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Integral [0,5] (w/(w-2)

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    integral [0,5] (w/(w-2))

    2. Relevant equations

    I just want to know if does not converge or not.

    3. The attempt at a solution


    ∫2(1(w-2)) dw + ∫dw

    = 2ln| w-2| + w from 0 to 5. I know about the discontinuity at w = 2.
    So is it still ok from me to just evaluate this as normal.
    I would get

    2ln|3| + 5 - 2ln|-2|

    ? Is this OK?
     
  2. jcsd
  3. Oct 8, 2013 #2

    LCKurtz

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    No. You must evaluate ##\int_0^2 + \int_2^5## and they must both converge.
     
  4. Oct 8, 2013 #3
    Confused to I evaluate ∫2(1(w-2)) dw + ∫dw from [0,2] then again from 2 to 5? The whole thing right?
     
  5. Oct 8, 2013 #4

    haruspex

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    If the range of an integral includes a point where the integrand is undefined then you must be careful. The integral as a whole is undefined if there is any subrange on which it is undefined.
     
  6. Oct 8, 2013 #5

    Mark44

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    Yes, if I understand your garbled question. This integral--
    $$\int_0^5 \frac{w}{w - 2}$$
    -- is an improper integral because the integrand is undefined at w = 2. You have to split the original integral into two improper integrals on the intervals [0, 2] and [2, 5], and evaluate each using limits. Both integrals have to converge so that the original integral converges.

    Also, try to be more careful with what you write. You started off with an integrand of w/(w - 2). In both of your last two posts you have somehow turned this into 2(1(w -2)) + 1, which is not the same.
     
  7. Oct 8, 2013 #6
    I just divided and I got 2/(w-2) +w is what it should say.

    So I got 2 (lim) ln|w-2| +w

    evaluate it between 0 and 2 and I got 2ln|0| +2 - (ln|-2| +2) I did not do the other one yet because since I have the ln of 0 does that mean that this doesn't converge? So I can stop here right?
     
  8. Oct 8, 2013 #7

    Mark44

    Staff: Mentor

    Yes. Neither of the log terms is defined. Since one integral diverges, the whole integral (on the interval [0, 5]) must also diverge.
     
  9. Oct 8, 2013 #8
    OK. Thanks a lot mark I appreciate it.
     
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