Obtaining the $\frac{1}{2\pi}$ factor in the Fourier transform This MathWorld page gives this definition of a Fourier transform: $$F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x}dx.$$  But, I wish to speak in terms of linear frequency $\nu$ and time $t$ rather than in terms of wavenumber $k$ and position $x$, so I will use the substitutions $k \rightarrow \nu$ and $x \rightarrow t$ rewrite this as: $$F(\nu) = \int_{-\infty}^{\infty} f(t) e^{-i2\pi \nu t}dt \; \; \; \text{(eq. 1).}$$  
Is this substitution valid, or did I miss a factor?
Now, angular frequency $\omega$ and linear frequency $\nu$ are related by $\omega = 2 \pi \nu$ so I can rewrite in terms of the angular frequency $\omega$: $$F({\omega\over2\pi}) = F(\nu) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t}dt \; \; \; \text{(eq. 2).}$$  However, my physics professor's distributed notes give this definition of the Fourier transform $F(\omega)$ of $f(t)$: $$F(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-i\omega t}dt \; \; \; \text{(eq. 3)}$$
How can I convert equation (2) to equation (3), to obtain the $\frac{1}{2\pi}$ factor?
 A: The answer is, as always, in wikipedia. In your derivation, the factor that is not coming up in the direct transformation, will show up in the inverse one. In your professor's, the factor that is in the direct transform will not be present in the inverse one. As long as you define them matching each other, you will be ok. It is also pretty common to split the difference and have a $1/\sqrt{2\pi}$ in front of each.
A: I have recently been thinking a lot about this.  You can say it is a question of taste, but why not think about taste, eh? 
First, in a nutshell, you cannot convert one equation to the other because the $F$ in one is numerically different from the other. Wikipedia is using the most natural, the most symmetrical, the most «unitary» conventions, as I will explain later.  But to transform to what your professor has, first do what you did, which is fine.  Then you have to realise the prof is using a different definition of F which is not equal to other people's.  What Jaime says is also correct.
Now for the in depth discussion.
Physically, the Fourier expansion (for example, of a periodic function) should not have any numerical factors in it since it is supposed to be an expansion in terms of a basis.  That is,
$$ \cos (t) = 0 + 0\sin t + 0\sin 2t + \dots + 1\cos t + 0\cos 2t + \dots .$$
Laurent Schwartz, for me, is the one who influenced a whole generation of mathematicians to say, well, the most basic period is period 1.  So the basis functions must be
$\cos 2\pi t$ etc. Okay, really he, like most scientists and engineers after the war, used complex exponentials.  So the basis for complex-valued functions on the real line with unit periodicity are $e^{i 2\pi nt}$ with $n$ an integer and $i$ a fixed choice of a square root of unity.  Don't use the unit circle since we want to allow $t$ to be time.
This will be an orthonormal basis if you choose the measure to be Lebesgue measure on the unit interval, again, the most natural choice.  (Forget the unit circle, the unit interval with total measure one is the most natural.  If you are going to use the unit circle, okay, as long as you use a measure with total mass unity.)
The other advantage here is that $n$ is the natural frequency $f$ rather than the angular frequency.  This is good for statisticians or anyone else who uses Hz or cps.  Angular frequency really is less natural.  I have even seen an author who changes the definition of $e$ to get rid of the factor of $2\pi$ in order to make the appearance of these formulae as natural as their reality.  One could alter $i$ by a scalar, too...
So, my point is, firstly, the inversion formula or expansion formula should not have any constants.  Secondly, it is possible to pick a very natural, «unitary» set of basis functions and frequencies.  This forces where the two and the pi go.
Now the Fourier transform is forced on you, i.e., the Fourier transform is how you get the coefficients from the given function, and the inverse transform is how you get the function from the coefficients, and these do not live on the same space.  It is good to make this distinction as clear as possible, since it crops up again for Lie Groups.  (It gets a little blurred for the real line.)
The Fourier transform really takes the easiest possible form, since the basis is orthonormal,
$$\hat x (n) = \int_0^1 x(t) e^{-i2\pi nt} dt$$
as we merely have to take the Hermitian inner product of $f$ with each basis element in order to get the expansion coefficient. (The negative $i$ is because it is the Hermitian inner product, not a real symmetric inner product.)
For the  case of a non-periodic function $x(t)$, the analogues look the same.  The transform to get the coefficients for each frequency $f$ is
$$\hat x (f) = \int_{-\infty}^\infty x(t) e^{-i2\pi ft} dt$$
and the inverse transform, that gives the expansion of $x$ in terms of each frequency, is
$$ x(t) = \int_{-\infty}^\infty \hat x(f) e^{i2\pi ft} df$$
and everything is easy to memorise: no coefficients in the expansion, unit periodicity, unit measure, reverse the sign of $i$ in the transform formula since it is a Hermitian form.
If for any reason you have to change units, like, days to years, since it's not periodic daily but only yearly, just change variables $t$ goes to $t'$ consistently in every formula but realise that now the frequency means in years, so of course what you now call $\hat x(f)$ is no longer equal to what you earlier called $\hat x(f)$ because $f$ means something quite different due to the change in units. $x(t)$ is the same, and the Fourier transform still means the same thing as before but its numerical value has changed since you changed the units.  Now that is exactly what your professor did: the change was from $dt$ to $\frac1{2\pi}dt$.  Unnatural, in a way, but there it is.  People who prefer angular frequency tend to do this.  It is analogous to a change of time units and has a similar effect on the formulas.
If you change the definition of the transform, you have to change the definition of the inverse transform.  In order to be completely clear, let's make this rival definition of the Fourier transform use a squiggly instead of a hat:
$$\tilde x(f) = \int_{-\infty}^\infty x(t) e^{-i ft} {dt\over 2\pi}.$$
What is the relationship of  $\tilde x$ to  $\hat x$?.  First, you must change the $e^{-i ft}$, which is appropriate when $f$ means angular frequency, to match the $e^{-i 2\pi ft}$ we use in our definition, which we use because it is normal frequency.
BTW, many people use the convention of capitalising the name of the function, $x$, to get its Fourier transform, $X$.  But statisticians can't do that since we already have the convention that lower case is a sample signal, and upper case is for the random variable or stochastic process.  So, like mathematicians but for the opposite reason, we prefer the hat or squiggle notation.
So change $f$ in the integral: write $\omega= {2\pi}f$.  (You could, instead, change $t$.  As I say, changing the $f$ to denote angular frequency is a lot like changing the time units.)  Then
$$\hat x({\omega\over2\pi}) = \int_{-\infty}^\infty x(t) e^{-i \omega t} dt.$$
But the right hand side is, in turn, equal to $2\pi\tilde x(\omega)$.
In short, $$\hat x(f) =  2\pi\tilde x(\omega),$$ where the relation between angular frequency and normal frequency has to be kept in mind, 
$$\omega= {2\pi}f,$$ i.e., $${\omega\over 2\pi}=f$$ and so $df = {d\omega\over2\pi}$. 
Therefore, substituting into the inversion formula we would use for $\hat x$, we get the appropriate one to use for the professor's definition:
$$x(t) = \int_{-\infty}^\infty 2\pi\tilde x(\omega) e^{i2\pi ft} df$$
$$= 2\pi\int_{-\infty}^\infty \tilde x(\omega) e^{i\omega t} {d\omega\over2\pi}= \int_{-\infty}^\infty \tilde x(\omega) e^{i\omega t} d\omega.$$
What has caused all the trouble is your professor's using $f$ for angular frequency.  Consistenly keeping track of the physical meaning of the different units actually helps one understand the numerical factors occurring in the Fourier transform and expansion formulas.  So now we, reluctantly, agree to write $f$ instead of $\omega$ for angular frequency in order to agree with your professor.  And we get...no, no, I won't do it...
A: As @jaime mentioned, the factor of $\frac{1}{2\pi}$ does show up when you derive the inverse relation. The inverse relation in terms of $f$ is given by:
$$
g(t)\ =\ \int_{-\infty}^\infty G(f) e^{j2\pi ft} df
$$ 
To convert this in terms of $\omega$, we need to do a substitution $ f = \frac{w}{2\pi}$ ; which implies $df = \frac{dw}{2\pi}$. Now substituting it into the above equation we get.
$$
g(t)\ =\ \int_{-\infty}^{\infty} G(\frac{w}{2\pi}) e^{jwt} (\frac{dw}{2\pi})
$$
We simply replace $G(\frac{w}{2\pi})$ by $G(w)$ which gives you the following result:
$$
g(t)\ =\ \frac{1}{2\pi}\int_{-\infty}^{\infty} G(w) e^{jwt} dw
$$
I have come across both the frequency$(f)$ and angular frequency $(w)$ forms in different texts but never the one your professor mentions.
