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.

7 thoughts on “Derivatives of the delta function

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