Understand $\delta $ function, why do me write $\int \delta (x)dx$ instead of $\int d\delta $ ? Since $\int \delta (x)dx$ should be $0$

Question

I have difficulties to understand delta function. We defined $\delta $ function as $$\delta (x)=\begin{cases}\infty&x=0\\ 0&\text{otherwise}\end{cases},$$ and $$\int_{\mathbb R}\delta (x)dx=1.$$

1) How can it be possible to have such definition since $\delta (x)=0$ a.e., and thus, the integral above should be $0$ ?

2) If we consider $\delta $ as a measure, then $\int_A f(x)d\delta(x) $ make completely sense for any set $A$. Now, I often see $$\int_A f(x)\delta (x)dx$$ for $$\int_A f(x)d\delta .$$

What ? This is not correct, is it ? Because $\int_A f(x)\delta (x)dx=0$ since $f(x-\delta (x)=0$ a.e. So why do we use this notation of $\int_A f(x)\delta (x)dx$ for $\int_{\mathbb R} f(x)d\delta (x)$ ? I'm really in truble with that.

Answer 1

why do me write $\int \delta (x)dx$ instead of $\int d\delta $ ?
Because we are physicists and not mathematicians!

You say you are a mathematician? Then you may write it correctly if you like. I agree that this notation is harmful to beginners and non-mathematicians. (Just see all the confused questions on it in math.se.) But usage among physicists is unlikely to change.

Answer 2

It is indeed as @GEdgar mentioned. But, one way, to answer your first question and to give a better idea about the delta function we consider $\delta$ to be defined as follow, $$\delta(x)=\lim\limits_{\varepsilon\rightarrow 0}\delta_\varepsilon(x),\quad \delta_\varepsilon(x)=\left\{ \begin{array}{lc} 0 & x<-\frac{\varepsilon}{2} \\ \frac{1}{\varepsilon} & -\frac{\varepsilon}{2}<x<\frac{\varepsilon}{2}\\ 0&\frac{\varepsilon}{2}<x \end{array} \right.$$

then we take the integral of $\delta_\varepsilon$ over $\mathbb{R}$, $$\forall\varepsilon>0,\quad \int_\mathbb{R}\delta_\varepsilon(x)dx=\int_{-\frac{\varepsilon}{2}}^{\frac{\varepsilon}{2}} \frac{1}{\varepsilon}dx=\frac{1}{\varepsilon}\bigg(\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\bigg)=1$$ Now we apply the limit when $\varepsilon\rightarrow 0$ and by applying the Dominated Convergence Theorem we get, $$ 1=\lim\limits_{\varepsilon\rightarrow 0}\int_\mathbb{R}\delta_\varepsilon(x)dx=\int_\mathbb{R}\lim\limits_{\varepsilon\rightarrow 0}\delta_\varepsilon(x)dx=\int_\mathbb{R}\delta(x)dx$$

It is one way to understand the first mentioned property delta function, but of course the "Distribution Theory" gives a more rigorous and well-suited definition of the delta distribution and its properties.