Why is $e^{-ix}$ an odd function?

Question

My physics prof pulled this out of the air to swap the limits on an integral:

$\int_{0}^{\infty} e^{-ix} \, dx = \int_{-\infty}^{0} e^{ix} \, dx$

"because the integrand is odd."

So far as I can tell:

$f(x) = e^{-ix}$

$$= \cos(-x) + i \sin(-x)$$

$$= \cos(x) - i \sin(x)$$

$f(-x) = e^{ix}$

$$= \cos(x) + i \sin(x)$$

$-f(x) = -e^{-ix} $

$$= - \cos(x) + i \sin(x), $$

therefore

$-f(x) \neq f(-x) $

and the function is not odd?

I'm guessing 9:1 I'm wrong, someone let me know why :')

Answer 1

Physicists often use a “free” language in the math domain.;-)

You are, of course, right — the integrand is not an odd function.

I guess that the author wanted to express something as this, having in mind that the identity ($x \mapsto x$) is an odd fuction:

• LHS integrand is $\ e^{-ix}$ and integration is for values in interval $ [0, \infty)$.

• Every $x$ we may express as $-(-x)\ $, or as $-y$, where $y = -x$. Then $$ e^{-ix} = e^{(-i)(-y)} = e^{iy}$$

• But if $x \in [0, \infty)$, then $y \in (-\infty, 0]$, so by the substitution $y = -x$ we obtain

$$\int_{0}^{\infty} e^{-ix} \, \mathrm dx = \int_{-\infty}^{0} e^{iy} \, \mathrm dy = \int_{-\infty}^{0} e^{ix} \, \mathrm dx$$

Answer 2

Ignoring any issues of the integral existing,

$$ \int_0^\infty f(x) dx = \int_{-\infty}^0 f(-x) dx $$

is true for every function $f$, not just odd functions. This is simply a variable substitution. Setting $u = -x$, we have

$$ \begin{align} \int_0^\infty f(x) dx &= \int_0^{-\infty}f(-u) (-1) du\\ &=-\int_{-\infty}^0 f(-u) (-1) du\\ &=\int_{-\infty}^0 f(-u) du\\ \end{align} $$