Integral of delta function multiplied by a Heaviside step function

[jht529100]

Homework Statement

consider two functions:ψ(x) which is eqaul to zero at a,that is ψ(a)=0
and f(x)=H(x-a)*β(x)+(1-H(x-a))*γ(x)
where H(x-a) is the heaviside step function and β(x),γ(x) is the continuous function.
it seems that the derivative of f(x) is not exist.
the question is whether ∫δ'(x-a)*ψ(x)*f(x)dx make sense?
δ'(x-a) is the derivative of the delta function.
if the above integral make sense,how to get the value of it?
Any hints will be grateful.:shy:

 

[lippyka]

Homework Equations The Attempt at a SolutionI think the integral make sense.Since ψ(x)=0 at x=a,so ∫δ'(x-a)*ψ(x)*f(x)dx will be zero.