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Something gone horribly wrong (Integration)

  1. Feb 5, 2004 #1

    Zurtex

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    Hi all, I've just recently been going on maths forums to find out why this problem occurs. I understand how to work out the solution but not why the problem occurs in the first place. Anyway here it is:

    Where y = cosh(x).e^x work out the general integral of y with respect to x.

    As integral(uv') = uv - integral(u'v)
    Say u = cosh(x), u'=sinh(x), v'=e^x, v=e^x

    So integral(cosh(x).e^x) = cosh(x).e^x - integral(sinh(x).e^x) : Equation 1

    Taking integral(sinh(x).e^x), u=sinh(x), u'=cosh(x), v'=e^x, v=e^x
    integral(sinh(x).e^x) = sinh(x).e^x - integral(cosh(x).e^x) : Equation 2

    Substituting equation 2 into equation 1

    integral(cosh(x).e^x) = cosh(x).e^x - sinh(x).e^x + integral(cosh(x).e^x)
    cosh(x).e^x - sinh(x).e^x = 0
    cosh(x).e^x = sinh(x).e^x
    cosh(x) = sinh(x)
     
  2. jcsd
  3. Feb 5, 2004 #2

    HallsofIvy

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    Didn't we have a post similar to this recently? You have to be very careful with indefinite integrals- remember that "constant of integration".

    What's true is not that ex(cosh x- sinh x)= 0 but that it is a constant.

    cosh x= (ex+e-x)/2 and
    sinh x= (ex-e-x)/2 so that
    (cosh x- siny x)= e-x.

    Then ex(cosh x- sinh x)= 1.
     
  4. Feb 5, 2004 #3

    Zurtex

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    Sorry I don't understand ho that relates to the problem, where exactly have I gone wrong please?
     
  5. Feb 5, 2004 #4
    There should be an "i" after the 2, I believe.
     
  6. Feb 5, 2004 #5

    matt grime

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    no i is missing from that statement. there is an h missing from somewhere above though (sinh sin mistake)

    as for the OP just write cosh in terms of exponentials, and do the resulting integral of

    integrate (1+e^{2x))/2 wrtx
     
  7. Feb 5, 2004 #6

    Zurtex

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    Your thinking of trig not hpyerbolic.
     
  8. Feb 5, 2004 #7

    Zurtex

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    Like I said I know the solution just not why my method end basically on the statement that 1 = -1
     
  9. Feb 5, 2004 #8
    Whoops, I thought that sinh x = sin(ix) for some reason.
     
  10. Feb 5, 2004 #9

    matt grime

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    That's because you're missing the constants of integration out, as someone as mentioned.

    I guess I was wondering why you'd use integration by parts whe it isnt' required, esp if you do it twice.
     
  11. Feb 5, 2004 #10

    Zurtex

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    Surely as I am dealing with the same integration the constants cancel each other out? Just like when you do an integral between limits.

    This is a point that has always confused me. Surely if that was true you could not write the integral[f(x)].integral[f(x)] = {integral[f(x)]}^2 because the 2 different constants would muck it all up.

    Very interested :smile:
    Err not done this yet (next topic) but doesn't sinh x = i.sin(ix)? Thus prooving osborns rule if you say ix = z, or summat...
     
  12. Feb 5, 2004 #11

    matt grime

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    Go through the integration with the lower limit as a, and the upper limit as x, and see what happens.

    And you aren't even doing the same integrals, so it cannot be the same integration.

    Also you're mixing up definite and indefinite things - the limits need some care.
     
  13. Feb 5, 2004 #12

    Zurtex

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    O.K, so I put limits in, I follow what I've done before and get to this point:

    integral[lim a,x](cosh(x).e^x) = cosh(x).e^x - sinh(x).e^x + integral[lim a,x](cosh(x).e^x)

    As I actually know cosh(x).e^x = [e^2x + 1]/2, I know this intergrates to [e^2x + 2x]/4 + C, between the limits x and a this goes to [e^2x + 2x]/4 - [e^2a + 2a]/4

    Substitution this into the equation we get:

    [e^2x + 2x]/4 - [e^2a + 2a]/4 = cosh(x).e^x - sinh(x).e^x + [e^2x + 2x]/4 - [e^2a + 2a]/4

    0 = e^x[cosh(x) - sinh(x)]

    As shown earlier e^x[cosh(x) - sinh(x)] = 1

    0 = 1

    I'm still confused exactly where this goes wrong.
     
  14. Feb 6, 2004 #13

    matt grime

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    you put the limits only in a few places, you have


    int vdu/dx = uv - int udv/dx


    it's the uv part where you're going wrong.

    you must evaulate uv at the limits x and a. By the way you're using x as a variable and a limit. DON'T DO THIS! Use a dummy variable inside the integral if you need to.


    in any case, when you evaluate uv

    what you get is


    e^x(coshx-sinhx) - e^a(cosha-sinha)=0

    which is reassuring, as you know both terms are 1.
     
  15. Feb 6, 2004 #14

    Zurtex

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    Right thanks.

    I think I am just about getting this. Help much appreciated, my teacher did not know why it went wrong (I'm only at A level at the moment).
     
  16. Feb 6, 2004 #15

    HallsofIvy

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    You are absolutely correct about THAT. If "integral[f(x)]dx" is and indefinite integral then it would be incorrect to write "integral[f(x)].integral[f(x)] = {integral[f(x)]}^2" because the two integrals on the left side are not necessarily the same. If however, the integral were "from a to b" for some specific a and b or "from a to x" for some specific a, then it would be correct.
     
  17. Feb 6, 2004 #16

    matt grime

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    Perhaps you should remember (it's not really taught at a-level) that the indefinite integral of f(x)dx can be treated as the integral from a to x of f(u)du with a an arbitrary constant. Constants don't quite match, though.

    Using integration by parts for indefinite integrals does take some care - and it always goes wrong cos of the limits thing.
     
    Last edited: Feb 6, 2004
  18. Feb 7, 2004 #17

    HallsofIvy

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    Actually saying " integral[f(x)].integral[f(x)] = {integral[f(x)]}^2" is not only wrong- it doesn't make sense. (As one famous mathematician {and I can't remember who) would say "It's not even wrong"!)
    integral[f(x)] in the sense of "indefinite" integral, without limits of integration, is not a number or function, it is a class of functions all differing from one another by a constant. "multiplication" of such classes is not defined.

    I think the basic error here is in thinking of "integral[f(x)]" as if it were a number or function when, in fact, it is not.

    [tex]\int_a^bf(x)dx[/tex] is a number.

    [tex]\int_a^xf(t)dt[/tex] is a function of x. (Choosing the lower limit, a, is the same as choosing a specfic constant.)
     
    Last edited: Feb 7, 2004
  19. Feb 8, 2004 #18

    Zurtex

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    Well thanks again. Having read over this again and having a bit of a think about I understand integration a little bit better :smile:. I am doing Maths and Further Maths at A level and it doesn't explain it very well.
     
  20. Feb 8, 2004 #19
    That would be Wolfgang Pauli. He said, "This isn't right. This isn't even wrong."

    I'd consider him a physicist.
     
  21. Feb 8, 2004 #20

    HallsofIvy

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    Thanks, meister (well chosen web-name!), I'll try to remember that.
    Darn that it's a physicist rather than a mathematician!
     
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