Why does integrating a complex exponential give the delta function?

Question

How come, when we integrate a complex exponential from $ -\infty $ to $ \infty $, we get a scaled delta function? $$ \begin{align} \int_{-\infty}^{\infty} e^{i k x} \; dk & = 2 \pi \delta \left ( x \right ) \end{align} $$ Specifically, why do we say that the integral converges for $ x \neq 0 $ to $ 0 $? Doesn't it just continue to oscillate?
Thanks!

Answer 1

Following Sylvain's comment look up the formulas for Fourier transform $F(w)=\int f(t) e^{-iwt}{\rm {d}}t$ and the inverse transform $ f(t)=1/{2\pi}\int F(w) e^{iwt} { \rm {d}}w$ and combine them to write \begin{equation} \begin{split} F(\hat w)=\int_{-\infty}^\infty f(t) e^{-i\hat{w}t}\rm {d}t & = \int_{-\infty}^\infty \frac{1}{2\pi}\int_{-\infty}^\infty F(w) e^{iwt} {\rm {d}} w\, e^{-i\hat{w}t}\,\rm {d}t \\ & = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)\,{\rm {d}} w\int_{-\infty}^\infty e^{iwt} \, e^{-i\hat{w}t}\,\rm {d}t\\ & = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)\,{\rm {d}} w\int_{-\infty}^\infty e^{i(w-\hat{w})t}\,\rm {d}t \end{split} \end{equation} Knowing the definition of the delta-function $f(y)=\int f(x) \delta(x-y) {\rm d} x$ one can see that in this case $$ F(\hat w) = \int_{-\infty}^{\infty} F(w) \delta (w-\hat w) {\rm d}w $$ and thus identifying the right integral with $$\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(w-\hat{w})t}\,\rm {d}t= \delta (w- \hat w) $$

Answer 2

The integral is not meant to be taken in the space of functions; it is meant to be taken over the space of distributions.

If I write $[f(x)]$ when I mean to view $f(x)$ as a distribution rather than a function, then this integral is saying

$$ \int_{-\infty}^{\infty} [e^{ikx}] \, \mathrm{d}k = 2 \pi \delta(x) $$

(note that $\delta(x)$ is a distribution, not a function, so I don't need to put brackets around it)

Two distributions are equal if and only if they have the same value when convolved with test functions. The above equation is asserting, for every test function $f(x)$, you have

$$ \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} [e^{ikx}] \, \mathrm{d}k \right) f(x) \, \mathrm{d}x = \int_{-\infty}^{\infty} 2 \pi \delta(x)f(x) \, \mathrm{d} x$$

What the test functions are precisely can vary with context; in this case they are probably meant to be the rapidly decreasing functions (aka Schwartz functions).

IIRC, distributions coming from functions of two variables satisfy

$$ \int_{-\infty}^{\infty}\left( \int_{-\infty}^{\infty} [g(x,y)] \, \mathrm{d}y \right)f(x) \mathrm{d}x = \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} g(x,y) f(x)\, \mathrm{d}x \right) \mathrm{d}y $$

(note the change of order of the integration variables)

Consequently, The above equation is asserting

$$ \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{ikx} f(x) \, \mathrm{d}x \right) \mathrm{d}k =2 \pi f(0)$$

Note, in particular, that the inner integral is that of ordinary functions, and the integrand is integrable.