Integrals with Dirac delta function, $\int\delta[(x-a)(x-b)]f(x)\, dx $

Question

I am struggling to find the result of the following integrals with dirac delta function.

Why are they true? For the second one, I thought $\delta(x_1-x_2)$ must be zero?

Answer 1

The first is answered on the wikipedia page

http://en.wikipedia.org/wiki/Dirac_delta_function#Composition_with_a_function

and the second one, informally, just sets $F(x):=\delta(x_2-x)f(x)$ and evaluates the other delta.

Answer 2

For example, if I define a function $\delta_{\Delta_0}(x)$ like this:

$$ \delta_{\Delta_0}(x) = \lim_{\Delta_1\to 0^+} \int\limits_{-\infty}^{\infty} \Big( \frac{1}{(\pi \Delta_1^2)^{1/2}} e^{-(x-x_1)^2/\Delta_1^2}\Big)\Big( \frac{1}{(\pi \Delta_0^2)^{1/2}} e^{-x_1^2/\Delta_0^2}\Big) dx_1 $$

then the following is true:

$$ \lim_{\Delta_0\to 0^+} \int\limits_{-\infty}^{\infty} \delta_{\Delta_0}(x-x_0) f(x_0)dx_0 = f(x) $$

Unlike the heuristic equations, these two equations are real equations which have the same property as the dirac delta function. When working or first learning this heuristic function, you can then work out the integrals using this equation and convince yourself the above are true.

Answer 3

Frim my viewpoint, the Dirac Delta function is de facto a Riemann-Stieltjes integral with integrator as a step function. (c.f. Rudin's Principle of Mathematical Analysis theorem 6.15)

we know that $\int_a^b f(x)\delta(\alpha(x)) = \int_a^b (f(x)\delta(\alpha(x))/dx)dt$ when $\alpha$ is differentiable and f*$\alpha'$ is Riemann integralble.(c.f. Rudin theorem 6.17) We generalise this idea, let h(x) be the step function, which equals 0 for positive x,1 for positiv x and 0.5 for x=0. it follows that h is differentiable for x not equals 0 and $\frac{dh}{dx}=0$ for non-zero t. let's pretend that h is differentiable on the entire real line, and $\frac{dh}{dx}$ at x=0 is equal to infinity. Now we get Dirac Delta function.

When f is continuous at x=0 (for a linear charge distribution, $\lambda=q*\delta(x)$,and writing as R-S integral gives $\int_a^b q dh(x)$ where q is constant, the charge of the point charge.)this R-S integral gives f(0) which is what we desire when using delta function.

And i think it's easy to handle the OP by R-S integral.

Edit: we can, too. view the function as the limit of convolution of the function and a family of "good kernels". the goddness of this notion is it can take the arithmetic average of the lhs limit and rhs limit, so to avoid of problem of the non-integrability.