Simple question, what is the answer to the following integral:
$\int_{-\infty}^\infty \exp[\delta(x-x_0)] dx$
Is it 1? Furthermore, is there an integral to a similar function, say,
$\int_{-\infty}^\infty \exp[\delta(x-x_0)]\exp[i(x-x_0)^2]dx$
Asked
Active
Viewed 267 times
1 Answers
2
There is no meaning to the expression $f(\delta(x-x_0))$ in general. One can assign meaning, however, to $\delta(f(x))$. I thought it might be useful to present a way to show the well-know result by regularizing the Dirac Delta. It is to that end we proceed.
The distribution $(\delta o f)(x)$, where $f$ is continuously differentiable with $f'(x)\ne 0$ can be defined as
$$\bbox[5px,border:2px solid #C0A000]{\delta(f(x))=\sum_{m=1}^M \frac{\delta(x-x_m)}{|f'(x_m)|}} \tag 1$$
where $f(x_m)=0$ for $m=1,2\cdots, M$.
Proof:
To show that the equality in $(1)$ is true in the sense of distributions, we regularize the Dirac Delta such that
$$\delta(x)\sim \lim_{n\to \infty}\delta_n(x)$$
where $\lim_{n\to \infty}\delta_n(x)=0$ for $x\ne 0$, $\lim_{n\to \infty}\delta_n(0)=\infty$, and for any test function $\phi(x)$,
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\int_{-\infty}^\infty \phi(x)\delta_n(x)\,dx=\phi(0)} \tag 2$$
Next, we choose a number $\epsilon$ so small that $f(x)$ is monotonic for $x\in [x_m-\epsilon,x_m+\epsilon]$. Then, we can write
$$\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \phi(x)\delta_n(f(x))\,dx&=\lim_{n\to \infty}\int_{-\infty}^{x_1-\epsilon}\phi(x)\delta_n(f(x))\,dx\\\\ &+\lim_{n\to \infty}\sum_{m=1}^M \left(\int_{x_m-\epsilon}^{x_{m}+\epsilon}\phi(x)\delta_n(f(x))\,dx\right)\\\\ &+\lim_{n\to \infty}\sum_{m=1}^{M-1}\left(\int_{x_m+\epsilon}^{x_{m+1}-\epsilon}\phi(x)\delta_n(f(x))\,dx\right)\\\\ &+\lim_{n\to \infty}\int_{x_M+\epsilon}^\infty \phi(x)\delta_n(f(x))\,dx\\\\ &=\lim_{n\to \infty}\sum_{m=1}^M \left(\int_{x_m-\epsilon}^{x_{m}+\epsilon}\phi(x)\delta_n(f(x))\,dx\right) \tag 3 \end{align}$$
where in arriving at $(3)$, we exploited the fact that $\lim_{n\to \infty}\delta_n(f(x))=0$ for $f(x)\ne 0$ (i.e., when $x\ne x_n$).
Now, we proceed by enforcing the change of variable $u=f(x)$ in $(1)$ to obtain
$$\begin{align} \lim_{n\to \infty}\int_{x_m-\epsilon}^{x_m+\epsilon}\phi(x)\delta_n(f(x))\,dx&=\lim_{n\to \infty}\int_{f(x_m-\epsilon)}^{f(x_m+\epsilon)}\phi(f^{(-1)}(u))\frac{\delta_n(u)}{\left.f'(x)\right|_{x=f^{(-1)}(u)}}\,du\\\\ &=\frac{\phi(x_m)}{f'(x_m)}\text{sgn}(f'(x_m)) \tag 4\\\\ &=\frac{\phi(x_m)}{\left|f'(x_m)\right|}\\\\ &=\lim_{n\to \infty}\int_{x_m-\epsilon}^{x_m+\epsilon} \phi(x)\frac{\delta_n(x-x_m)}{\left|f'(x_m)\right|}\,dx\tag 5 \end{align}$$
where we used $(2)$ to arrive at $(4)$ and $(5)$.
Putting $(3)$ and $(5)$ together yields
$$\lim_{n\to \infty}\phi(x)\delta_n(f(x))\,dx=\lim_{n\to \infty}\int_{-\infty}^\infty \phi(x)\,\sum_{m=1}^M\frac{\delta_n(x-x_m)}{|f'(x_m)|}\,dx$$
which can be expressed in terms of the distributional equality
$$\delta(f(x))=\sum_{m=1}^M\frac{\delta_n(x-x_m)}{|f'(x_m)|} $$
which is the expression in $(1)$ that was to be shown.
Mark Viola
- 179,405