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Help with integrating step function

  1. Jul 11, 2006 #1
    Dear All,

    I have a moderate knowledge of mathematics and need help on an integration question. How would I go about integrating a step function:
    H(K - Z), when the integral is from K=Z to infinity.

    Please advise, your help would be much appreciated.

  2. jcsd
  3. Jul 11, 2006 #2
    A step function is simply a constant that turns on at time Z in your case. The intergral of a constant can be written as C*t + C0 which is a linear linear function. For the case of a step function the initial value is zero at time t thus the equation takes the form C*t, however this also really does not jump up until time Z, so if we evaluate the integral from Z to some arbirary time T, then we have y = C*(T-Z) where C is the magnitude of the step function, T is some arbitrarty point in time, and Z is when the step function turns on. In your case the integral will tend to infinity because your step function never turns off.
  4. Jul 11, 2006 #3
    Additional help with integration

    Thanks for the comment, I kinda understand what you are saying but the full integral is as follows:

    Integrate H(e^(a+b*epsilon) -z)*(e^-0.5*epsilon^2) with respect to epsilon. H is the step function. How can this be done?

    Please advise. Thank u.
  5. Jul 12, 2006 #4


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    Let [itex]u= e^{(a+b\epsilon)-z}[/itex] so [itex]du= -u= e^{(a+b\epsilon)-z}= -udx[/itex] and so [itex]dx= -\frac{du}{u}[/itex]. When z= K, [itex]u= e^{(a+b\epsilon)-K}[/itex] and as z goes to infinity, u goes to 0. Letting L = [itex]u= e^{(a+b\epsilon)-K}[/itex] and using the negative to swap the direction of integration, your integral becomes
    Since H(u) is 1 for u> 0, this is simply
    That should be easy. (except for the minor detail that it "blows up" at 0.)
  6. Jul 12, 2006 #5


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    basic rule, a step function has piecewise constant height, hence piecewise linear integral.

    if your step function is constant from K to infinity, then its graph is linear from K to infinity, so the graph of the integral has form H.(X-K), for X > K, where H is the height. but it has value zero for X<K.
  7. Jul 12, 2006 #6
    If you integrate from zero to infinity...Fundamental Theory of Calculus talked about that, or the teacher did during that class...well it will certainly have an X in the answer, and you can ignore the lower boundry...
    Do you know your upper lmiit? does it have x in it?

    I mean from what I know...an integral can be defined finite (from a to b) or infinite (just the symbol...so assumed all area existent under curve.).

    So to go from something to infinity...that "infinity" must be something with X in it.

    To integrate, you take the Derivative of the upper boundry times the antiderivative of the function you integrate and you plug in the upper boundry.

    Okay, can someone check me on that? This is what I got wrong every single time I had to do in integration with bonudries w/ x in them. Do I andtiderivate, plug in the bonudry and multiply by the derivative fo the boundry? minus of course the same process for the lower boundry?

    :D plz help!
  8. Jul 12, 2006 #7


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    you need to distinguish definite from indefinite integrals. i assumed he meant an indefinite integral, i.e. a function.
  9. Jul 12, 2006 #8


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    Then you don't know the Fundamental Theory (theorem) of Calculus and you didn't listen to your teacher. A definite integral of a function of x, even if from 0 to infinity, is a number not a function of x. And you certainly cannot "ignore the lower boundary".
    [tex]\int_{1}^\infty \frac{dx}{x^2}= 1[/tex]
    [tex]\int_{2}^\infty \frac{dx}{x^2}= \frac{1}{2}[/tex]
    The integral here depends only on the lower boundary!

    ?? I seen many definitions of "infinity". I have never seen one that was a functionof x.

    No, you don't "take the derivative of the upper boundary". Since, typically, the upper boundary is a constant, that makes no sense.

    Were you responding to the question or asking another question? You make a series of assertions and then, at the very end, ask if they are correct or not. Okay- every one of your assertions is wrong (or, worse, meaningless). As every calculus book will tell you, to evaluate a definite integral, find the anti-derivative, evaluate at the "upper boundary" and "lower boundary" and subtract the latter from the former. That is one-half of the "fundamental theorem of calculus": If F'(x)= f(x) then
    [tex]\int_a^b f(x)dx= F(b)- F(a)[/tex]

    (The other half, no relevant here, is: If
    [tex]F(x)= \int_a^x f(t)dt[/tex]
    then F'(x)= f(x).)
  10. Jul 12, 2006 #9
    [tex]\int_0^x f(x)dx[/tex]

    This is what I understood from the orriginal question that the orriginal poster was talking about. maybe i'm terribly wrong. I never met an itegral going to infinity...with the actual "infinity" sign in it.

    Now what I said...let's take this example:

    [tex]\int_0^x f(x)dx[/tex] where f(x)=1/x

    Okay I can't get it to show what I want but the top bound i want 2x not plain x.

    By ignoring the lower bound (I feel like an idiot because my example itself is the abstraction to my own thought) I meant in a polynomial...most of the times if you plug in the zero...if you don't have a +C you don't have anything but a zero. It wasn't math, it was a "commodity" that I know it's not correct...I just wanted a shortcut that doesn't always work. Sorry about that confusion.

    so in that case, you'd go [tex]2 ln(2x) - ln(0)[/tex] correct?

    I realise now that I chose a very very idiotic example since Ln(0) is undefined but...I did a lot of boo-boos in this topic already. hehe.
    Last edited: Jul 12, 2006
  11. Jul 13, 2006 #10


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    If you mean
    [tex]\int_0^{2x} \frac{1}{t} dt [/tex]
    [tex]\int_0^{2x} \frac{1}{t} dt= ln(2x)- ln(0) [/tex]
    which, as you point out, doesn't exist because the ln(0) is not defined.

    In general,
    [tex]\int_a^b \frac{1}{t}dt= ln(b)- ln(a)= ln(\frac{b}{a})[/tex]
    whenever the logarithms exist.

    [tex]\int_1^{2x} \frac{1}{t}dt= ln(2x)- ln(1)= ln(2x)[/itex]
    No, you do not "multiply by the derivative of the upper bound".
  12. Jul 13, 2006 #11

    Either I was sleeping trough that class or my teacher was. I'll check it really fast...I'll psot results.

    Okay...i'm taking the follwoing example in my ti-83:

    [tex]\int_1^{2x} \frac{1}{t} dt[/tex]

    So I graph

    y=(fnint(1/x,x,1,2x))/(x>1) because I can't have xgo trough 0 or I get error.

    Yeah. I seem to be terribly wrong. Looks like I got some studying to do. Thank you for poniting out this...I'd never have figured it out that I learned it wrong.

    Edit: A little embarassing but does anynoe have a good link? Wikipedia only has integral from a to x explained...and I don't know what to ask in google...and wolfram mathematics seems to not cover this or I can't find it. I don't know how to describe it so a search engine can find something of the kind.

    Thank you in advance.
    Last edited: Jul 13, 2006
  13. Jul 13, 2006 #12

    matt grime

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    One more thing, Robokapp, since the integrand is a step function, and therefore not continuous, citing the FTC is both unhelpful and wrong since that requires the integrand to be continuous.
  14. Jul 13, 2006 #13
    Search for "the (second) fundamental theorem of calculus". Probably omit the parentheses.
  15. Jul 13, 2006 #14
    Not trying to suck the attention away from the orriginal question but..."integrand" is a term i never heard of in my life. logaritmig function is continuous though...past 0.

    to D-Lett I'll try. thank you!

    Edit: I think what I remember is maybe from taking the derivative of the antideirvative with a a boundry of 2x or so...That sort of makes sense.

    Edit2: yep. Doing F'(X) is what i was thinking of. here's the link in case someone wonders...It actually makes sense.
    Last edited: Jul 13, 2006
  16. Jul 14, 2006 #15


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    Then I suggest that you consult a calculus book- it's a very fundamental term. The "integrand" is the function f(x) in [itex]\int f(x)dx[/itex], the function being integrated.

    And the fact that the logarithm is continuous is irrelevant to matt grime's point- the step function is not. The original post (did you read it?) said nothing about a logarithm.
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