one hard integral for Maple: int(exp(cos(x))*cos(x-sin(x)),x=0..2*Pi);

Discussion in 'Maple' started by Nasser M. Abbasi, Dec 6, 2009.

  1. FYI,

    Maple 12

    Can do this integral numerically, but not analytically:
    f:=exp(cos(x))*cos(x-sin(x));
    int(f,x=0..2*Pi);
    ------> returns unevaluated
    f:=exp(cos(x))*cos(x-sin(x));
    evalf(Int(f,x=0..2*Pi));

    6.283185307

    OK.

    I do not have Maple13, but was wondering if it can do it in Maple13?
    --Nasser
     
    Nasser M. Abbasi, Dec 6, 2009
    #1
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  2. Nasser M. Abbasi

    Axel Vogt Guest


    Plotting and playing suggests f:= x -> exp(cos(x))*cos(x-sin(x)) is
    periodic and symmetric like cos, actually both f(Pi+x) - f(Pi-x) and
    f(2*Pi+x) - f(x) are zero, so the task is 2*Int(f(x), x= 0 .. Pi)

    Smells like a Fourier thing ... playing with exp(cos(x)+I*sin(x)) and
    differentiating gives

    Int(f(x),x=0..Pi); expand(%);


    Pi
    /
    |
    | exp(cos(x)) cos(x) cos(sin(x))
    |
    /
    0

    + exp(cos(x)) sin(x) sin(sin(x)) dx


    And exp(cos(x)+I*sin(-x)); evalc(Im(%)); diff(%, x);


    -exp(cos(x)) cos(x) cos(sin(x)) + exp(cos(x)) sin(x) sin(sin(x))


    Now -exp(cos(x))*sin(sin(x)) is 0 in 0 and in Pi and an antiderivative,
    hence the task is

    Int(2*exp(cos(x))*cos(x)*cos(sin(x)),x = 0 .. Pi)

    which Maple 12 evaluates to be Pi.

    Done.
     
    Axel Vogt, Dec 6, 2009
    #2
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  3. hi Alex;

    But Pi is the WRONG answer. It should be 2 Pi?

    --Nasser
     
    Nasser M. Abbasi, Dec 6, 2009
    #3
  4. Maple 11, 12, 13 can do it if we expand f first:

    int(expand(f), x=0..2*Pi);

    The result is 2*Pi.
     
    Rouben Rostamian, Dec 6, 2009
    #4
  5. Nasser M. Abbasi

    Axel Vogt Guest

    As said: your original integral is twice of it. Axel
     
    Axel Vogt, Dec 6, 2009
    #5
  6. Nasser M. Abbasi

    Axel Vogt Guest

    Much better than my way ...
     
    Axel Vogt, Dec 6, 2009
    #6
  7. Alex, With the transformation you did, I was not sure if Pi you showed above
    is the answer to the same problem I had.

    The analytical answer to

    f:=exp(cos(x))*cos(x-sin(x));
    int(f,x=0..2*Pi);

    should be 2 Pi, not Pi.

    --Nasser
     
    Nasser M. Abbasi, Dec 6, 2009
    #7
  8. Nasser M. Abbasi

    Fırat Guest

    I want to ask new integral such as

    f:=(t-a)^c/(b-t)^d; #a,b,c,d are contant

    How can I found analytical integral solution

    Firat
     
    Fırat, Dec 8, 2009
    #8
  9. Nasser M. Abbasi

    G. A. Edgar Guest

    Well, int(t^c*(1-t)^d,t=0..1); is the Beta function.
    Change variables x=(t-a)/(b-a) to reduce yours to the case a=0, b=1.
    Then since yours is the indefinite integral, it would be called an
    "incomplete Beta function" I suppose. Maple reports
    Int(x^c*(1-x)^(-d), x) = x^(c+1)*hypergeom([d, c+1], [2+c], x)/(c+1)
     
    G. A. Edgar, Dec 9, 2009
    #9
  10. Nasser M. Abbasi

    Fırat Guest


    Thanks for your answer. But I have some problem. I wrote my maple file
    as you say in the following form:

    with(student):
    F:=Int((t-a)^c/(b-t)^d,t);

    / c
    | (t - a)
    F := | -------- dt
    | d
    / (b - t)
    / c
    | (x b - x a) (b - a)
    | -------------------- dx
    | d
    / (b - a - x b + x a)
    / c
    | (x b - 1. x a) (b - 1. a)
    | -------------------------- dx
    | d
    / (b - 1. a - 1. x b + x a)

    but it does not give any answer. Main problem is: this integral form
    appear in my iteration statement and I have to use this integral value
    for new iteration after previous iteration. I want to integrate this
    function and get numerical value.

    Is this possible?

    Firat
     
    Fırat, Dec 10, 2009
    #10
  11. Nasser M. Abbasi

    Ray Vickson Guest

    You cannot get a numerical answer unless you specify numerical values
    for a, b, c and the integration limits. Maple cannot read your mind.
    You can write answer_above = (b-a)^(c+1-d) * int(x^c/(1-x)^d,x), and
    then express this last integral in terms of an incomplete beta
    function (or a hypergeometric function, as Maple seems to prefer).
    Of course not. You have not said what are the values of a, b, c and
    the integration limits.

    R.G. Vickson
     
    Ray Vickson, Dec 10, 2009
    #11
  12. Nasser M. Abbasi

    Fırat Guest

    In my previous message I said the power c and d are non-integer value.
    I give some values as you say but it does not evulate integral for
    example
     
    Fırat, Dec 10, 2009
    #12
  13. Nasser M. Abbasi

    Ray Vickson Guest

    Here is what I get using Maple 9.5:
    restart;params:={c=0.3,d=0.5};

    params := {c = 0.3, d = 0.5}
    (c + 1)
    x hypergeom([d, c + 1], [2 + c], x)
    J := ------------------------------------------
    c + 1

    J1:=subs(params,J);

    1.3
    J1 := 0.7692307692 x hypergeom([0.5, 1.3], [2.3], x)

    Of course, if we give numerical values for the integration limits we
    will get a numerical evaluation:
    J2:=subs(params,int(x^c/(1-x)^d,x=0 .. .97)): value(J2);
    1.362551855


    R.G. Vickson
     
    Ray Vickson, Dec 11, 2009
    #13
  14. Nasser M. Abbasi

    Fırat Guest


    Thanks for your kindly help and understanding. I am beginner in Maple.
    I want to ask a question. I want to use the first example in Maplesoft
    webpage

    http://www.maplesoft.com/support/he...Student/NumericalAnalysis/InitialValueProblem

    I wrote as follow
    `Error, NumericalAnalysis is not a command in the Student package\n`

    and give above error message.

    I want write this in Maple 10 classical worksheet. Please let me know
    where I did wrong.

    Firat
     
    Fırat, Dec 11, 2009
    #14
  15. I hope what you wrote was actually

    with(Student[NumericalAnalysis]);
    The NumericalAnalysis subpackage was first introduced in Maple 13. It
    does not exist in Maple 10.
     
    Robert Israel, Dec 11, 2009
    #15
  16. Nasser M. Abbasi

    Bill Guest

    Actually, a little simplify will produce the desired result directly:

    y := exp(cos(x))*cos(x-sin(x)):
    ti := int(y, x=0..2*Pi):
    simplify(ti, [trig]);
    2 Pi

    Bill
     
    Bill, Dec 11, 2009
    #16
  17. Nasser M. Abbasi

    Fırat Guest


    Thanks for your answer
     
    Fırat, Dec 12, 2009
    #17
  18. Nasser M. Abbasi

    Archimedes Guest

    Using simplify with the trig option is square brackets is equivalent
    to issuing the command "expand" first, and then simplifying the
    output. So simplify(ti, trig); is not the same as simplify(ti,
    [trig]); Is this documented somewhere? I did not see this in the
    help pages for simplify, or simplify/siderels

    Regards,
    Georgios
     
    Archimedes, Dec 13, 2009
    #18
  19. Nasser M. Abbasi

    Bill Guest

    I agree that the help pages don't explain this well. If you want to
    see what the difference is, try executing
    and then run the other commands. It's quite a bit different. Also try
    'expand. to see how it calls 'simplify.'

    Bill
    Mele Kalikimaka to all
     
    Bill, Dec 14, 2009
    #19
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