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Tricky Intregral for numerical quadrature

  1. Jan 21, 2015 #1
    Hi - I have just started 'Computational Physics' by Koonin & Meredith, - through distance learning.
    Exercise 1.3 needs a program to evaluate an integral - I'm finding myself kinda rusty on integrals. The hint says - split range of integration into parts, use different change of variable in each part... but I've spent most of a day trying different things (I suspect a trig substitution), so would appreciate a strong hint here - the actual exercise is to write the program :) Equation is ∫(t-2/3)(1-t)-1/3dt between 0 and 1.

  2. jcsd
  3. Jan 21, 2015 #2


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    The problem is that [itex]t^{-2/3}[/itex] goes to infinity as x goes to 0 and [itex](1- t)^{-1/3}[/itex] goes to infinity as x goes to 1. The suggestion is that you separate the interval into "0 to a" and "a to 1" where "a" is some number between 0 and 1, 1/2, say. Then, set x be equal to some function of t such that the function goes to infinity as t goes to 0.
  4. Jan 21, 2015 #3
    Hi, yes I was that far, my 'brick wall' is that I cant find any suitable substitutions to make - from which I would be able to figure out the transformed limits. I tried Integrating by parts and various substitutions, all of which complicated rather than simplified.... could you suggest a suitable split or substitution please?

  5. Jan 24, 2015 #4


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    $$t=\sin^3(x) \text{ or } t=\cos^3(x)$$

    that will remove the singularity on both sides
    you may or may not then wish to use
  6. Jan 24, 2015 #5
    I had tried t=sin3(x), probably I am making a mistake so would appreciate your checking as far as I got:
    dt/dx = 3sin2(x)cos(x) ⇒ dt=3sin2(x)cos(x)dx
    integral becomes ∫3cos(x) dx/(1-sin3(x))1/3)
    ...which still leaves me stumped....
  7. Jan 24, 2015 #6


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    That is right why are you stumped? The integral is very well behaved.
  8. Jan 24, 2015 #7
    Split integral into two, $$\int_0^1 t^{-2/3} (1-t)^{-1/3} dt = \int_0^{1/2} t^{-2/3} (1-t)^{-1/3} dt + \int_{1/2}^1 t^{-2/3} (1-t)^{-1/3} dt.$$ In the first integral the term ##(1-t)^{-1/3}## is bounded, so you only need to take care of the term ##t^{-2/3}##. Integrating ##\int t^{-2/3} dt = 3 t^{1/3} +C## you can see that using substitution ##u=t^{1/3}## gives you an integral of a bounded function and over a bounded interval, which you can then evaluate numerically. In fact, any substitution ##u = t^a##, ##0<a\le 1/3## works. but I feel the substitution ##u=t^{1/3}## works best.

    For the second integral you can act similarly, a substitution ##u =(1-t)^a##, ##0<a\le 2/3## works. Again I think ##a=2/3## would be best.
  9. Jan 25, 2015 #8
    Hi Lurflurf, it looks good to me too - until I try and resolve it. If it was 1-sin2 in the bracket or some simpler power ... then I could see a way forward, but I'm just fiddling without any intuition ... I will try Hankeye's approach tomorrow, getting late here - but thanks both.
  10. Jan 25, 2015 #9


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    as I mentioned above you can use

    $$\int_0^{\pi/2} \! \dfrac{\cos(x)}{\sqrt[3]{1-\sin^3(x)}}\, \mathrm{d}x=\int_0^{\pi/2} \! \sqrt[3]{\dfrac{\cos^3(x)}{1-\sin^3(x)}}\, \mathrm{d}x=\int_0^{\pi/2} \! \sqrt[3]{\cos(x)\dfrac{1+\sin(x)}{1+\sin(x)+\sin^2(x)}}\, \mathrm{d}x$$

    I am not sure that is better though
  11. Jan 25, 2015 #10


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    one other idea

    $$\int_0^1 \! t^{-2/3}(1-t)^{-1/3} \, \mathrm{d}t=\int_0^1 \! (t^{-2/3}+(1-t)^{-1/3}) \, \mathrm{d}t-\int_0^1 \! (t^{-2/3}+(1-t)^{-1/3}-t^{-2/3}(1-t)^{-1/3}) \, \mathrm{d}t$$
  12. Jan 25, 2015 #11
    Hi - Hawkeye's was easiest for me, please check my outcome:
    1st part: 3 ∫ (1 - u3))-1/3 du from u=0 to u= 0.125
    2nd part: -3/2 ∫ (1 - v3/2)-2/3 dv from v= 1 - (1/2)3/2 to 0.0
    I have checked this and it seems right, but would just appreciate confirmation - the program I wrote using this isn't giving the expected result...
  13. Jan 25, 2015 #12
    Just correcting my typos above, sorry:
    1st part: 3 ∫ (1 - u3)-/13 du - from u=0 to u= 0.125
    2nd part: -3/2 ∫ (1 - v3/2)-2/3 dv - from v= (1/2)3/2 to v = 0.0
    I have checked this and it seems right, but would just appreciate confirmation - the program I wrote using this isn't giving the expected result...
  14. Jan 26, 2015 #13


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    I got
    $$3\int_0^{2^{-1/3}} \! (1-u^{3})^{-1/3} \, \mathrm{d}u=\dfrac{\pi}{\sqrt{3}}+\log(2) \\ \dfrac{3}{2}\int_0^{2^{-2/3}} \! (1-u^{3/2})^{-2/3} \, \mathrm{d}u=\dfrac{\pi}{\sqrt{3}}-\log(2)$$
  15. Jan 26, 2015 #14
    Upper bound in the first part should be ##2^{-1/3}##, probably that is why you are getting an unexpected answer. The integrals should be as in the https://www.physicsforums.com/threads/tricky-intregral-for-numerical-quadrature.793526/members/lurflurf.19666/ [Broken]'s post above: try to use them in your program.

    PS. If you plug the integrals in Mathematica (or something similar) you get the answers as in the https://www.physicsforums.com/threads/tricky-intregral-for-numerical-quadrature.793526/members/lurflurf.19666/ [Broken]'s post. Or, if you plug the original integral, you get ##\frac{2\pi}{\sqrt 3}##. But the exact integration is quite complicated, the antiderivative of the original function involves hypergeometric functions, and I will not discuss it here.

    I understand that you do not need an exact answer, you need to compute the integral numerically. Applying standard numerical methods to integrals in the https://www.physicsforums.com/threads/tricky-intregral-for-numerical-quadrature.793526/members/lurflurf.19666/ [Broken]'s post you should get the result. The reason that you need to transform the integral before using a numerical method is that standard numerical methods do not work well when functions have singularities (when functions are unbounded).
    Last edited by a moderator: May 7, 2017
  16. Jan 26, 2015 #15
    Hi - I think I have the same bounds, first part with u=t1/3 I have the upper as (1/2)1/3 - which is the same as 2-1/3 - but the latter does look 'better'. Similarly the 2nd upper bound I have 1/23/2 which is the same as 2-3/2. I assume this will make no difference? ..... so - now to recheck the program....
  17. Jan 26, 2015 #16
    While I am checking the body - I am using the composite simpsons rule - here are the functions I coded for each part ...
    Function FUNC_1(u) !calc 3* [(1-(u**3))**(-1/3)]
    integer, parameter :: dp = selected_real_kind(15, 307)
    real(kind=dp) :: u, FUNC_1, temp, out
    ! print 10, 'u= ',u, ' temp= ', temp, ' out= ', out !debug
    Func_1= out
    10 format (3(A,F10.7))
    End Function

    Function FUNC_2(v) !calc (3/2)* [(1-(v**(3/2))**(-2/3)]
    integer, parameter :: dp = selected_real_kind(15, 307)
    real(kind=dp) :: v, FUNC_2, out
    ! print 10, 'v= ',v, ' out= ', out !debug
    Func_2= out
    10 format (2(A,F10.7))
    End Function
  18. Jan 30, 2015 #17
    Hi again, haven't managed to find the problem, my program looks OK but doesn't output values close enough to the actual answer to be correct; must be something wrong. Also it doesn't converge with smaller step size. The above functions are coded in fortran95, can anyone see an error in the way I have implemented the 2 integrals?
  19. Mar 18, 2015 #18
    Been going over both the math and the program and still can't find the error. To summarize the math:

    The problem I'm trying to solve numerically, is the integral wrt t, between 0 and 1, of f(t)=(t(-2/3))*(1-t)(-1/3).
    Both the limits give singularities, so I have to split the integral into 2 parts, each part deals with 1 of the singularities. Arbitrarily split them at 0 -> 1/2, 1/2 -> 1

    Part A) Let u=t(1/3). Then 3du=t(-2/3)dt and t=u3
    Limits: 0 ≤ t ≤ 1/2, gives 0 ≤ u ≤ (1/2)(1/3)

    So 1st part becomes Integral wrt u, between (a in the program) 0 and (b) (1/2) (1/3), of g(u)=3*(1-u3)(-1/3)

    B) Let v=(1-t)(2/3). Then -3/2dv=(1-t)(-1/3)dt and t=1-v(3/2)
    Limits (1/2)(3/2) ≤ t ≤1, gives (1/2)(3/2) ≤ v ≤ 0

    So 2nd part becomes integral wrt v, of g(v) =(-3/2)*(1-v(3/2))(-2/3)
    I then swapped the limits to remove the - in front of the integral, so limits are c=0 to d=(1/2)(3/2)
    Then the program should estimate the 2 integrals using simpsons method.
    h is the step size, I use the same number of steps in each part, which means a different step size; I'm pretty sure this is OK.
    Some of my coding is from me trying to find the error, so its become a bit verbose to help me check through it using debugger - the program seems to do exactly what I expect from manual calcs - so after many hours at this I just can't see where I have gone wrong - math or program.....

    I have posted the program here - https://drive.google.com/file/d/0ByNoaXX7FNqRSVpfY1hKekR4Vk0/view?usp=sharing, click and it should display correctly in a browser.
  20. Mar 18, 2015 #19
    You got the upper limit in the second integral wrong, it should be ##(1/2)^{2/3}##, not ##(1/2)^{3/2}##, see the post # 13 for the correct integrals. And there are coefficient in front of each integral, did you include them? I could not see in your program where you add 2 integrals.
  21. Mar 18, 2015 #20
    Many thanks about the limit, dont know how I kept mixing that up. (I included the coefficients in the 2 functions (Eg out=3*out) )
    I've just tried the program and it behaves much better, never completely diverges though - the best error I get is 0.0000000026 for N=96 (once it was working I improved the exact calc with a full _DP value for pi).
    Do you know if the simpson method ever can converge exactly? (I suspect not)
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