Prove functions $e^{ikx}$ are orthonormal on $\mathbb{R}$ I need to show the following identity $$\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(k-k')x}dx = \delta(k-k'),$$ where $\delta$ is the Dirac function. In the text it says that the functions $u_k = \frac{1}{\sqrt{2\pi}}e^{ikx}$ represent an orthonormal system and in order to show this it provides the identity above. In case $k=k'$ the right and left hand side of the equation coincide, i.e. $\delta(k') = \infty.$ The quetsion is then why $u_k$ would be orthonormal. In case $k \neq k'$ I do not know how the left hand side will vanish. Can somebody provide a comment or an explanation of how to check that $u_k$ is orthonormal and why there are the difficulties I raised above ?
Many thanks.
 A: One needs to assign a meaning to the symbol $\frac1{2\pi}\int_{-\infty}^\infty e^{i(k-k')x}\,dx$.
Suppose $\psi\in \mathbb{S}$ (i.e. $\psi$ is a Schwartz function). Then, we define the distribution denoted by the symbol $\frac1{2\pi}\int_{-\infty}^\infty e^{i(k-k')x}\,dx$ in terms of the following:
$$\begin{align}
\lim_{L\to\infty} \int_{-\infty}^\infty \psi(k') \frac1{2\pi}\int_{-L}^L e^{i(k-k')x}\,dx\,dk'&=\lim_{L\to\infty} \frac1\pi\int_{-\infty}^\infty \psi(k') \frac{\sin(k-k')L}{k-k'}\,dk'\tag1
\end{align}$$
I showed in THIS ANSWER that the limit on the right hand side of $(1)$ is equal to $\psi(k)$.  Therefore, we find that in distribution
$$\lim_{L\to\infty}  \frac1{2\pi}\int_{-L}^L e^{i(k-k')x}\,dx=\delta(k-k')$$
as was to be shown.
A: What you say the book says is simply nonsense; none of the integrals even exist.
The functions are not "orthornomal on $\Bbb R$"; they don't have a chance to be since they're not square-integrable. Of course they are sort of analogous to orthonormal in some sense. Heh, they are orthonormal on the Bohr compactification...
