I have a question about this integral with a Dirac delta

$$ \displaystyle \int_{-\infty}^{+\infty} \delta'(x-3)e^{x^2}dx $$

by integration by parts I get;

$$ \displaystyle \delta(x-3)e^{x^2}\biggr\rvert_{-\infty}^{+\infty} - 2\int_{-\infty}^{+\infty} \delta(x-3)xe^{x^2}dx = -2(3e^{3^2})= -6e^{9}$$

My question is:

Why is the first term $= 0$?

What I Think is $\delta(\infty)= 0$, but $\delta(\infty)\cdot e^{\infty}=$ ?

  • $\begingroup$ The integral is not defined (as your computation shows). $\endgroup$
    – A.S.
    Commented Feb 29, 2016 at 3:00

6 Answers 6



The Dirac Delta and the Unit Doublet (the so-called "derivative" of the Dirac Delta) are not functions. Rather, they are Generalized Functions, also known as Distributions.

Distributions are linear Functionals that map test functions (smooth functions) into numbers, whereas a function maps numbers into numbers. For the Dirac Delta, the functional definition is given as

$$\langle f,\delta_a\rangle =f(a) \tag 1$$

where $f$ is a suitable test function.

Now, in practice, we often write the functional notation in $(1)$ formally as

$$\langle f,\delta_a\rangle=\int_{-\infty}^{\infty}f(x)\delta(x-a)\,dx \tag 2$$

But the object on the right-hand side of $(2)$ is actually not an integral. And the evaluation of $\delta (x-a)$ as a function is meaningless. In practice, we often see the Dirac Delta defined at points by

$$\delta(x)=\begin{cases}0&,x\ne 0\\\\\infty&,x=0\end{cases}$$

but this is obvious nonsense. Rather, the interpretation here can be made physical through a regularization of the Dirac Delta wherein there is a family of functions $\delta_n(x)\in C^\infty_C$ for which

$$\lim_{n\to \infty}\delta_n(x)= \begin{cases} 0&, x\ne 0\\\\ \infty&,x=0\end{cases}$$


$$\lim_{n\to \infty}\int_{-\infty}^\infty f(x)\delta_n(x-a)\,dx=f(a)$$

for all suitable test functions $f$. We can formally write this regularization of the delta function as

$$\delta(x)\sim\lim_{n\to \infty}\delta_n(x)$$

Therefore, we interpret the integral notation for the functional relationship in $(2)$ as

$$\int_{-\infty}^{\infty} f(x) \delta(x-a)\,dx=\lim_{n\to \infty}\int_{-\infty}^\infty f(x)\delta_n(x-a)\,dx$$

For more on the Dirac Delta, see THIS ANSWER, THIS ONE, THIS ONE, and THIS ONE.


The Unit Doublet $\delta'$ is defined as in terms of the Dirac Delta as

$$\langle f,\delta_a'\rangle=-\langle f',\delta_a\rangle=-f'(a)$$

It is, therefore, a functional that maps a test function $f$ into $-f'$. We can formally, write

$$\langle f,\delta_a'\rangle=\int_{-\infty}^{\infty}f(x)\delta'(x-a)\,dx=-\int_{-\infty}^{\infty}f'(x)\delta(x-a)\,dx=-f'(a)$$

and proceed heuristically as with the Dirac Delta.

Note that while $f(x)=e^{x^2}$ is not a test function, since $\delta'_a$ has compact support on $\{a\}$,we can proceed. For $a=3$, we have immediately that

$$\langle e^{x^2},\delta'_3\rangle =-\left.\frac{de^{x^2}}{dx}\right|_{x=3}=-6e^9$$

  • 2
    $\begingroup$ This brings light to what most miss. The delta function is utilized as a "function" for quick evaluations, but needs to have a background known as a distribution. Presented here is that background to learn from and utilize in future problems. $\endgroup$
    – Leucippus
    Commented Aug 29, 2015 at 18:17
  • $\begingroup$ @leucippus Yes. Each time I see issues arise on the Dirac, I try and post something that illustrates the Generalized Function essence. I hope these writings are useful for others. Thanks for your comment! Very appteciative! +1 $\endgroup$
    – Mark Viola
    Commented Aug 29, 2015 at 18:23
  • $\begingroup$ @Dr.MV Thanks for your very helpful answer. I think this shows clearly what is really going on for these types of problems, so I have changed my answer to yours! $\endgroup$ Commented Aug 31, 2015 at 8:56
  • $\begingroup$ A big problem with your computation (based either one of your definitions) is that $f(x)=e^{x^2}$ is unbounded ($\to \infty$) and I don't see a way to meaningfully define $\langle f,\delta_3'\rangle$ as a result. $\endgroup$
    – A.S.
    Commented Feb 29, 2016 at 0:37
  • $\begingroup$ @A.S. Are you familiar with Generalized Functions? And I gave exactly a meaningful way to define it. $\endgroup$
    – Mark Viola
    Commented Feb 29, 2016 at 1:21

Using the shifting property $f(x) \, \delta(x-a) = f(a) \, \delta(x-a)$ then \begin{align} \int \delta^{\prime}(x-a) \, f(x) \, dx &= \left[ f(x) \, \delta(x-a) \right] - \int \delta(x-a) \, f^{\prime}(x) \, dx \\ &= \left[ f(a) \, \delta(x-a) \right] - f^{\prime}(a) \end{align} For the case of $f(x) = e^{x^2}$ then $f^{\prime}(x) = 2 \, x \, e^{x^2}$. For the limits $(-\infty, \infty)$ the $\delta$ function is zero at the end points being evaluated. Now, \begin{align} \int_{-\infty}^{\infty} \delta^{\prime}(x-a) \, e^{x^{2}} \, dx &= \left[ e^{a^{2}} \, \delta(x-a) \right]_{-\infty}^{\infty} - 2 \, a \, e^{a^{2}} \\ &= - 2 \, a \, e^{a^{2}}. \end{align}

Shifting property: $$ \int \delta(x-a) \, f(x) \, dx = f(a) = \int f(a) \, \delta(x-a) \, dx $$ or $$ \int \left[ f(x) \, \delta(x-a) - f(a) \, \delta(x-a) \right] \, dx = 0.$$ In order for the general integral to have a zero result then the integrand must be zero which leads to $$f(x) \, \delta(x-a) = f(a) \, \delta(x-a)$$

  • $\begingroup$ Thank you for your help. I'm not sure I understand where $f(x)\delta (x-a) = f(a) \delta(x-a) $ comes from. I don't think I've come across it so I wouldn't like to use it without being able to prove it. $\endgroup$ Commented Aug 29, 2015 at 16:08
  • $\begingroup$ @EdinburghDruid The shifting property is now demonstrated. $\endgroup$
    – Leucippus
    Commented Aug 29, 2015 at 16:42
  • $\begingroup$ @Leucippus I provided an answer that goes into what is really going on here. I value your opinion, so please, if you have a moment, let me know your thoughts on the post. $\endgroup$
    – Mark Viola
    Commented Aug 29, 2015 at 17:14
  • 1
    $\begingroup$ @Leucippus Thanks for showing how to prove the property. I have changed my chosen answer to Dr. MV's one, but I do appreciate your help also! $\endgroup$ Commented Aug 31, 2015 at 8:58

The simple answer is that the $\delta(x-3)$ and all its derivatives are supported at $x=3$. Even if the integral were $$ \begin{align} \int_2^4\delta'(x-3)e^{x^2}\,\mathrm{d}x &=\left[\delta(x-3)e^{x^2}\right]_2^4-\int_2^4\delta(x-3)2xe^{x^2}\,\mathrm{d}x\\ &=0-6e^9 \end{align} $$ the boundary terms vanish. That is, away from $x=3$, $\delta(x-3)$ can be represented by the zero function.

From Comments

A.S. mentions that $e^{x^2}$ does not have the proper decay at $\infty$ to be a standard test function. However, $\delta'$ has compact support, i.e. $\{0\}$, and a test function applied to such a distribution can be modified outside the compact support and not alter the value obtained.

So to be technically correct, we should write the integral as $$ \int_{-\infty}^\infty\varphi(x-3)\delta'(x-3)e^{x^2}\,\mathrm{d}x $$ where $\varphi\in C_C^\infty$, $\varphi(x)=1$ on a neighborhood of $\{0\}$. Then $\varphi(x-3)\delta'(x-3)=\delta'(x-3)$. Furthermore, $\varphi(x-3)e^{x^2}$ is a standard test function.

  • $\begingroup$ This can't be correct because $f(x)=e^{x^2}$ is unbounded on $\mathbb R$ (and $\notin L^1$) hence the integral is simply undefined. Changing domain of integration from $\mathbb R$ to $[a,b]$ effectively changes $f$ to $f1_{[a,b]}$ and the latter is compactly supported which makes the new the integral well-defined (unlike the original one). The problem OP ran into is valid and is indicative of the integral not being defined. $\endgroup$
    – A.S.
    Commented Feb 29, 2016 at 1:01
  • $\begingroup$ @A.S.: There are different definitions of an integral. For a Riemann Integral, the integral in the question is improper. It actually means $$\lim_{\substack{L_1\to-\infty\\L_2\to+\infty}}\int_{L_1}^{L_2}\delta'(x-3)\,e^{x^2}\,\mathrm{d}x$$ As you mention, on each of the intervals in the limit, the integral is defined, and the limit is $-6e^9$. $\endgroup$
    – robjohn
    Commented Feb 29, 2016 at 7:47
  • $\begingroup$ Had the integral been $\int_{\mathbb R}\delta(x) e^{x^2}$, we could indeed view it as Riemann-Stieltjes improper integral $\int_{\mathbb R} e^{x^2}\,d\theta(x)=1$. In other words, we'd view $\delta(x)$ as a measure - not as a distribution. But $\delta'$ is NOT a measure, it is strictly a distribution (generalized function), so you cannot call in R-S improper integral. $e^{x^2}$ simply lies outside of the class of functions $\delta'$ can be applied to. Otherwise, I could come up with a $\varphi_n\to 0$ s.t. $\int_{\mathbb R}\varphi_n e^{x^2}=\infty$ which I'm pretty sure is not kosher. $\endgroup$
    – A.S.
    Commented Feb 29, 2016 at 17:31
  • $\begingroup$ @A.S.: I've taken the liberty of fixing the typo in my previous comment ($\delta\to\delta'$). The support of $\delta'$ is $\{0\}$. While $e^{x^2}$ is not a standard test function because it does not display the proper decay at $\infty$, for any distribution with compact support, the "test" function can be modified outside the compact support and not change the application of the distribution. $\endgroup$
    – robjohn
    Commented Feb 29, 2016 at 17:52
  • $\begingroup$ Then you're restricting consideration to compactly supported distributions only (e.g. take $d=\sum_{n\in\mathbb Z} a_n\delta'_n$ with proper $a_n>0$ - we simply cannot use your rule to apply $d$ to $e^{x^2}$) and forgoing continuity (as I mentioned above). Also, strictly speaking, compactness of a distribution ($\delta'$ in particular) is based on its action on test-functions and hence cannot be extended to its action on non-test-functions. I do see how we can assign value to the integral in question, but it seems unstable and I question usefulness of this extension. What would it be? $\endgroup$
    – A.S.
    Commented Feb 29, 2016 at 18:42

The best way to study the Dirac function is in the context of measures or distributions, but if you are willing to accept as a definition for $\delta $ the functional equation

$\tag1\int_{x_{0}-\epsilon}^{x_{0}+\epsilon}f(x)\delta (x-x_{0})dx=f(x_{0}),\quad \forall \epsilon>0$ then it follows easily that

$\tag2\int_{-\infty}^{\infty}f(x)\delta' (x-x_{0})dx=-\int_{-\infty}^{\infty}\frac{d f(x)}{d x}\delta (x-x_{0})dx$

so with $f(x)=e^{x^{2}}$, you get, using $(1)$ and $(2)$,

$\tag3\displaystyle \int_{-\infty}^{+\infty} \delta'(x-3)e^{x^2}dx=-6e^9$


The expression written is ill-defined, but look at the improper integral as the limit

$$ \int_{-\infty}^{\infty} \delta '(x-3) e^{x^2} dx = \lim_{L, \Lambda \to \infty}\int_{-\Lambda}^{L} \delta '(x-3)e^{x^2} dx \\ =\lim_{L, \Lambda \to \infty}\left( \delta (x-3) e^{x^2}\biggr\rvert_{-\Lambda}^{+L} - 2 \int_{-\Lambda}^{L} \delta (x-3)xe^{x^2}dx \right) = -6e^{9} $$ And for $L$ large enough, $L>3$, we can be sure that the first term vanishes when evaluated on the limits, before taking the limits, hence the result you have follows from the remaining term

  • $\begingroup$ Do you mean that because $\delta(x) \cdot e^{x^2} = 0, $ almost everywhere it has to vanish? Or something else? Sorry if it is simple. $\endgroup$ Commented Aug 29, 2015 at 15:00
  • $\begingroup$ Indeed, the point is that improper integrals (those of unbounded functions over bounded intervals or unbounded functions over bounded intervals) are defined as the limit of proper integrals with finite boundaries. For each of these finite boundaries $[-\Lambda, L]$ you have $e^{x^2}$ finite but $\delta (x-3) =0$ (when $L >3$, i.e. large enough), so they are 0, taking the limit $L, \Lambda \to \infty$ will still give 0. $\endgroup$ Commented Aug 29, 2015 at 15:20
  • $\begingroup$ A word of caution, as you surely know the $\delta(x)$ is not an actual function but a distribution, i.e. it must be thought of as a smooth function depending on some parameter, then in some limit the parameter makes the smooth function singular, yet its integral remains well defined, over nice enough functions. There are subtleties when manipulating these expressions, which when not taken into account seem to lead to paradoxes. The theory of distributions studies all these things. $\endgroup$ Commented Aug 29, 2015 at 15:23
  • $\begingroup$ I have expanded my answer, I hope it is now of some help. $\endgroup$ Commented Aug 29, 2015 at 15:33
  • $\begingroup$ Thank you very much, that has cleared it up very well for me. $\endgroup$ Commented Aug 29, 2015 at 16:05

Here is a brief comment on the previous answers: Let $X \subset \Bbb{R}^d$ be open. Then

  • Distributions are elements of $\mathcal{D}'(X)$, where $\mathcal{D}'(X)$ is the dual of $C_c^{\infty}(X)$ equipped with a suitable topology.

  • $\mathcal{E}'(X)$ is the dual of $C^{\infty}(X)$ equipped with a suitable topology.

There is a natural way of identifying $\mathcal{E}'(X)$ as the space of compactly supported distributions. Consequently, people are often sloppy about making a clear distinction between these two notions.

In this problem, derivatives of $\delta$ are compactly supported and thus the problem leads us to two different interpretation:

  • The problem does not make sense when $\delta'$ is understood merely as distribution.

  • It does make sense when $\delta'$ is understood as an element of $\mathcal{E}'(\Bbb{R})$. Then the integral can be computed as @robjohn pointed out: introduce any cut-off function $\varphi \in C_c(\Bbb{R})$ such that $\varphi \equiv 1$ near $3$ and then replace $e^{x^2}$ by $e^{x^2}\varphi(x)$.

  • $\begingroup$ For completeness, you should include tempered distributions that act on Schwartz functions (Fourier transform and all) which basically limits distribution's growth to polynomial at most and puts them in-between the two cases you mentioned. Is restriction to compactly supported distributions frequently used and if so, where? $\endgroup$
    – A.S.
    Commented Feb 29, 2016 at 19:46
  • $\begingroup$ @A.S. I am afraid I have little idea on your question since my knowledge on this toplc is rectricted to the textbook level. I would be glad to hear about it If anyone knows it. $\endgroup$ Commented Feb 29, 2016 at 20:21

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