$\int_{-\infty}^\infty e^{ikx}dx$ equals what?

Integral $$\int_{-a}^{a} e^{ikx}dx=\frac{e^{ikx}}{ik}|_{-a}^{a}=\frac{e^{ika}-e^{-ika}}{ik}=\frac{2i\sin{ka}}{ik}=2a\frac{\sin{ka}}{ka}$$ Now if $a\to\infty$, the term $\frac{\sin{ka}}{ka}=\delta(ka)=\frac{1}{a}\delta(k)$ where $\delta$ is Dirac-delta function.

$$ \begin{align} \int_{-\infty}^\infty e^{ixy}\overbrace{\ \ \ e^{-\epsilon x^2}\ \ \ }^{\to1}\,\mathrm{d}x &=e^{-\frac{y^2}{4\epsilon}}\color{#C00}{\int_{-\infty}^\infty e^{-\epsilon\left(x-\frac{iy}{2\epsilon}\right)^2}\,\mathrm{d}x}\tag{1}\\ &=\underbrace{\color{#C00}{\sqrt{\frac\pi\epsilon}}\,e^{-\frac{y^2}{4\epsilon}}}_{\to2\pi\delta(y)}\tag{2} \end{align} $$ Using Cauchy's Integral Theorem, the red integral in $(1)$ is simply $\int_{-\infty}^\infty e^{-\epsilon x^2}\,\mathrm{d}x=\sqrt{\frac\pi\epsilon}$ .

As $\epsilon\to0$, we get that $(2)$ approximates $2\pi\delta(y)$. That is, the integral of $(2)$ is $2\pi$ for all $\epsilon$, and as $\epsilon\to0$, the main mass of the function is squeezed into a very small region about $0$.

Draks is right on this (+1) it makes sense as (up to a constant) a representation of the Dirac delta distribution (it is divergent from other points of view!).

More exactly : $$2\pi \delta(k)=\int\limits_{-\infty}^\infty e^{ikx}dx$$