[enriquto]

You can also eliminate the constant in some of the integral formulas by using đx instead of dx. I'm surprised the author does not propose this.

However, some constants will still remain. Most conspicuously, the 2π constant in the very definiton of the Fourier transform. I once took a personal crusade to eliminate all such constants in the elementary Fourier formulas (plancherel-parseval, convolution theorems, commutation with derivatives), and it turns out to be possible by using the Lebesgue measure divided by sqrt(2π) in all the integrals. Thus it may seem that defining đx=dx/sqrt(2π) can be a better choice.

[mgunyho]

> You can also eliminate the constant in some of the integral formulas by using đx instead of dx. I'm surprised the author does not propose this.

In the post I propose doing that for Gauss' theorem and Cauchy's formula, because there it's convenient, heh. But to me it feels better to use Θ^ix than a scale factor in front, since the 2pi is always present in the exponential, while the prefactor can be avoided in Fourier transforms if you keep the 2pi in the exponential (or hide it inside Θ). Does this not apply also to the elementary formulas you mention?

[enriquto]

> Does this not apply also to the elementary formulas you mention?

Oh, you are right! I disliked the 2pi factor in the exponential because it messes with derivatives. But if you define your scaled derivative then the factor disappears again. So cool!

[scythe]

If you're really slick about it, you even "fix" the Gaussian integral this way.

Let é = e^sqrt(2pi), déx = dx/sqrt(2pi), and we have

int_{R}(é^(int_0^x(t dét)) déx)

= int_{R}(e^(sqrt(2pi) x^2/(2 sqrt(2pi))) dx/sqrt(2pi))

= 1/sqrt(2pi) int_{R}(e^(x^2/2) dx)

= sqrt(2pi) / sqrt(2pi)

= 1

[enriquto]

> exp(sqrt(2pi))

Now, this is a number that I don't recall having seen before. The letter é seems strangely fitting for it

é = 12.2635111...

[BlueTemplar]

A bit similar to how pi is only half of a turn, so you need to apply the transformation twice (so square root) to get back where you were(ish) ?