Evaluate the integral of e^(i*omega*x) with respect to omega from minus infinity to infinity, where x is a parameter.

Step-by-step solution

1. 1

   Integrate over a symmetric finite window first. An improper integral over $(-\infty,\infty)$ is defined through limits of finite pieces, so start with $\int_{-A}^{A}$. For $x\ne 0$ the antiderivative in $\omega$ is $\dfrac{1}{ix}e^{i\omega x}$, hence

   $\int_{-A}^{A}e^{i\omega x}\,d\omega=\frac{1}{ix}\left(e^{iAx}-e^{-iAx}\right)$

2. 2

   Turn the exponentials into a sine. Using $e^{i\theta}-e^{-i\theta}=2i\sin\theta$ with $\theta=Ax$:

   $\int_{-A}^{A}e^{i\omega x}\,d\omega=\frac{2i\sin(Ax)}{ix}=\frac{2\sin(Ax)}{x}$

   This is $2A\,\operatorname{sinc}(Ax)$ — a spike of height $2A$ at $x=0$ whose total area stays fixed.

3. 3

   Take $A\to\infty$ pointwise, and note the failure. For fixed $x\ne 0$ the factor $\sin(Ax)$ keeps oscillating between $-1$ and $1$ and never settles, so $\dfrac{2\sin(Ax)}{x}$ has no limit. For $x=0$ the integrand is the constant $1$ and $\int_{-A}^{A}1\,d\omega=2A\to\infty$. Either way the integral diverges in the ordinary (Riemann or Lebesgue) sense.

4. 4

   Test the family against a smooth function instead. Although $\dfrac{2\sin(Ax)}{x}$ has no pointwise limit, its area is constant:

   $\int_{-\infty}^{\infty}\frac{2\sin(Ax)}{x}\,dx=2\pi\qquad\text{for every }A>0$

   and as $A$ grows the mass concentrates ever more tightly at $x=0$ (the oscillations away from the origin cancel when weighted against any smooth test function $\varphi$). That is precisely the defining behaviour of a delta sequence:

   $\lim_{A\to\infty}\int_{-\infty}^{\infty}\frac{2\sin(Ax)}{x}\varphi(x)\,dx=2\pi\varphi(0)$

5. 5

   State both answers, and say which is meant where. As an ordinary improper integral the expression does not converge. As a distribution — the sense used in Fourier analysis, where it is the inverse transform of the constant function $1$ — it is

   $\int_{-\infty}^{\infty}e^{i\omega x}\,d\omega=2\pi\,\delta(x)$

   The $2\pi$ is convention-dependent: with the symmetric transform $\tfrac{1}{\sqrt{2\pi}}$ convention the same identity reads $\int e^{i\omega x}\,d\omega=2\pi\delta(x)$ still, but the transform pair carries $\tfrac{1}{\sqrt{2\pi}}$ on each side.