How to prove this simple fact without using distribution theory? Suppose function $f(x) $ is normalized to unity, i.e., 
$$ \int dx |f(x)|^2 =1 . $$ 
Now consider the Fourier transform of $f$, i.e.,
$$ F(k) = \int d x f(x) e^{-i k x}  . $$
Here we assume that $f $ is a very well-behaved function (say, a Gaussian function) so that $F(k)$ is well defined and also well-behaved. 
It is well-known that 
$$ \int d k |F(k)|^2 =2 \pi .  $$
But how to prove it without using the identity 
$$ \int dk e^{i k x } = 2 \pi \delta (x)  . $$ 
 A: $\int dk |F(k)|^2= \int dk \int dx' \int dx'' f(x')f^*(x'')e^{-ik(x'-x'')}$ because $|F(k)|^2=F(k)F^*(k)$. Integrating over $k$ yields:
$\int dk |F(k)|^2 = \int dx' \int dx'' f(x')f^*(x'') \frac{[e^{-ik(x'-x'')}]_{- \infty}^\infty}{i(x''-x')}$.
Substitution: $z=x''-x'$. Now the Tools from complex Analysis can be used: The integral over the whole real axis can be Extended to a contour integral over the upper-half circle and the real axis (the path is $C$). The half-circle integral vanishes because $z^{-1} \rightarrow 0$ as $z \rightarrow \infty$. Therefore one is left with the following contour integral:
$\int dx' \int dx'' f(x')f^*(x'') \frac{[e^{-ik(x'-x'')}]_{- \infty}^\infty}{i(x''-x')} = \int dx' f(x') \oint_C dz f^*(x'+z) \frac{[e^{ikz}]_{- \infty}^\infty}{iz}$.
Using the residue Theorem (Attention: The pole lies on $C$!) and elementary trigonometry you have:
$\int dx' f(x') \oint_C dz f^*(x'+z) \frac{[e^{ikz}]_{- \infty}^\infty}{iz} = \int dx' f(x') f^*(x') \pi i \frac{\lim_{z \rightarrow 0, k \rightarrow \infty} 2sin(zk)}{i} =$ 
$2 \pi \int dx' |f(x')|^2 \lim_{z \rightarrow 0, k \rightarrow \infty} sin(zk) = 2 \pi \lim_{z \rightarrow 0, k \rightarrow \infty} sin(zk)$
A: Assume that $f$ is real, smooth, absolutely integrable with absolutely integrable derivative. Then $f \in L^{1}\cap L^{2}$ follows. Define $g(x)=f(-x)$. Then, 
$$
           g^{\wedge}(s) =\overline{f^{\wedge}(s)}.
$$
The convolution $f\star g$ has Fourier transform $\sqrt{2\pi}|f^{\wedge}(s)|^{2}$. Consequently,
$$
  (f\star g)(x)=\int_{-\infty}^{\infty}e^{isx}|f^{\wedge}(s)|^{2}ds.
$$
Evaluating at $x=0$ gives
$$
        \int_{-\infty}^{\infty}f(x)^{2}dx = (f\star g)(0)=\int_{-\infty}^{\infty}|f^{\wedge}(s)|^{2}ds.
$$
