Chapter 11: Delta Functions

### The Exponential Representation of the Delta Function

As discussed in § {Representations of the Dirac Delta Function}, the Dirac delta function can be written in the form $$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}\, dk .\\$$ We outline here the derivation of this representation.

In order to evaluate the integral, we introduce a regularization factor, $e^{-k\epsilon}$, as follows: \begin{eqnarray} \int_{-\infty}^{\infty} e^{ikx}\, dk &= \int_{-\infty}^0 e^{ikx}\, dk + \int_0^\infty e^{ikx}\, dk \nonumber\\ &= \int_0^\infty e^{-ikx}\, dk + \int_0^\infty e^{ikx}\, dk \nonumber\\ &= \int_0^\infty (e^{ikx} + e^{-ikx})\, dk \nonumber\\ &= \lim_{\epsilon\to0^+}\int_0^\infty (e^{ikx} + e^{-ikx}) \,e^{-k\epsilon}\, dk \nonumber\\ &= \lim_{\epsilon\to0^+}\int_0^\infty (e^{ik(x+i\epsilon)} + e^{-ik(x-i\epsilon)}) \,dk \nonumber\\ &= \lim_{\epsilon\to0^+} \left[ \frac{e^{ik(x+i\epsilon)}}{i(x+i\epsilon)} + \frac{e^{-ik(x-i\epsilon)}}{-i(x-i\epsilon)} \right]_0^\infty \nonumber\\ &= \lim_{\epsilon\to0^+} \left( 0 + 0 - \frac1{i(x+i\epsilon)} - \frac1{-i(x-i\epsilon)} \right) \nonumber\\ &= \lim_{\epsilon\to0^+} \left( \frac{i}{x+i\epsilon} - \frac{i}{x-i\epsilon} \right) \nonumber\\ &= \lim_{\epsilon\to0^+} \frac{2\epsilon}{x^2+\epsilon^2} = \cases{0 & $x\ne0$ \cr \infty & $x=0$ \cr} \end{eqnarray} where we have used the fact that, for $\epsilon>0$, $e^{-k\epsilon}$ goes to $0$ as $k$ goes to $\infty$. 1)

It remains to show that the final expression has the correct normalization. But $$\int_{-\infty}^\infty \frac{2\epsilon}{x^2+\epsilon^2} \,dx = 2 \arctan\left(\frac{x}{\epsilon}\right) \Bigg|_{-\infty}^\infty = 2\pi ,$$ which is independent of $\epsilon$. Thus, \begin{eqnarray} \frac{1}{2\pi} \int_{-\infty}^\infty f(x) \,e^{ikx}\, dk &=& \lim_{\epsilon→0^+} \frac{1}{2\pi} \int_{-\infty}^\infty \frac{2\epsilon f(x)}{x^2+\epsilon^2} \,dx \nonumber\\ &=& \lim_{\epsilon→0^+} \frac{1}{2\pi} \int_{-\infty}^\infty \frac{2\epsilon f(0)}{x^2+\epsilon^2} \,dx \nonumber\\ &=& f(0) , \end{eqnarray} which is the defining property of the delta function.

1) The use of such regularization factors is quite common. Rigorous mathematical justification can be given, but informal arguments along the lines above are usually sufficient—if the informal argument works, the formal derivation should also, for an appropriate, reasonable class of functions.