Euler's Gem 2

Here is a sketch of a second derivation of Euler's famous formula:

eiθ = cosθ + i sinθ

as presented by William Dunham in his book Euler, The Master of Us All. First post here.

The first step is to recall a standard trigonometric substitution in calculus:

y = sin x x = sin-1 y √(1 - y2) = cos x

We're interested in the integral:

∫ dy / √(1 - y2)

Substituting with x we see that:

dy = cos x dx

And the integral is

∫ (1/cos x) cos x dx = ∫ dx = x x = ∫ dy / √(1 - y2)

Now Euler makes a complex change of variable:

y = iz x = ∫ dy / √(1 - y2) = ∫ i dz / √(1-(iz)2) = i ∫ dz / √(1 + z2) = i ln [√(1 + z2) + z]

The last step is another standard result from calculus which I will assume for the time being (more here).

Undo the substitution:

z = y/i = sin x / i z2 = -sin2 x √(1 + z2) = √(1 - sin2 x) = cos x x = i ln (cos x + sin x / i)

We will use two identities involving i:

u / i = - i u 1 / (cos u - i sin u) = (cos u + i sin u)

(For the second one, see the previous post). Now:

x = i ln (cos x + sin x / i) x = i ln (cos x - i sin x) ix = - ln (cos x - i sin x) = ln [ 1 / (cos x - i sin x) ] = ln (cos x + i sin x)

Just eponentiate:

eix = cos x + i sin x

Wow, again!