Fourier transform convention: $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)e^{\pm ikx}dx $? I've come across the Fourier transform being defined as:
$$\tilde{f}(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)e^{ikx}dx$$
But this convention is not present in the Wikipedia article. The one given there, under "Fourier transform: unitary, angular frequency" has a minus sign in the exponent. Are the two equivalent? Switching variables from $x$ to $-x$ wouldn't work because I would get $f(-x)$. Is it perhaps something to do with the symmetric nature of $e^{ikx}$ if expressed in trigonometric form?
 A: The conventions each have a purpose, and there is a relationship between them all.  A general relation which covers all of the standard conventions is (see here for details)
$$\hat{f}(k) = \sqrt{\frac{|b|}{(2 \pi)^{1-a}}} \int_{-\infty}^{\infty} dx \, f(x) \, e^{i b k x}$$
$$f(x) =  \sqrt{\frac{|b|}{(2 \pi)^{1+a}}} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-i b k x} $$
Note that the "Physics" convention has $(a,b)=(1,1)$, while the "Mathematics" convention has $(a,b)=(1,-1)$. I was also exposed to an "electrical engineering" convention that has $(a,b)=(0,2 \pi)$.
The question of whether they are equivalent is tricky.  Of course they are, but one must be careful in defining the scale of one's frequency space before blindly expecting equality. 
A: This is usually the inverse Fourier transform. The usual conventions on $\mathbb R^n$ are
$$\mathcal Ff(\xi) = \hat{f}(\xi) = (2\pi)^{-\frac n2} \int_{\mathbb R^n} e^{-i\xi\cdot x} f(x) dx$$
and
$$\mathcal F^{-1}g(x) = \check{g}(x) = (2\pi)^{-\frac n2} \int_{\mathbb R^n} e^{i\xi\cdot x} g(\xi) d\xi$$
Note that $\hat{\hat f}(x) = f(-x)$ similar to your idea and thus, the inverse FT and the FT share a lot of properties (such as the convolution thm.)
