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 $F^{-1}$.