# How do you do an integral involving the derivative of a delta function?

17,679

## Solution 1

The $\delta$ function is not continuous, so it's a priori not differentiable. In fact, it's not even well-defined as an ordinary real-valued function, but can be made so in terms of distributions - linear maps on a space of test functions given by $f\mapsto\int\delta f=f(a)$.

It's possible to sensibly define derivatives of distributions by looking at representations as limits of functions:

If $\delta_i$ is a family of functions so that $\lim_{i\rightarrow\infty}\int\delta_i(x) f(x)\mathrm dx=f(a)$ for any test function $f$, then it can be considered a representation of the Dirac delta. Now, if we take the family of derivatives $\frac{\mathrm d}{\mathrm dx}\delta_i$ we arrive at $$\int\left[\frac{\mathrm d}{\mathrm dx}\delta_i(x)\right]f(x)\mathrm dx=-\int\delta_i(x)\left[\frac{\mathrm d}{\mathrm dx}f(x)\right]\mathrm dx$$ through integration by parts and using the fact that $f$ has by definition compact support (which makes the boundary term vanish).

As the derivative is linear as well, this defines another linear map $f\mapsto-\int\delta f'$ on the space of test functions, which we call the derivative of our distribution.

Symbolically, $$\left[\frac{\mathrm d}{\mathrm dx}\delta(x-a)\right]f(x)=-\delta(x-a)f'(x)$$ which you can just plug in into your formula above without any need for actual computation as it holds true by definition.

## Solution 2

The Dirac delta function is often defined as the following distribution:

$$\int_a^b \delta(x - x_0) F(x)\mathrm{d}x = \begin{cases}F(x_0), & a < x_0 < b \\ 0, & \text{otherwise}\end{cases}$$

where $F$ is a suitable test function. Its derivative is then defined as

$$\int_a^b \delta'(x - x_0) F(x)\mathrm{d}x = -\int_a^b \delta(x - x_0) F'(x)\mathrm{d}x$$

which is also the result one would get from naively applying integration by parts. You can use this result directly to calculate your integral by setting $F(x) = \frac{y(x)}{x}$ - no need to split the integral or take any limits.

## Solution 3

So, the properties of the derivative of the delta function can be shown relatively quickly though the following ansatz: Consider a function $\delta(x)$ such that $\delta(x) = \frac{1}{a^{2}}(x+a)$ if $-a<x<0$ and $\delta(x) = \frac{1}{a^{2}}(a-x)$ if $0<x<a$, and $\delta(x) = 0$ elsewhere. It is easy to see that $\delta(x)$ has area 1 irrespective of the value of $a$, so we can consider $\delta(x)$ to be the dirac delta function in the limit $a\rightarrow0$.

Now, consider the derivative of our putative delta function. It will be $\frac{1}{a^{2}}$ for $-a<x<0$ and $-\frac{1}{a^{2}}$ for $0<x<a$. Let's integrate a function $f(x)$ against $\delta^{\prime}(x)$:

\begin{align} \int \delta^{\prime}(x)f(x)dx &= \int_{-a}^{0}\frac{f(x)}{a^{2}}dx - \int_{0}^{a}\frac{f(x)}{a^{2}}dx\\ &=\int_{0}^{a}\frac{f(-x)}{a^{2}}dx-\int_{0}^{a}\frac{f(x)}{a^{2}}dx\\ \end{align}

Extracting the $a^{2}$ out of the integral, and taking the limit $a\rightarrow0$, we find, after applying L'Hopital's rule once, and then using the definition of the derivative:

$\int \delta^{\prime}(x)f(x) = -f'(0)$

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### nagendra

Updated on August 01, 2022

• nagendra 10 months

I got an integral in solving Schrodinger equation with delta function potential. It looks like

$$\int \frac{y(x)}{x} \frac{\mathrm{d}\delta(x-x_0)}{\mathrm{d}x}$$

I'm trying to solve this by splitting it into two integrals

$$\int_{-\infty}^{x_0 - \epsilon} \frac{y(x)}{x} \frac{\mathrm{d}\delta(x-x_0)}{\mathrm{d}x} + \int_{x_0 + \epsilon}^{\infty} \frac{y(x)}{x} \frac{\mathrm{d}\delta(x-x_0)}{\mathrm{d}x}$$

and then do the limit $\epsilon\to 0$. Could you tell me how to solve this integral please? I used Mathematica, it gave out a weird result.

• Jerry Schirmer over 10 years
Are you trying to evaluate $\int \left(\frac{y(x)}{x}\frac{d\delta(x-x_{0})}{dx}\right)dx$ for an arbitrary $y(x)$? I don't understand your statement about the limits, either, are you evaluating two separate integrals with limits $\int_{-\infty}^{x_{0}-\epsilon}$ and $\int_{x_{0}+\epsilon}^{\infty}$?
• nagendra over 10 years
Thanks Jerry. Yes it is. Basically it is a part of the radial part of my Schrodinger equation and y[x] is radial component and delta function is my potential function. There is a derivative of the potential function. I am trying to solve the equation for the delta function barrier about xo.Finally I can take the limit of e->0.
• David Z over 10 years
I wonder if perhaps this would be better off at Mathematics? (I'll migrate it if that is the case)
• Jerry Schirmer over 10 years
@DavidZaslavsky: it is primarily a mathematics question, but of the type that physicists will care about more than mathematicians. I'll answer this soon.
• David Z over 10 years
@Jerry yes, but I still think those sorts of questions should be sent to math.SE.
• nagendra over 10 years
Thanks Jerry. Hope it would work for the limiting point xo. Let me check it with my problem.
• nagendra over 10 years
Dear Christoph. Just to clarify that the derivatives is only for the DiracDelta function. What about in that case?
• Christoph over 10 years
@Nagendra: added some parens to clarify - I believe this is the case you're interested in