My teacher wrote on the board for Fourier Transforms for $f(x)$ from $-\infty<x<\infty$:
$$F(\mu)=\int_{-\infty}^{\infty}f(x)e^{i\mu x}dx$$ and for the inverse transform $$f(x)=\int_{-\infty}^{\infty}F(\mu)e^{-i\mu x}\frac{d\mu}{2\pi}$$
Notice where I have the negatives in the exponents... My confusion is that I see the following all over online:
$$F(\mu)=\int_{-\infty}^{\infty}f(x)e^{-i\mu x}dx$$ and for the inverse transform $$f(x)=\int_{-\infty}^{\infty}F(\mu)e^{i\mu x}\frac{d\mu}{2\pi}$$ with the non-negative in the inverse transform. But I can't find the version that my teacher wrote.
1) I am trying to figure out if they are equivalent somehow or if there was a mistake in class.
2) Also, if there was not a mistake, would both forms give the same answer in the end of problems? Or is there some kind of change of variable that happens?
Thank you.
The two forms are equivalent.
But the second form is indeed the preferred one at 99.9 %. I see at least two reasons for that :
$$\langle f,g \rangle=\int_{-\infty}^{\infty}f(x)\overline{g(x)}dx$$ where $g(x):=e^{i\mu x}$.