# Problem with integrating heaviside function.

1. Sep 24, 2011

### perpich08

1. The problem statement, all variables and given/known data
I need to find the integral of the following equation..

integral from 0 to 8 of {(2H[x-0]+2H[x-4])*(x/8)}dx

3. The attempt at a solution
Im am not sure what the integral of the heaviside function is?
is it 1 or 0?

Any help would be appreciated!
Thanks

2. Sep 24, 2011

### LCKurtz

Write your function as a piecewise defined function and you will see how to integrate it.

f(x) = ? if 0 < x < 4
f(x) = ? if 4 < x < 8

Remember H(x) = 0 if x < 0 and H(x) = 1 if x > 0.

Also drawing the graph might help.

Last edited: Sep 24, 2011
3. Sep 24, 2011

### HallsofIvy

Staff Emeritus
The integral of the Heaviside function, H(x), is just xH(x)+ C.

4. Sep 24, 2011

### Dickfore

Use integration by parts and the fact that $h'(x - x_{0}) = \delta(x - x_{0})$:

$$\int_{a}^{b}{h(x - x_{0}) \, dx} = \left. x \, h(x - x_{0}) \right|^{b}_{a} - \int_{a}^{b}{x \, \delta(x - x_{0}) \, dx}$$

Next, the integral with the Dirac delta-function can be evaluated using its property:

$$\int_{a}^{b}{x \, \delta(x - x_{0}) \, dx} = x_{0} \, h(x_{0} - a) \, h(b - x_{0})$$

where the product of the two Heaviside functions ensures that $x_{0}$ is inside the segment $x_{0} \in \left[a, b \right]$. The integrated-out part is:

$$b \, h(b - x_{0}) - a \, h(a - x_{0})$$

Combining everything together gives:

$$\int_{a}^{b}{h(x - x_{0}) \, dx} = b \, h(b - x_{0}) - a \, h(a - x_{0}) - x_{0} \, h(x_{0} - a) \, h(b - x_{0})$$

5. Sep 24, 2011

### Dickfore

As a check, let us differentiate it w.r.t. $b$:

$$\frac{\partial}{\partial b} \left( b \, h(b - x_{0}) \right) = 1 \cdot h(b - x_{0}) + b \cdot \delta(b - x_{0}) = h(b - x_{0}) + x_{0} \, \delta(b - x_{0})$$

$$\frac{\partial}{\partial b} \left( -a \, h(a - x_{0}) \right) = 0$$

$$\frac{\partial}{\partial b} \left( -x_{0} \, h(b - x_{0}) h(x_{0} - a) \right) = -x_{0} \, \delta(b - x_{0}) \, h(x_{0} - a) = -x_{0} \, \delta(b - x_{0}) \, h(b - a) = -x_{0} \delta(b - x_{0})$$

where again we used the derivative of the Heaviside step function and a consequence of the properties of the Dirac delta-function ($x \, \delta(x - x_{0}) = x_{0} \, \delta(x - x_{0})$) as well as the assumption that $b > a$. Combining everything together and substituting $b \rightarrow x$. we see that we get the integrand. Also, if $b = a$, the first two terms cancel and the last cancels because $h(a - x_{0}) h(x_{0} - a)$ is always zero. So, according to the fundamental theorem of calculus, we should get the correct result.

Last edited: Sep 24, 2011
6. Sep 24, 2011

### Dickfore

Of course, you can always write:

$$\int_{0}^{8}{\left( 2 h(x) + 2 h(x - 4) \right) \frac{x}{8} \, dx} = \frac{1}{4} \, \left[\int_{0}^{8}{x \, h(x) \, dx} + \int_{0}^{8}{x \, h(x - 4) \, dx}\right]$$

$$\frac{1}{4} \, \left[ \int_{0}^{8}{x \, dx} + \int_{4}^{8}{x \, dx}\right]$$

and do the remaining elementary integrals :)

7. Sep 25, 2011

### perpich08

Thanks guys! I think I can figure it out now

8. Sep 25, 2011

### HallsofIvy

Staff Emeritus
I would consider using integration by parts for integrating the Heaviside function "overkill".

If a<0, then H(x)= 0 for all $x\le 0$ so $\int_{-\infty}^a H(x)dx= 0$. For $a\ge 0$, $\int_{-\infty}^a H(x)dx= \int_0^a H(x)dx$$= \int_0^a dx= a$.

For the "indefinite integral" $\int H(x)dx$ is 0+ C for x< 0, x+ C for $x\ge 0$.