# 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

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, or ${a.

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

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 ${x=2}$.

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