# Simplifying the integral of dirac delta functions

1. Oct 21, 2009

### sleventh

hello all,
i am unaware of how to handle a delta function. from what i read online the integral will be 1 from one point to another since at zero the "function" is infinite. overall though i dont think i know the material well enough to trust my answer. and help on how to take the integral of a dirac delta function would be much appreciated. the kind of problems im dealing with are similar to
$$\int$$e^(-x^2) $$\delta$$(x+1)[1-cos(5 (pi/2) x)] dx from -infinity to 0
for this i solved the delta function for when x+1=0 since that is the only time when the function will have a value. i then subbed -1 for all x's and took the integral, assuming the delta function became 1.
thank you very much for help

2. Oct 21, 2009

### tiny-tim

hello sleventh!

(have a delta: δ )
no, there's no integral in the result …

δ is defined by the property ∫ δ(x) f(x) dx = f(0) for any f.

Put F(x) = f(x-1).

Then ∫ δ(x+1) f(x) dx = ∫ δ(x) f(x-1) dx = ∫ δ(x) F(x) dx = F(0) = f(-1).

Do you see how the δ function gets rid of the integral?

3. Oct 21, 2009

### sleventh

hey tiny-tim. thank you very much. that clears up a lot. so if you were to be taking the integral of the above problem from limits 0-∞ (thanks for the copy and paste) would the problem have the same solution as -∞ -0 because you are still subbing in 0 to the f(x-1) adaptation of the function?

4. Oct 21, 2009

### tiny-tim

Hi sleventh!

Any limits have to strictly include the zero of what's inside the δ (which in this case is -1) …

so it can be ∫-∞ or ∫-∞0 or ∫-20,

but not ∫0 or even ∫-1 or ∫-∞-1