TITLE: $\int_{-\infty}^\infty e^{ikx}dx$ equals what?
QUESTION [15 upvotes]: What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral? 
I saw this in the Fourier transform, and am unsure how to solve this.

REPLY [3 votes]: I just want to clarify what is meant by saying that "as a distribution",
$$
2\pi\delta(k) = \int_{-\infty}^\infty e^{ikx}\,dx. 
$$
The right hand side certainly doesn't make sense, and the left hand side can only make sense for $k\not=0$.  What we mean though is that for any smooth function $\phi$ with rapid decay,
$$
2\pi \phi(0) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{ikx} \phi(k) \,dk\,dx.
$$
Now both sides make sense.  This isn't trivial to prove, and in fact it implies the Fourier inversion formula (using the relationship between modulation and translation under the Fourier transform).  The proof often goes through a calculation similar to what @robjohn's answer describes.