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Can anyone help me solve this Integration of three terms?

  1. Mar 2, 2015 #1
    • Member warned about posting without the template and with no effort
    I have been trying to solve an integration that i have

    upload_2015-3-3_0-54-32.png
    I am not even sure if it's possible. Here, A, m, alpha, a these are constants. I have tried few methods, but couldn't find any way out. I would appreciate any help.
     
  2. jcsd
  3. Mar 2, 2015 #2

    Ray Vickson

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    Please show us first some of the methods you HAVE tried, whether or not they failed. (Those are PF requirements.)
     
  4. Mar 2, 2015 #3
    Thank you so much for replying.
    Like is said, i tried with multiple integration method but individual terms were also complicated, like


    upload_2015-3-3_1-45-55.png

    then i don't know where to go.
     
  5. Mar 2, 2015 #4

    Ray Vickson

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    If I were doing it I would simplify it first, by re-defining some of the variables and introducing some new constants. If we start with
    [tex] f(t) = t^2 e^{-t^2/a} \frac{(A+mt)^{1/m}}{1+\alpha (A + mt)^{1/m}}, [/tex]
    and assume that ##a > 0##, we can let ##t/\sqrt{a} = v##, ##C = a^{3/2} A^{1/m}##, ##k = m \sqrt{a}/A##, ##b = \alpha A^{1/m}## and, finally, ##p = 1/m##. We then have the cleaner-looking result that
    [tex] \int_0^{\infty} f(t) \, dt = C \int_0^{\infty} v^2 e^{-v^2} \frac{(1+k v)^p}{1 + b(1+k v)^p} \, dv [/tex]
    If we further let ##1 + kv = y## (assuming ##k > 0##) we can get
    [tex] \text{integral} = \frac{C e^{1/k}}{k^3} \int_1^{\infty} \frac{(y-1)^2 y^p}{1+b y^p} e^{-y/k} \, dy [/tex]
    I doubt that there is a closed-form formula for this, but one can tackle it numerically if you are given the parameter values.
     
    Last edited: Mar 3, 2015
  6. Mar 2, 2015 #5

    Dear Ray

    Thank you so much.

    I already see some lights.

    Now as you said, if the parameter values were given, then it could be solved. Now, you were right in assuming a>0, because a = 0.1, m = 1, 2, 3, 4, 5 upto 20. So, yes k>0. But I want to understand how did you come up with all the right assumptions?? By the way, so now i have the integration, how do we do the integration ? Still i have multiple terms in the integration.
     
    Last edited: Mar 2, 2015
  7. Mar 2, 2015 #6

    Ray Vickson

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    Looking at the original integral, we need a > 0 in order to have a convergent integral. Why is that? Well, the integrand is not defined at all if a = 0, and the integral diverges exponentially if a < 0. Next, we need A > 0 in order that (A + mt) be ≥ 0 for all t > 0. This is needed because if (A + mt) < 0 in a range of t-values, then (A + mt)^(1/m) will involve roots of negative numbers, so will involve complex numbers, etc. Finally, we need α ≥ 0 in order to not have the denominator go through 0 at t increases up from 0. If the denominator did go through 0, we would need to worry about the possible divergence of the integral, and whether to interpret as a principle-value integral, or whatever. All these issues are avoided if we just assume all your original parameters a, A,m,α are > 0.

    As to how to integrate: NUMERICALLY! Or, perhaps you can expand things in convergent infinite series and then integrate term-by-term.
     
  8. Mar 3, 2015 #7
    Dear Sir

    Thank you so much for you reply and thorough analysis.

    Sir, in the 4th line, should it be exp(-v^2) ?
    And sir, i tried with MATLAB, it's showing values. I don't know how they got the value.

    and what did you mean by "expand things in convergent infinite series", how do i expand that term, especially that y^p/(1+b*y^p) format?

    Thanks in advance.
     
    Last edited: Mar 3, 2015
  9. Mar 3, 2015 #8

    Ray

    I mean, as i said earlier m>1, so that means p is fractional. How do i expand that fractional power thing?
     
  10. Mar 3, 2015 #9
    How can y be a function of t if the integral is over t?

    Chet
     
  11. Mar 3, 2015 #10
    Does the answer have to be closed form, or can it be expressed as an infinite series?

    Chet
     
  12. Mar 3, 2015 #11

    Ray Vickson

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    Yes, the 4th line should have exp(-v^2), not exp(-v). I have edited out this typo.

    Exactly what input did you use with MATLAB, and what type of answer did MATLAB deliver?
     
  13. Mar 3, 2015 #12
    Dear Chet

    The answer should be just a value, for the given parameters, so it should be in closed form, but now i doubt if it's possible. So, finite or infinite solution, please do enlighten me.
     
    Last edited: Mar 3, 2015
  14. Mar 3, 2015 #13
    Ray

    I simply wrote the function, integrated using built-in integration function.

    bt=0.02;
    s=1;
    g=3;
    a=0.1;
    k=g*s/bt;
    p=1/g;
    b=a*bt^p;
    C=2*s*bt^p;

    fun=@(y) (C/k^3)*((y-1).^2.*exp(-(y-1)./k).^2.*y.^p)./(1+b.*y.^p);
    m=integral(fun,0,Inf);

    the m is now 0.6839.
     
  15. Mar 3, 2015 #14

    Ray Vickson

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    So, did MATLAB give you a numerical answer, or did it give you a "formula"?
     
  16. Mar 3, 2015 #15

    MATLAB gave me a numerical answer i need to know the how and the formula.
     
  17. Mar 3, 2015 #16

    Ray Vickson

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    I suspect that there is NO possible formula. I did say---more than once---that I guessed you would need to do numerical integration when looking at a numerical example (that is, with numerical values of the inputs). MATLAB probably tried some "analytical" methods, found it could not be done, and then applied one of its built-in numerical integration routines.
     
  18. Mar 3, 2015 #17

    Ray

    Surely your reply helped me a lot, now i understand the problem at least. Thank you so much for your time.
    Using MATLAB, i would probably get the numerical value but i wont be able to explain how i got that.
    That's why i needed the explanation. I will share if i can find anything.

    Thank you all.
     
  19. Mar 3, 2015 #18

    Ray Vickson

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    Please give the original parameters in your first post. Here you have bt, s and g, none of which were present in the original post. Just tell us the values of m, a, A, and α, as they appeared in post #1. I want to check your numerical value, but to do that I need to know exactly what you computed.

    Anyway, it looks like you have entered (in part) exp(-(y-1)/k)^2 , which is = exp(-2(y-1)/k), instead of the correct form exp(-[ (y-1)/k ls]^2) = exp(-(y-1)^2/k^2), which is very different.
     
  20. Mar 3, 2015 #19


    Dear Ray


    upload_2015-3-3_23-31-27.png
    Another constant can be multiplied sometimes which is (2/a^4), this is optional.

    There is a little change from the 1st post, it's a^2 instead of a. It's a constant, so doesn't make any difference.
    Other values are like alpha = 0.1, a=1, m=3 (usually >1), A = 0.02.

    You are right about the exponential term, my mistake.
     
  21. Mar 3, 2015 #20
    Shouldn't those x's be t's?

    Chet
     
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