 at 15:29
    
sorry integral limits are $\int_{-\pi}^{\pi}$ –  juaninf Nov 13 '12 at 15:30
    
Ah, then in English, we call this expression the "Fourier series" for $X$. While it is related to the Fourier transform, it is usually denoted "Fourier series" to make it a distinct concept. –  Thomas Andrews Nov 13 '12 at 15:33
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2 Answers

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 $c_k$.

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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.

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I am found that $x(m) = x(-m)$ this is true? –  juaninf Nov 13 '12 at 18:40
    
That would make sense for the Laplace transform. But $i$ vs $-i$ has nothing to do with convergence. –  user Aug 20 '13 at 1:25
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