How Do You Compute the Integral of a Delta Function with a Quadratic Argument?

[thenewbosco]

let f(y)=\int_0^2 \delta(y-x(2-x))dx. Find f(y) and plot it from -2 to 2.

I know how to calculate \delta (g(x)) but i am not sure how to treat it with the y. I thought possibly to solve the quadratic in the delta function to find what x will equal for the roots in terms of y and got 1+(1-y)^1/2 and 1-(1-y)^1/2. i am not sure though how to find f(y) with the integral. any suggestions?

 

[arildno]

Treat the cases separately:
1. The singularity lies in the open interval between the end points
2. The singularity lies outside the interval
3. The singularity lies at an end point

 

[HallsofIvy]

The singularity arildno refers to is where y- x(2-x)= 0. In other words when y= 2x- x². Find the value of that integral, f(y), for y in the positions arildno listed and that's your function.

 

[thenewbosco]

so what youre saying is to form the integral f(y)=\int 2x-x^2 dx?
if its outside the interval do i let the limits be (-infinity to 0) and (2 to infinity)
for inside 0 to 2 but at the endpoints what are the limits?

 

[arildno]

No, why do you think that?

 

[HallsofIvy]

What is \int_a^b \delta(x)dx for any a, b?

 

[D H]

What is \int_a^b \delta(x)dx for any a, b?

Even more appropriately,

What is \int_a^b \delta(x)f(x)dx for any a, b?

One way to solve the problem at hand is to convert the original problem,
f(y)=\int_0^2 \delta(y-x(2-x))dx
to the form
f(y)=\int_{u_0}^{u_1} \delta(u)f(u)du

Note that the integrand may have two singularities between the integration limits, depending on the value of y.