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

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:

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

Author by

nagendra

Updated on August 01, 2022