# On the scaling of the Fourier transform integral

Lately, I have been reading about Fourier transforms and integrals and I came across the inverse Fourier transform of a function, say, $$f(w)$$, which was given as

$$F^{-1}(f(w))=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(w)e^{iwt}dw$$

but at other place it was just

$$F^{-1}(f(w))=\int_{-\infty}^{\infty}f(w)e^{iwt}dw$$

i.e., with a missing coefficient of $$\frac{1}{2\pi}$$. Can somebody please explain which one is correct?

However, the Fourier Transform was simply given as

$$F(f(t))=\int_{-\infty}^{\infty}f(t)e^{-iwt}dt$$

As comments say, it's just a convention. The idea is that $$F \circ F^{-1}$$ should be identity. I.e. this should hold: $$F^{-1} \{F\{f\}\}=F\{F^{-1}\{f\}\}=f$$
So, if you write $$F\{f\}(t)=\int_\mathbb{R} f(w)e^{-iwt}dw$$ and $$F^{-1}\{f\}(t)=\int_\mathbb{R} f(s)e^{ist}ds$$, you get: $$F^{-1} \{F\{f\}\}(t)=\int_\mathbb{R} ds \cdot e^{ist}\int_\mathbb{R} dw \cdot e^{-iws}f(w) = \\ \int_\mathbb{R} ds \cdot \int_\mathbb{R} dw \cdot e^{is(t-w)}f(w) = |\theta=t-w|= \\ \int_\mathbb{R} ds \cdot \int_\mathbb{R} d\theta \cdot e^{i\theta s}f(t-\theta) = \int_\mathbb{R} d\theta \cdot f(t - \theta) \int_\mathbb{R} ds \cdot e^{i\theta s}= \\ \int_\mathbb{R} d\theta \cdot f(t-\theta) \cdot 2\pi \delta(\theta) = 2\pi f(t)$$ So as you see $$F^{-1}\{F\{f\}\}=2\pi \cdot f$$, which is something we don't really want to have. You can use $$1/2\pi$$ as a "normalizing constant" in $$F^{-1}$$ to get things working. If you like symmetry you could use $$1/\sqrt{2\pi}$$ in both $$F$$ and $$F^{-1}$$.