# Step Function

1. Aug 4, 2006

Suppose $$U(x)$$ is the step function and $$\delta(x)$$ is its derivative. Find $$\int^{6}_{-2}(x^{2}-8)\delta(x)\dx$$. I know $$\delta(x) = 0$$ for all $$x$$ except $$x = 0$$. So at $$x = 0$$ $$v(x) = - 8$$. After this step, how do we get $$\int_{-2}^{2}(x^{2}-8)\delta(x)\dx +\int^{6}_{2}(x^{2}-8)\delta(x)\dx = -8 + 0$$. How do you get the limits of integration?

Thanks

2. Aug 5, 2006

### benorin

If by step function you mean

$$U(x)=\left\{\begin{array}{cc}0,&\mbox{ if }x\leq 0\\1, & \mbox{ if } x\geq 0\end{array}\right.$$​

(a.k.a. the Heaviside step function) then $$\frac{d}{dx}U(x) = \delta (x)$$ is a Dirac delta function which has the so-called snifting property, that is

$$\int_{-\infty}^{\infty}f(x)\delta (x) \, dx = f(0)$$​

so the limits of integration don't matter (so long as they contain zero) and the value of the integral is -8.

3. Aug 5, 2006

Remember that you can split up the integral. So,

$$\int^{6}_{-2}(x^{2}-8)\delta(x)\dx = \int_{-2}^{0}(x^{2}-8)\delta(x)\dx = \int_{0}^{6}(x^{2}-8)\delta(x)\dx$$

$$\int^{6}_{-2}(x^{2}-8)\delta(x)\dx = \int_{-2}^{0}(x^{2}-8)\delta(x)\dx = \int_{0}^{1}(x^{2}-8)\delta(x)\dx =\int_{1}^{6}(x^{2}-8)\delta(x)\dx$$

or however you want (within the bounds of the original limits, etc...).

So the integral equals 0 for some limits of integration, it will hit 0 and the sifting property kicks in, and then the integral equals 0 again.

Last edited: Aug 5, 2006