Dirac delta function – simple examples

Required math: calculus

Required physics: none

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.23.

Here are a few simple examples of integrals involving the Dirac delta function. The delta function is defined by the two conditions:

\displaystyle   \delta(x) \displaystyle  = \displaystyle  0\;\mathrm{if}\; x\ne0\ \ \ \ \ (1)
\displaystyle  \int_{-\infty}^{\infty}\delta(x)dx \displaystyle  = \displaystyle  1 \ \ \ \ \ (2)

Since it is zero everywhere except at {x=0} it follows that

\displaystyle  \int_{-\infty}^{\infty}f(x)\delta(x)dx=f(0) \ \ \ \ \ (3)

for any ‘ordinary’ function {f(x)}. A simple extension of this is

\displaystyle  \int_{a}^{b}f(x)\delta(x-k)dx=\begin{cases} f(k) & a<k<b\\ 0 & \mathrm{otherwise} \end{cases} \ \ \ \ \ (4)

This follows by making the substitution {y=x-k}. Then {dx=dy} and we get

\displaystyle  \int_{a}^{b}f(x)\delta(x-k)dx=\int_{a-k}^{b-k}f(y+k)\delta(y)dy \ \ \ \ \ (5)

This integral is {f(y=0)=f(k)} provided the limits of integration include {y=0}, that is, {a-k<0<b-k}, or {a<k<b}.

For example

\displaystyle   \int_{-3}^{1}f(x)\delta(x+2)dx \displaystyle  = \displaystyle  \int_{-3}^{1}\left(x^{3}-3x^{2}+2x-1\right)\delta(x+2)dx\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  f(-2)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -25 \ \ \ \ \ (8)

Another example:

\displaystyle   \int_{0}^{\infty}f(x)\delta(x-\pi)dx \displaystyle  = \displaystyle  \int_{0}^{\infty}\left[\cos\left(3x\right)+2\right]\delta(x-\pi)dx\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  f(\pi)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  1 \ \ \ \ \ (11)

And a final example:

\displaystyle  \int_{-1}^{1}e^{\left|x\right|+3}\delta(x-2)dx=0 \ \ \ \ \ (12)

since the limits of integration don’t include {x=2}.

4 thoughts on “Dirac delta function – simple examples

  1. Dami

    Hi, Please could you explain the last example better? I don’t understand the statement “since the limits of integration don’t include {x=2}. ” and how it applies to the solution. Thanks

    Reply
    1. gwrowe Post author

      Since {\delta\left(x-2\right)} is zero everywhere except at {x=2}, the integral {\int_{-1}^{1}f\left(x\right)\delta\left(x-2\right)dx=0} for any function {f\left(x\right)}, since the range of integration (from {x=-1} to {x=+1}) doesn’t contain the point {x=2}.

      Reply
    2. manoj

      r=2 I.e
      This value may be lies between lower limit and upper limerotherwise it should be zero
      If it lies between limit ex limit is -4 to 4
      Then we put r=2in f2f and get answer

      Reply

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