Fourier transform convention: $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)e^{\pm ikx}dx $?

Question

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?

Answer 1

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.

Answer 2

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.)