# Thread: Dirac Delta Function Integral

1. ## Dirac Delta Function Integral

Hi.

Im having some trouble with the following question:

"Evaluate the integral: e^x [dirac(x + 1) − dirac(x − 2)] dx from -ve infinity to 1"

Am i supposed to use a Laplace transformation? In which case I split it into 2 integrals:

e^x * dirac(x + 1)

minus....
e^x * dirac(x − 2) dx

So for the first integral I get exp(p-1), and for the second exp(2-2p). So overall exp(p-1) - exp(2-2p).

However as the original integral was from -ve infinity to 1, the second dirac integral (at x=2) would be out of this boundary. So would I disregard this?

2. Why in the world would you use a "Laplace Transform"? $\int_{-\infty}^1 e^x[\delta(x+ 1)- \delta(x- 2)]dx= \int_{-\infty}^1 e^x\delta(x+1) dx- \int_{-\infty}^1 e^x\delta(x- 2)dx$
The definition of the Dirac delta function is that it is the functional that assigns to every function f(x) the value $\int_a^b f(x)\delta(x)dx= f(0)$ if 0 is in the interval [a, b], 0 otherwise. A simple change of variable extends that to $\int_a^b f(x)\delta(x- q)dx= f(q)$ if q is in the interval [a, b], 0 otherwise.
The first integral involves $\delta(x+ 1)= \delta(x-(-1))$ and -1 is in the interval $(-\infty, 1]$. The integral is $e^{-1}$.
The second integral involves $\delta(x- 2)$ but 2 is NOT in the interval $(-\infty, 1]$. The integral is 0.