Why for dirac function $\int_{-\infty}^{\infty}\delta(x)\ dx=1$

Question

The dirac function is defined as $\delta(x)=\infty$ when $x=0$, $\delta(x)=0$ otherwise. I am wondering why we can derive $\int_{-\infty}^{\infty}\delta(x)\ dx=1$, or this is just a definition

Answer 1

That isn't how dirac is defined; that is how it is conceptualized.

The definition of dirac delta distribution is that it is a distribution with the property:

$$\int_{-\infty}^{\infty} \delta(x - k)f(x) {\rm d}x = f(k)$$

So if you take the case $k = 0$ and $f(x) = 1$ then you get your result.

The conceptualization that $\delta(x) = \begin{cases} \infty \text{ for } x=0\\ 0 \text{ for } x \ne 0 \end{cases}$ is very useful in disciplines like electrical engineering and DSP where it is sometimes called a "spike" and approximations of it can be seen on an oscilloscope. But due to ambiguity, "$\infty$" is almost never used as an object just for the reason you state, it wouldn't have properties that let us solve the integral.

Answer 2

The idea that "$\delta(0)=\infty$ and $\delta(x)=0$ otherwise" is not a definition of the Dirac delta function, it is just a convenient graphical intuition. Although there is a sense in which one can approximate $\delta$ by ever-sharper ever-longer needle curves jutting out at $x=0$. I'll explain momentarily.

Often in life we have weighted averages. For instance, my grade in my history class was comprised of three parts: attendance was 25 points, midterm was 100 points, and final exam 125 points, for a total of 250 points. Thus, if my scores on attendance, midterm and final are $A,B,C$ then my final grade can be computed as the weighted average $\frac{25}{250}A+\frac{100}{250}B+\frac{125}{250}C$. The coefficients of the unknowns are the weights - they specify to what extent each score contributes to my final grade.

Another example is mass and density. Volume of a region in space is just the volume integral over the region, $\int_RdV$, in which every point in space contributes equally to the measure. But mass is different, and is $\int_R\rho dV$, where $\rho$ is the density at a particular point in space. The center of mass of an object of uniform density would be $\int_R\vec{x}dV$, i.e. the vectorial "average" of the points within the region. But if the object does not have uniform density, then the center is $\int_R\rho\vec{x}dV$, so that how much a point contributes to the vectorial average is proportional to the density at that point in space. If an object is lopsided with higher density in one part of the space it occupies then the center of mass will be more biased towards that direction than it otherwise would be.

Other examples exist too. If $X$ is a random variable, how do we compute the expected value of the new random variable $f(X)$ using $X$'s probability distribution? Same idea: $\int f(x)p_X(x)dx$, an average of $f$'s values weighted by the probability of the corresponding inputs. Many times in physics and PDEs, integral transforms weigh functions against kernels to get new functions, generalizing matrix multiplication (if we think of a function as a coordinate vector, where each possible input is an index and the output is the coordinate at that index). For instance, heat kernels and Fourier methods use integral transforms. (Delta is also used this way; see Green functions.)

The takeaway is this: sometimes the use of a "weighing" function is in integrating other functions against it. The act of weighing functions $f$ against a given weight function $w$, i.e. the assignment given by $f\mapsto \int wf$, is a linear map from the space of (suitable) functions to scalars, which inspires the idea: what if we speak more generally of such linear functionals, and not necessarily ones that can be obtained by integrating $f$ against a weight function? These are distributions, also known as "generalized functions."

Dirac delta is the distribution $f\mapsto f(0)$, which is obviously linear. For convenience, even though this distribution cannot be obtained by a bona fide function, we "pretend" (notationally, at least) that it does and write $\int f(x)\delta(x)dx=f(0)$. There is of course no function with this property, and the notation is tricky because, as Mariano says, there are many things one can do to integrands that one cannot do to this make-believe integrand $\delta(x)$, and in any case it inevitably confuses newcomers. And yet there are many manipulations of $\delta$ that work even though it's not an integrand, or even give us more power. For instance, by invoking by-parts integration, we can speak of so-called weak solutions to PDEs by transforming PDEs into (logically weaker) integral equations and then reinterpreting "integrating function against" as "applying distribution to."

There are certainly families of "spike" functions $\delta_\epsilon(x)$ which grow an ever thinner and longer spike at $x=0$ (as $\epsilon\to0^+$) so that, while $\lim_{\epsilon\to0^+}\delta_\epsilon(x)$ would converge pointwise to a function which is $0$ for all $x\ne0$ but not defined at $x=0$, nonetheless $f(0)=\lim_{\epsilon\to0^+}\int \delta_a(x)f(x)dx$ (note the limit is on the outside of the integral, not on the inside). This justifies the intuition that $\delta$ is $0$ outside of $x=0$ and an infinite spike at $x=0$. Wikipedia provides the following example:

$\hskip 2.3in$

These are the functions $\delta_\epsilon(x)=e^{-(x/\epsilon)^2}/(\epsilon\sqrt{\pi})$. In fact, each is a probability distribution, so $\delta(x)$ is a certain "weak limit" of probability distributions. Anyway, because of this infinite spike nature of $\delta$, it is used in physics to model impulses; this is essentially an idealization in which we let the region of space a pulse acts on tend to a zero-dimensional point, forcing us to up the amplitude to infinity along the way as compensation, so the resulting dynamics approach a meaningful limit.

Answer 3

This is a conventional notation, as this integral does not exist.

Anyway, if it did, it could be evaluated as the difference of its antiderivative at $\pm\infty$. This hypothetic antiderivative would be constant everywhere (zero derivative), except with a discontinuity at $0$ (infinite derivative).

Such a function is known as the Heaviside step, defined as $0$ for negatives and $1$ for positives.

$$\int_{-\infty}^{\infty}\delta(x)\ dx=h(x)\Big|_{-\infty}^{\infty}=1.$$

A $\delta$ function such as this integral would yield $0$ or $\infty$ would be of little interest as it would generate degenerate equalities. And for the sake of simplicity, the area is defined to be $1$ (in principle, it can be attributed any value, as scaling it has no effect).

The $\delta$ function is often presented as the limit of a peak function becoming narrower and narrower, that acts as an averaging weight. A weighted average is obtained as $$\overline f_w=\frac{\int_{-\infty}^{\infty}w(x)f(x)\ dx}{\int_{-\infty}^{\infty}w(x)\ dx}.$$ The weights are said to be normalized when $$\int_{-\infty}^{\infty}w(x)\ dx=1,$$ and the average reduces to $$\overline f_w=\int_{-\infty}^{\infty}w(x)f(x)\ dx.$$ You can see $\delta$ as a perfectly concentrated weighting function (all weight at $0$), which is normalized, so that $$\overline f_\delta=\int_{-\infty}^{\infty}\delta(x)f(x)\ dx=f(0).$$

Answer 4

Let $X = C^0(\mathbb{R}) \cap L^2(\mathbb{R})$ be the space of continuous functions $f:\mathbb{R} \to \mathbb{R}$ with $\int_{-\infty}^\infty |f|^2< \infty$. (We can also work with the space of compactly supported $f$, etc.) Consider the continuous linear map $L:X \to \mathbb{C}$ defined by $L(f) = f(0)$. By the Riesz representation theorem, any continuous linear map $H\to \mathbb{R}$ with $H$ a Hilbert space must be of the form $x \to (x, y)$ for some fixed $y\in H$. Our space $X$ is not a Hilbert space; it's just a vector space with inner product $(f, g) = \int_{-\infty}^\infty fg$, under which it is not complete. If it were complete, though, we could write $ \int_{-\infty}^\infty f\delta = L(f) = f(0)$ for some function $\delta\in X$.

That's what the $\delta$ function is: It's the functional $f \to f(0)$, written as $$f \to \int_{-\infty}^\infty f(x) \delta(x)\, dx = f(0)$$ by the analogy above. In particular, taking $f\equiv 1$ gives $\int_{-\infty}^\infty \delta = 1$. There certainly isn't a continuous function $\delta$ that satisfies the previous equation. On the other hand, we can for example define bump functions $$ h_n(x) = \begin{cases} n & \text{if $x\in [-\frac{1}{2n}, \frac{1}{2n}]$}; \\ 0 & \text{otherwise} \end{cases}$$ and note that $$\lim_{n\to\infty} \int_{-\infty}^\infty f(x)h_n(x)\, dx = f(0) = \int_{-\infty}^\infty f(x) \delta(x)\, dx$$ for reasonable functions $f$. We can therefore vaguely identify $\delta$ with $\lim_{n\to \infty} h_n$ and make statements like $$\delta(x) = \lim_{n\to\infty} h_n(x) = \begin{cases} \infty & \text{if $x = 0$}; \\ 0 & \text{otherwise}. \end{cases}$$ Of course, $\delta$ isn't even a function (nor are the $h_n$ uniquely defined), and so the previous equation doesn't make any sense. It's occasionally a convenient of thinking about the issue, though, in the same way that a derivative $dy/dx$ is not literally a quotient but often behaves as if it were.

Answer 5

Probably one of the best ways to approach this problem is to use measure theoretic definition. That is $$\delta_{x_0}(A)=\cases{1,\quad x_0\in A\\0\quad x_0\not\in A }$$ Summing over all the real line $$P(A)=\int_{\mathbb{R}}\delta_{x_0}(A)\mathrm{d}x_0$$ and letting $A=\mathbb{R}$ leads to $P((-\infty,\infty))=1$.