1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with integrating heaviside function.

  1. Sep 24, 2011 #1
    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. jcsd
  3. Sep 24, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
  4. Sep 24, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The integral of the Heaviside function, H(x), is just xH(x)+ C.
     
  5. Sep 24, 2011 #4
    Use integration by parts and the fact that [itex]h'(x - x_{0}) = \delta(x - x_{0})[/itex]:

    [tex]
    \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}
    [/tex]

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

    [tex]
    \int_{a}^{b}{x \, \delta(x - x_{0}) \, dx} = x_{0} \, h(x_{0} - a) \, h(b - x_{0})
    [/tex]

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

    [tex]
    b \, h(b - x_{0}) - a \, h(a - x_{0})
    [/tex]

    Combining everything together gives:

    [tex]
    \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})
    [/tex]
     
  6. Sep 24, 2011 #5
    As a check, let us differentiate it w.r.t. [itex]b[/itex]:

    [tex]
    \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})
    [/tex]

    [tex]
    \frac{\partial}{\partial b} \left( -a \, h(a - x_{0}) \right) = 0
    [/tex]

    [tex]
    \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})
    [/tex]

    where again we used the derivative of the Heaviside step function and a consequence of the properties of the Dirac delta-function ([itex]x \, \delta(x - x_{0}) = x_{0} \, \delta(x - x_{0})[/itex]) as well as the assumption that [itex]b > a[/itex]. Combining everything together and substituting [itex]b \rightarrow x[/itex]. we see that we get the integrand. Also, if [itex]b = a[/itex], the first two terms cancel and the last cancels because [itex]h(a - x_{0}) h(x_{0} - a)[/itex] is always zero. So, according to the fundamental theorem of calculus, we should get the correct result.
     
    Last edited: Sep 24, 2011
  7. Sep 24, 2011 #6
    Of course, you can always write:

    [tex]
    \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]
    [/tex]

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

    and do the remaining elementary integrals :)
     
  8. Sep 25, 2011 #7
    Thanks guys! I think I can figure it out now
     
  9. Sep 25, 2011 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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

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

    For the "indefinite integral" [itex]\int H(x)dx[/itex] is 0+ C for x< 0, x+ C for [itex]x\ge 0[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problem with integrating heaviside function.
  1. Heaviside function (Replies: 4)

Loading...