# When are definite integrals undefined?

We have $$\int_{-1}^{1} \dfrac{1}{x} \, dx$$ as undefined and then we have $$\int^1_{-1} f(x)\delta(x) = f(0)$$

assuming $f(x)$ is continuous everywhere and $$\delta(x) = \begin{cases} 0 & x\ne 0, \\ \infty & x = 0. \end{cases}$$

In both cases the integrand is infinite for $x = 0$, then why second integral is not undefined?

• $\delta$ is not a function and your definition makes no sense. It is a distribution. You can integrate $1/\sqrt x$ even though it is "infinite" at the origin. – user223391 Jul 8 '17 at 15:28
• @ZacharySelk I did not define it. I quoted my physics book. – user8277998 Jul 8 '17 at 15:30
• Your physics book is wrong! The symbol "$\int_{-1}^1 f(x)\delta(x)\,dx$" is NOT an integral. – Mark Viola Jul 8 '17 at 15:30
• @MarkViola But it has an integration sign :((. – user8277998 Jul 8 '17 at 15:39
• It's a symbol only. It is a linear functional and has properties that are shared with integrals. ;-)) – Mark Viola Jul 8 '17 at 15:40

As mentioned in the comments, $\delta$ is not a function in the classical sense, but a distribution. For simplification you can think of it as a definition that $\int f(x)\delta(x)\,dx = f(0)$.

But be careful because in the if you define a classical function $$\hat \delta(x) = \begin{cases} 0 & x\ne 0 \\ \infty & x = 0{}\end{cases}$$ as a classical function then for all integrable functions $f$ we get: $\int f(x)\hat\delta(x)\,dx = 0$ since $\hat\delta$ is zero almost everywhere and therefore $f\cdot\hat\delta$ is zero almost everywhere.

To further understand why you could say that this equation is kind of valid for the dirac distribution, you need a fundamental understanding of measure theory and functional analysis.

If you have a basic background in measure theory, you can understand the delta distribution as the extended radon-nikodym-derivative of the dirac measure and the lebesgue measure.

• Oh I did not know any of this. – user8277998 Jul 8 '17 at 15:54

In the physics-y spirit of the question: The "$\delta$-function" (which, as Zachary comments, is not a real-valued function) has finite integral even if "the value at one point is infinite".

By contrast, the integral $\int_{-1}^{1} \frac{1}{x}\, dx$ is undefined (as an improper Riemann integral) because the limits $$\lim_{a \to 0^{-}} \int_{-1}^{a} \frac{1}{x}\, dx = \lim_{a \to 0^{-}} \ln(-a),\qquad \lim_{b \to 0^{+}} \int_{b}^{1} \frac{1}{x}\, dx = -\lim_{b \to 0^{+}} \ln b$$ formally give the indeterminate expression $$\int_{-1}^{1} \frac{1}{x}\, dx = \infty - \infty.$$ Geometrically, the region under the graph $y = \frac{1}{x}$ for $0 \leq x \leq 1$ has infinite area, and the region above the graph $y = \frac{1}{x}$ for $-1 \leq x \leq 0$ has infinite area.

Though "the reciprocal function and the $\delta$-function are both infinite at one point", their respective behaviors differ with respect to integration over an arbitrarily small neighborhood of $0$.

Don't believe everyone telling you that you need to know functional analysis or measure theory... you just need to use the appropriate definition for $\delta$:

\begin{align*} \delta(x) &= \frac{1}{2}\,\operatorname{sgn}'(x) \\ \int_{-1}^{1} f(x)\,\delta(x)\,dx &= \int_{-1}^{1} f(x)\,\frac{\operatorname{sgn}'(x)}{2}\,dx \\ &= \left.f(x)\operatorname{sgn}(x)\right|_{-1}^{+1} - \int_{-1}^{1} \frac{\operatorname{sgn}(x)}{2}\,f'(x)\,dx \\ &= \frac{f(1)+f(-1)}{2} + \int_{-1}^{0} \frac{f'(x)}{2}\,dx - \int_{0}^{1} \frac{f'(x)}{2}\,dx \\ &= \frac{f(1)+f(-1)}{2} + f(0) - \frac{f(-1)+f(1)}{2} \\ &= f(0) \end{align*}

Note the following:

• You don't actually need $f$ to be differentiable. If it can be approximated arbitrarily well by a differentiable function, the same arguments here work.

• You don't need $\operatorname{sgn}$ to be the sign function. Just pretend it's a smooth approximation whose behavior only changes very close to the zero. Again, the proof holds just fine in that case, and as the approximation gets better, the error vanishes.

You don't need to know measure theory or functional analysis to understand any of this intuitively.

• "... and as the approximation gets better the error goes to $0$". Indeed, that's the spirit. But/and one might occasionally worry how to show such a thing. Don't get me wrong, the heuristics are great (since, done well, they do show the correct outcome), but all the more motivation to be able to relax and know that they always work (if they do). – paul garrett Jul 9 '17 at 0:20
• @paulgarrett: right, I wasn't suggesting the theory is useless, I was just saying you don't need it to understand this result and the intuition behind it... kind of like how you don't need to know game theory to intuitively understand why lying to people will make them stop believing you, etc. – user541686 Jul 9 '17 at 0:25
• True, one doesn't need to understand thermodynamics to drive a car, or quantum mechanics to use a cell phone. If anything, there is quite a lot of noise from slightly naive people obsessed with "rigor" who seem to have a grudge against "the Dirac delta function" and such. People using such things should, instead, be aware that the slick manipulations are justifiable, even if the justifications are at a different level of sophistication (and are not obligatory before one is allowed to use the reality). – paul garrett Jul 9 '17 at 0:30
• Oh, thank you very much. I felt a little worried about crediblity of my Physics book after reading answer here :)). – user8277998 Jul 9 '17 at 4:41
• @123: You're welcome. Your physics book is most likely fine. Just remember that physicists (and engineers, etc.) are not always admired by mathematicians for their rigor. ;) – user541686 Jul 9 '17 at 5:18

\begin{align} & \int_{-1}^0 \frac{dx} x = -\infty, \\[10pt] & \int_0^1 \frac{dx} x = +\infty, \\[10pt] & \lim_{\varepsilon \, \downarrow \, 0} \left( \int_{-1}^{-\varepsilon} \frac{dx} x + \int_\varepsilon^1 \frac{dx} x \right) = \lim_{\varepsilon\,\downarrow\,0} \left( \log_e \varepsilon -\log_e \varepsilon \right) = 0, \tag 1 \\[10pt] & \lim_{\varepsilon\,\downarrow\,0} \left( \int_{-1}^{-\varepsilon} \frac{dx} x - \int_{2\varepsilon}^1 \frac{dx} x \right) = \lim_{\varepsilon\,\downarrow\,0} (\log_e \varepsilon - {}\log_e (2\varepsilon)) = \log_e \frac 1 2 \ne 0. \tag 2 \end{align} When the positive and negative parts are both infinite, then "rearrangements" of this sort can alter tha value of the integral. The result in $(1)$ is the "Cauchy principal value" of the integral from $-1$ to $1.$

$$\int_{-\infty}^\infty \delta(x)\,dx = 1.$$ With the delta function, the positive part of the value of the integral is finite and the negative part is zero -- hence finite.

But notice that $$\int_{-\infty}^\infty \Big( 3\delta(x) \Big) \, dx = 3,$$ so if one says that $\delta(0) = \infty,$ then one would have $\delta(0) = 3\cdot\infty = \text{what?}$ And here the answer is that the delta function is not a "function" in the sense of something that returns an output for each input; rather it is characterized by the values of the integrals of products $f(x)\delta(x)$ where $f$ actually is a function in that sense.

• What does down arrow limit mean ? – user8277998 Jul 8 '17 at 17:23
• @123 : It means approaching $0$ from above. Somethimes you'll see $\text{“} {u\to0+}\text{''}$ as opposed to $\text{“} {u \to 0-} \text{''}. \qquad$ – Michael Hardy Jul 8 '17 at 17:26