# Derivatives of the delta function

As the Dirac delta function is essentially an infinitely high spike at a single point, it may seem odd that its derivatives can be defined. The derivatives are defined using the delta function’s integral property

 $\displaystyle \int_{-\infty}^{\infty}f\left(x\right)\delta\left(x\right)dx$ $\displaystyle =$ $\displaystyle f\left(0\right) \ \ \ \ \ (1)$

Consider the integral involving the ${n}$th derivative ${\delta^{\left(n\right)}\left(x\right)}$ and apply integration by parts:

 $\displaystyle \int_{-\infty}^{\infty}f\left(x\right)\delta^{\left(n\right)}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \left.f\left(x\right)\delta^{\left(n-1\right)}(x)\right|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}f^{\prime}\left(x\right)\delta^{\left(n-1\right)}\left(x\right)dx \ \ \ \ \ (2)$

The integrated term is taken to be zero, since the delta function itself is constant (at zero) for all ${x\ne0}$, so all its derivatives are zero except at ${x=0}$. Therefore ${\delta^{\left(n-1\right)}(x)=0}$ at the limits ${-\infty}$ and ${\infty}$. We’re therefore left with

 $\displaystyle \int_{-\infty}^{\infty}f\left(x\right)\delta^{\left(n\right)}\left(x\right)dx$ $\displaystyle =$ $\displaystyle -\int_{-\infty}^{\infty}f^{\prime}\left(x\right)\delta^{\left(n-1\right)}\left(x\right)dx \ \ \ \ \ (3)$

Since this is true for all functions ${f\left(x\right)}$, the integrands must be equal, so we get

$\displaystyle f\left(x\right)\delta^{\left(n\right)}\left(x\right)=-f^{\prime}\left(x\right)\delta^{\left(n-1\right)}\left(x\right) \ \ \ \ \ (4)$

A common case is the first derivative, which satisfies

$\displaystyle f\left(x\right)\delta^{\prime}\left(x\right)=-f^{\prime}\left(x\right)\delta\left(x\right) \ \ \ \ \ (5)$

If ${f\left(x\right)=x}$, we get the relation

$\displaystyle x\delta^{\prime}\left(x\right)=-\delta\left(x\right) \ \ \ \ \ (6)$

By iterating 4, we get

$\displaystyle f\left(x\right)\delta^{\left(n\right)}\left(x\right)=\left(-1\right)^{n}\delta\left(x\right)f^{\left(n\right)}\left(x\right) \ \ \ \ \ (7)$

Example 1 Suppose ${f\left(x\right)=4x^{2}-1}$. Then

 $\displaystyle \int_{-\infty}^{\infty}\left(4x^{2}-1\right)\delta^{\prime}\left(x-3\right)dx$ $\displaystyle =$ $\displaystyle -\int_{-\infty}^{\infty}8x\delta\left(x-3\right)dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -24 \ \ \ \ \ (9)$

Example 2 With ${f\left(x\right)=x^{n}}$ we have, using 7

$\displaystyle x^{n}\delta^{\left(n\right)}\left(x\right)=\left(-1\right)^{n}n!\delta\left(x\right) \ \ \ \ \ (10)$

Another use of the derivative of the delta function occurs frequently in quantum mechanics. In this case, we are faced with the integral

$\displaystyle \int\delta^{\prime}\left(x-x^{\prime}\right)f\left(x^{\prime}\right)dx^{\prime} \ \ \ \ \ (11)$

where the prime in ${\delta^{\prime}}$ refers to a derivative with respect to ${x}$, not ${x^{\prime}}$. Thus the variable in the derivative is not the same as the variable being integrated over, unlike the preceding cases. In this case, since only ${x}$ (and not ${x^{\prime}}$) is visible outside the integral, we can move the derivative outside the integral and get

 $\displaystyle \int\delta^{\prime}\left(x-x^{\prime}\right)f\left(x^{\prime}\right)dx^{\prime}$ $\displaystyle =$ $\displaystyle \frac{d}{dx}\int\delta\left(x-x^{\prime}\right)f\left(x^{\prime}\right)dx^{\prime}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle f^{\prime}\left(x\right) \ \ \ \ \ (13)$

Notice that in this case, there is no minus sign attached to the ${f^{\prime}}$ unlike in 5.