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When I try to evaluate the following integral, it runs for a much longer time than I think is should without producing a result. What might be the problem? Can anybody please point it out?

h[s_, t_] := HeavisideTheta[s] HeavisideTheta[t - s]
Integrate[Abs[(b*(1 - e^((-b)*s)))/b - h[s, t]], {s, 0, Infinity}]
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closed as too localized by Jens, m_goldberg, belisarius, István Zachar, Simon Woods Apr 8 '13 at 8:35

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1 Answer

Granting the beginner's mistake of confusing e with E (not really the culprit here), let's look at your integrand:

Simplify[Abs[(b*(1 - e^((-b)*s)))/b - h[s, t]]]

(* ==> Abs[1 - e^(-b s) - HeavisideTheta[s] HeavisideTheta[-s + t]] *)

it becomes clear that the integral over $s$ to $\infty$ is not convergent for $b\neq 0$. So you should go back and check your math. When you encounter such integration problems, it's a good idea to test the expression for some reasonable numerical choice of the parameters (b and t in particular)!

In this case you perhaps confused multiplication by the HeavysideTheta terms with subtraction.

Because I'm guessing that this is a math error on the user end, I'm proposing to close this question as too localized.

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It may still be good to have an answer here, since people who encounter similar problems should be encouraged to do some sanity checking before posting to StackExchange... –  Jens Apr 8 '13 at 3:55
    
well, thank you for your response, but i evaluated the integral by hand too, and there is no problem, it is convergent. I am trying to find out what is wrong with the mathematica code. I am correcting e -> E, however, i am not following your comment on heavisidetheta –  No Leaf Clover Apr 8 '13 at 4:09
    
@jens Are you sure about convergence? I'm not on my pc so can't check this, but I believe the integral converges for positive b. –  Sjoerd C. de Vries Apr 8 '13 at 5:26
    
@SjoerdC.deVries The integrand goes to 1 outside the unit box, so it doesn't converge. –  Jens Apr 8 '13 at 5:52
    
@NoLeafClover Of course I can't really guess what your goal is. But the step functions have no effect for s > t, and the remaining terms don't go to zero, so the infinite integral goes to infinity. –  Jens Apr 8 '13 at 6:00
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