No, [itex]\int_{-\infty}^{\infty} \delta (a- x)f(x)dx= -f(a)[/itex]. Use the substitution u= a-x.pivoxa15 said:Homework Statement
int[d(x-a)f(x)dx]=f(a) is the dirac delta fn
but is int[d(a-x)f(x)]=f(a) as well? If so why?
That's a very rough way of putting it. More correctly the dirac delta function (actually a "generalized function" or "distribution") is DEFINED by the property that [itex]\int \delta(x) f(x)dx = f(0)[/itex] as long as the integration include x= 0.The Attempt at a Solution
Is it because at x=a, d(0)=infinite and integrate dirac delta over a region including x=0 when d(0) is in the value in the integral will produce 1 hence f(a).
The Dirac Delta Function is a mathematical function that is defined as zero everywhere except at a single point, where it is infinite. It is commonly represented by the symbol δ(x) or δa(x), where a is the point where the function is non-zero.
The Dirac Delta Function is often used in physics to model point masses or point charges. It is also used to represent impulsive forces or impulses in physical systems.
The integral of the Dirac Delta Function at a point a is equal to 1. This is because the function is defined as zero everywhere except at the point a, where it is infinite.
No, the Dirac Delta Function cannot be graphed in the traditional sense because it is not a continuous function. However, it can be represented using a spike or impulse at the point a.
The Kronecker Delta Function is a discrete version of the Dirac Delta Function. It is defined as 1 at the point a and 0 at all other points. In the limit as the spacing between the points approaches zero, the Kronecker Delta Function becomes the Dirac Delta Function.