Not too long ago, I attended a lecture given by William Dunham about Leonhard Euler. Prior to this lecture, I thought the only way to derive Euler's identity used Taylor expansions, or that there was some complex method for deriving it which I was unable to comprehend at the time. Yet there is an absolutely beautiful proof for this identity which Dunham showed us, and which I will now show my fellow E2ers.

To understand the following proof for Euler's identity, e^{i x} = cos(`x`) + i sin(`x`), you need to know the following tidbits of The Calculus:

### OMFG IT'S A PROOF!!!11

FIRST: We have the following integral:

∫ dz / √(1 + `z`^{2})

NEXT: Since we are mathematicians, and want to save time, we look this up in a table of integrals. (N.B. if you feel so compelled, you could evaluate the integral using trig substitution.)

∫ dz / √(1 + `z`^{2}) = ln (`z` + √(1 + `z`^{2}))

*

MORE NEXT: We'll perform a simple variable substitution (obviously, i = √-1 here) by letting `z` = i `y`:

`z` = i `y`

dz = i dy

∫ dz / √(1 + `z`^{2}) = ln ( `z` + √(1 + `z`^{2}) )

⇔

∫ i dy / √(1 + (i `y`)^{2}) = ln ( i `y` + √(1 + (i `y`)^{2}) )

i ∫ dy / √(1 - `y`^{2}) = ln( i `y` + √(1 - `y`^{2}) )

**

THENNILY: we perform trigonometric substitution by letting `y` = sin(`x`)

`y` = sin(`x`)

dy = cos(`x`) dx

N.B.: 1 - sin^{2}(`x`) = cos^{2}(`x`)

i ∫ dy / √(1 - `y`^{2}) = ln( i `y` + √(1 - `y`^{2}) )

⇔

i ∫ cos(`x`) dx / √(1 - sin^{2}(`x`)) = ln( i sin(`x`) + √(1 - sin^{2}(`x`)) )

i ∫ cos(`x`) dx / √(cos^{2}(`x`)) = ln ( i sin(`x`) + √(cos^{2}(`x`)) )

i ∫ ~~cos(~~`x`) dx / ~~cos(~~`x`) = ln ( i sin(`x`) + cos(`x`) )

i ∫ dx = ln ( i sin(`x`) + cos(`x`) )

i `x` = ln ( i sin(`x`) + cos(`x`) )

We're almost there....

FINALLY: Raise e to the power of both sides:

e^{i x} = e^{ln( i sin(x) + cos(x) )}

e^{i x} = i sin(`x`) + cos(`x`)

†

quod erat demonstrandum, beeyatch.

There you have it: a proof of Euler's identity that doesn't involve Taylor Series. It's so purty.

#### Sources

- William Dunham's lecture, given on 11 May, 2005
- Wikipedia (For the table of integrals)
- My internalized Calculus notes

This proof has been brought to you by the letter e, the number i, and the person Leonhard Euler.

* You'll notice that I'm missing the constant of integration. That's because we don't really need it here. Besides, Leonhard Euler didn't use it! Why should we? (N.B. this excuse does not work with your Calculus teacher, and it doesn't work with College Board. Don't try this at home, kids.)

** Yes, we can treat i as a constant here. In Euler's time, i was considered a bit of a taboo number (hence the somewhat derogatory term, "imaginary number"). Euler made i hip, though. He felt that if you couldn't treat i as a constant, you'd be destroying the foundation of mathematics.

† You don't need to worry about absolute value signs when you're dealing with these numbers. When you have an integral whose answer involves an ln (usually from u substitution), putting on absolute value signs really doesn't change much, except make it "prettier" when dealing with real numbers only. Since we're not dealing with real numbers, they have no need. Besides, the only difference between ln(x) and ln(-x) is a constant factor. Especially where integrals are involved, it's just like treating it as a constant of integration, which can "drop out," anyway.