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Any suggestions on how to solve the next integral?

  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the value of the integration:

    2. Relevant equations
    Mod note: Edited the TeX to render properly at this site.
    Here is the integration written using MAketex:
    ##\frac{.5}{\sqrt{\pi}}\int_0^{\infty}\exp(-z(1+1.5/v))z^{L-.5} \frac{1}{(\frac{z}{v}+1)^n (\frac{z}{v}+.5)^q} dz##


    3. The attempt at a solution
    I tried solving it using the binomial coeffiecent but when i tried to plot it, the plot was wrong, so any other suggestions of how to solve it?
     
  2. jcsd
  3. Mar 11, 2017 #2

    Ray Vickson

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    Do you need an actual formula, or is doing it numerically (for given numbers v, L, n ,q) good enough?

    If you want an ugly and almoset-intractable formula, you can get one. Here is how.

    In different notation, you can express your integral in terms of ##\int_0^{\infty} f(z) \, dz,## where
    $$f(z) = \frac{\exp(-az) z^b}{(z+r)^n (z+s)^q}.$$
    This appears to be non-elementary, but if you want an infinite series of non-elementary (but known functions) you can expand ##(z+s)^{-q}## as an infinite (binomial) series, then get a series involving terms of the form ##\int_0^{\infty} g(z) \, dz,## where
    $$g(z) = \frac{\exp(-az) z^m}{(z+r)^n}, \; m = b, b+1, b+2, \ldots.$$
    For ##a,m,r > 0##, Maple is able to perform this integral in terms of generalized Laguerre functions, but it is not pretty.

    Probably just using numerical integration is the way to go.
     
  4. Mar 11, 2017 #3
    I've already tried the binomial coefficient replacement you suggested, and tried to plot it and the plot was so wrong! can i ask why? if the binomial coefficient is correct and i'm pretty sure that the derivation is correct, why am i getting wrong output? what is wrong about this binomial coefficient? does it have any conditions?

    How can i do numerical integration? i use Matlab and Mathematica for my codes.
     
  5. Mar 11, 2017 #4

    Ray Vickson

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    So, did you get an infinite series of generalized Laguerre functions and things? I cannot understand what you did, or what you got. You need to tell us more details.

    I don't have either Matlab of Mathematica, but I am sure you can perform the integration numerically in Mathematica, just by giving the appropriate instruction. In Maple (which is like Mathematica in most respects) we can get a numerical evaluation just by saying evalf(Int(f,z=0..infinity)). (If you want 30-digit accuracy, just say "evalf[30](int(f,z=0..infinity))", or "evalf[100]... for 100-digit accuracy, etc.) That command will call on Maple's built-in numerical integration schemes, which in turn, may try and test out different schemes to achieve the required accuracy---but all of that takes place in the background. There are also ways of testing different numerical integration schemes "manually"---but still with no coding needed. I suppose something similar can be done in Mathematica.
     
  6. Mar 11, 2017 #5
    Fine, i will try it now, i have a question, can we plot the result of the numerical integration? i need to plot the function, it is more important than finding the result.
     
  7. Mar 11, 2017 #6

    Ray Vickson

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    You have four input parameters, v, L, n, q. Do you want to hold L,n,q constant (at some numerical values) and plot the integral as a function of v, or do you want to get a 3d-plot in which you hold (say) n and q constant and plot surface z = I(v,L) of integral values as a function of the two variables (v,L), or what? It is meaningless to ask about plotting until you say what, exactly, you want to plot.
     
  8. Mar 11, 2017 #7
    before thinking of numerical integration, i wanted to plot the function vs v, since for sure the variable "z" is no longer available, is this changes if it is a numerical integration?
     
  9. Mar 11, 2017 #8

    Ray Vickson

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    No, integration is integration, numerical or otherwise. The variable z is "integrated out", so will not appear in the answer. The answer will depend on v, L, n and q, and not on anything else.

    Anyway, I have been asking you a series of questions, and you have not answered any of them.
     
  10. Mar 11, 2017 #9
    ok,
    about the series, i tried to replace the term (1/(1+z/v))^n by its series representation which is the binomial coefficient http://mathworld.wolfram.com/BinomialCoefficient.html, please see (8), and i did the same for the other term, which is(1/(.5+z/v))^q, and then i continued with the integration.
     
  11. Mar 11, 2017 #10
    L,n,q are all indices in the constant c, v can take any value.
     
  12. Mar 11, 2017 #11

    Ray Vickson

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    I think I see the problem: you made a fatal error, as did I also in post #2.

    The problem lies with expanding ##(z/v+1)^{-n} (z/v + 1/2)^{-q}## using binomial series. For ##x, a > 0##, the series for ##(1+x/a)^{-p}## converges only if ##x < a##, so integrating it out to ##\infty## is not allowed, and probably produces nonsensical results. We can write ##(z/v + 1/2)^{-q} = 2^q (1 + z /(v/2))^{-q},## so we can expand it in a binomial series if ##z < v/2##. If ##z > v/2## we have ##(z/v + 1/2)^{-q} = 2^q [z/(v/2)]^{-q} (1 + (v/2)/z )^{-q},## which can be expanded in a binomial series. So, if ##C(u,v)## denotes the binomial coefficient we have
    $$(z/v+ 1/2)^{-q} = \begin{cases}
    2^q \sum_{j=0}^{\infty} C(-q,j) (2/v)^j z^j&\text{if} \;\; 0 < z < v/2 \\
    v^q \sum_{j=0}^{\infty} C(-q,j) (v/2)^j z^{-j-q} &\text{if} \;\; v/2 < z < \infty
    \end{cases}
    $$
    Similarly,
    $$ (z/v+1)^{-n} = \begin{cases}
    \sum_{k=0}^{\infty} C(-n,k)/v^k z^k & \text{if} \;\; 0 < z < v\\
    v^n \sum_{k=0}^{\infty} C(-n,k) v^k z^{-k-n} & \text{if} \;\; v < z < \infty
    \end{cases}$$

    Therefore, we need to split up the integration region into three parts: ##0 < z < v/2##, ##v/2 < z < v## and ##z > v##. We have different series expansions in each part, and each part involves an integral like ##\int_a^b \exp(-cz) z^{p} \, dz##. If ##L \geq 1/2## the part for ##0 < z < v/2## involves only positive powers ##p##, so those integrals converge (that is give no problem at the lower limit ##z = 0##). The other two parts can involve some terms with ##p < 0##, but since ##z > v/2 > 0## there is no difficulty with convergence of those integrals either. For all these reasons our integrals ##\int_a^b \exp(-cz) z^{p} \, dz## can be evaluated in terms of the (non-elementary) incomplete gamma function. Therefore, the final answer involves three separate doubly-infinite series of incomplete gamma functions!

    This is bordering on the un-doable, so again, numerical analysis is the preferred way to go.
     
    Last edited: Mar 12, 2017
  13. Apr 4, 2017 #12
    Hello, sorry for the delay, but i need to ask you something, the integral from v/2 to v , can i solve it as:https://wikimedia.org/api/rest_v1/media/math/render/svg/c3221b60dd5db844ea4835922394ea7cd1b1cb35 ?? from : https://en.wikipedia.org/wiki/Incomplete_gamma_function, under the topic: integral representation
    Integral representation
     
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