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Sometimes the way physicists/engineers do math really drives me crazy.

If you try to search, as I did, thinking I saw a proof once, for why the continuous fourier transform and its inverse

F[g](\phi) = {1\over\sqrt{2\pi}} \int_{-\infty}^{\infty} g(t) e^{-i t \phi} dt

F^{-1}[G](t) = {1\over\sqrt{2\pi}} \int_{-\infty}^{\infty} g(t) e^{i t \phi} d\phi

need those normalization factors 1\over\sqrt{2\pi}, you get a wide range of non-answers that are mostly about the choice of shunting that 2pi around asymetrically to one side or the other, or the fact that you can make it vanish by stuffing it in the e^{-2 \pi i t \phi}... without ever explaining why it's there in the first place, or showing why changing the wave you're integrating against affects it the way it does.

So I had to sit down and actually crunch out some good ol' leibniz sums starting from the discrete fourier transform

DF[f] = \lambda s. {1\over\sqrt{2M}} \sum_{n = -M}^{M-1} f(n)e^{-\pi ins/M}

DF^{-1}[F] = \lambda n . {1\over\sqrt{2M}}\sum_{s = -M}^{M-1} F(s)e^{\pi ins /M}

for which you absolutely can check that the normalization constants are correct, since you get 2M copies of 1 as the frequencies exactly cancel when you do the computation of DF o DF^{-1} or DF^{-1} o DF.

Following that, cleverly choosing A = \sqrt{M / 2k} and d = 1 / \sqrt{2kM} means that you can take an integral like

\int_{-A}^{A} g(t) e^{-2 \pi k i t \phi} dt

and Leibniz it up into

\sum_{n = -M}^{M-1} d g(dn) e^{-2\pi k i d n \phi}

and with some algebra you can see that

F[g](\phi) = \sqrt{k} \int_{-A}^{A} g(t) e^{-2 \pi k i t \phi} dt

F^{-1}[G](t) = \sqrt{k} \int_{-A}^{A} g(t) e^{2 \pi k i t \phi} d\phi

are approximated by

\lambda \phi. DF[\lambda n . g(dn)](\phi/d)

\lambda t . DF^{-1}[\lambda s . G(ds)](t/d)

and these are easily seen to be inverses of each other.

This isn't a proof of course, I just wave my hands and say "pick k to match whatever convention you like, let M be really big, then d is small and A is big, and therefore it's an integral from -infinity to infinity" but at least it's an argument, not just "well you need a 2pi there because the book says so".

If you try to search, as I did, thinking I saw a proof once, for why the continuous fourier transform and its inverse

F[g](\phi) = {1\over\sqrt{2\pi}} \int_{-\infty}^{\infty} g(t) e^{-i t \phi} dt

F^{-1}[G](t) = {1\over\sqrt{2\pi}} \int_{-\infty}^{\infty} g(t) e^{i t \phi} d\phi

need those normalization factors 1\over\sqrt{2\pi}, you get a wide range of non-answers that are mostly about the choice of shunting that 2pi around asymetrically to one side or the other, or the fact that you can make it vanish by stuffing it in the e^{-2 \pi i t \phi}... without ever explaining why it's there in the first place, or showing why changing the wave you're integrating against affects it the way it does.

So I had to sit down and actually crunch out some good ol' leibniz sums starting from the discrete fourier transform

DF[f] = \lambda s. {1\over\sqrt{2M}} \sum_{n = -M}^{M-1} f(n)e^{-\pi ins/M}

DF^{-1}[F] = \lambda n . {1\over\sqrt{2M}}\sum_{s = -M}^{M-1} F(s)e^{\pi ins /M}

for which you absolutely can check that the normalization constants are correct, since you get 2M copies of 1 as the frequencies exactly cancel when you do the computation of DF o DF^{-1} or DF^{-1} o DF.

Following that, cleverly choosing A = \sqrt{M / 2k} and d = 1 / \sqrt{2kM} means that you can take an integral like

\int_{-A}^{A} g(t) e^{-2 \pi k i t \phi} dt

and Leibniz it up into

\sum_{n = -M}^{M-1} d g(dn) e^{-2\pi k i d n \phi}

and with some algebra you can see that

F[g](\phi) = \sqrt{k} \int_{-A}^{A} g(t) e^{-2 \pi k i t \phi} dt

F^{-1}[G](t) = \sqrt{k} \int_{-A}^{A} g(t) e^{2 \pi k i t \phi} d\phi

are approximated by

\lambda \phi. DF[\lambda n . g(dn)](\phi/d)

\lambda t . DF^{-1}[\lambda s . G(ds)](t/d)

and these are easily seen to be inverses of each other.

This isn't a proof of course, I just wave my hands and say "pick k to match whatever convention you like, let M be really big, then d is small and A is big, and therefore it's an integral from -infinity to infinity" but at least it's an argument, not just "well you need a 2pi there because the book says so".