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Integral of dn/(n^2 - 4) using trig substitution with sine

  1. Oct 2, 2016 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{}^{∞} \frac{1}{n^2 - 4} dn [/tex]
    2. Relevant equations
    I'm trying to do this a way that it isn't usually done. Normally this is done with partial fractions. I'm trying to do it by using trig substitution using sine, which requires some algebraic manipulation. For some reason, no matter what I do I end up with a different function from what various integral calculators give me.

    The reason I'm doing this is this integral represents the infinite sum of the same function from n=3 to infinity, and I'm trying to show whether or not it converges. My hunch is that it converges. The reason I'm trying to do it this way is just for practice. But sadly I hit a wall. Somewhere I'm making a mistake I'm not seeing.

    3. The attempt at a solution
    I'll show every step so hopefully it will be easy to see where I mess up.

    **I'm editing this to put numbers next to each step so it is easy to refer to. That is the only change I'm making**

    [tex]\mbox{1. }\int_{}^{∞} \frac{1}{n^2 - 4} dn [/tex]
    [tex]\mbox{2. }\int_{}^{∞} \frac{1}{-4(1 - \frac{n^2}{4})} dn [/tex]
    [tex]\mbox{3. }-\frac{1}{4}\int_{}^{∞} \frac{1}{(1 - \frac{n^2}{4})} dn [/tex]
    [tex]\mbox{4. }-\frac{1}{4}\int_{}^{∞} \frac{1}{(1 - [\frac{n}{2}]^2)} dn [/tex]
    [tex]\mbox{5. }\mbox{Let u =} \frac{n}{2}. \mbox{Then 2du = dn} [/tex]
    [tex]\mbox{6. }-\frac{1}{4}\int_{}^{∞} \frac{1}{(1 - u^2)} 2du [/tex]
    [tex]\mbox{7. }-\frac{1}{2}\int_{}^{∞} \frac{1}{(1 - u^2)} du [/tex]
    Now it's in the form I recognize for substituting with sine.
    [tex]\mbox{8. }\mbox{Let u =} \sin{θ} \mbox{. Then du =} \cos{θ}dθ[/tex]
    [tex]\mbox{9. }-\frac{1}{2}\int_{}^{∞} \frac{\cos{θ}}{(1 - \sin^2{θ})} dθ[/tex]
    [tex]\mbox{10. }-\frac{1}{2}\int_{}^{∞} \frac{\cos{θ}}{(\cos^2{θ})} dθ[/tex]
    [tex]\mbox{11. }-\frac{1}{2}\int_{}^{∞} \frac{1}{(\cos{θ})} dθ[/tex]
    [tex]\mbox{12. }-\frac{1}{2}\int_{}^{∞} \sec{θ} dθ[/tex]
    I just looked up the antiderivative of secant because it won't highlight the problem I'm having.
    [tex]\mbox{13. }-\frac{1}{2}\ln{|\sec{θ}+\tan{θ}|} [/tex]
    [tex]\mbox{14. }-\frac{1}{2}\ln{|\frac{1}{\cos{θ}}+\frac{\sin{θ}}{\cos{θ}}|} [/tex]
    [tex]\mbox{15. }-\frac{1}{2}\ln{|\frac{1+\sin{θ}}{\cos{θ}}|} [/tex]
    [tex]\mbox{16. }-\frac{1}{2}(\ln{|1+\sin{θ}|}-\ln{|\cos{θ}|}) [/tex]
    [tex]\mbox{17. }\mbox{Noting that u =} \sin{θ} \mbox{ which means}\cos{θ}=\sqrt{1-u^2}[/tex]
    [tex]\mbox{18. }-\frac{1}{2}(\ln{|1+u|}-\ln{|\sqrt{1-u^2}|})[/tex]
    [tex]\mbox{19. }-\frac{1}{2}(\ln{|1+u|}-\frac{1}{2}\ln{|1-u^2|}) [/tex]
    [tex]\mbox{20. }-\frac{1}{2}\ln{|1+u|} + \frac{1}{4}\ln{|1-u^2|}) [/tex]
    [tex]\mbox{21. }\mbox{Noting that u =} \frac{n}{2} [/tex]
    [tex]\mbox{22. }-\frac{1}{2}\ln{|1+\frac{n}{2}|} \mbox{ } +\mbox{ } \frac{1}{4}\ln{|1-\frac{n^2}{4}|} [/tex]

    and the upper bound is infinity.

    Now, here's the thing. This seems to be pretty close to the right answer, which according to integral calculators is:

    [tex]\frac{1}{4}\ln{|n-2|} \mbox{ } -\mbox{ } \frac{1}{4}\ln{|n+2|} [/tex]


    I know this is not the optimal method to try to solve this, but hopefully someone will see my mistake. Thanks for reading through this. I know it's a lot.
     
    Last edited: Oct 2, 2016
  2. jcsd
  3. Oct 2, 2016 #2
    I'm not sure how you get from
    −1/2 ln|1+u| - 1/2 ln|1−u2|)
    to
    −1/2 ln|1+u| + 1/4 ln|1−u2|)

    I think you should factorize (1-u2) to (1+u)(1-u) and continue from there...
     
  4. Oct 2, 2016 #3
    I just distributed the
    -1/2 in

    [tex]-\frac{1}{2}(\ln{|1+u|}-\frac{1}{2}\ln{|1-u^2|})[/tex]


    I don't know how to make larger parentheses yet, which would make it easier to read. Not seeing it anywhere. But thanks I will look into the difference of squares.
     
  5. Oct 2, 2016 #4
    Okay, so upon factoring as you suggested, I got even closer.



    [tex]-\frac{1}{2}(\ln{|1+u|}-\frac{1}{2}\ln{|1-u^2|})[/tex]
    [tex]-\frac{1}{2}(\ln{|1+u|}-\frac{1}{2}\ln{|(1+u)(1-u)|})[/tex]
    [tex]-\frac{1}{2}(\ln{|1+u|}-\frac{1}{2}\ln{|(1+u)|} - \frac{1}{2}\ln{|(1-u)|})[/tex]

    Then I distribute the -1/2 and get:
    [tex]-\frac{1}{2}\ln{|1+u|}\mbox{ }+\mbox{ }\frac{1}{4}\ln{|(1+u)|} + \frac{1}{4}\ln{|(1-u)|}[/tex]
    and combining like terms gives me

    [tex]-\frac{1}{4}\ln{|1+u|}\mbox{ }+ \frac{1}{4}\ln{|(1-u)|}[/tex]

    or

    [tex]\frac{1}{4}\ln{|(1-u)|}\mbox{ }-\frac{1}{4}\ln{|1+u|}[/tex]

    And this is so very close. The problem is when I sub u= n/2 back in, I get:


    [tex]\frac{1}{4}\ln{|(1-\frac{n}{2})|}\mbox{ }-\frac{1}{4}\ln{|1+\frac{n}{2}|}[/tex]

    I mean I could multiply by 2 inside each log, but I can't just pull that factor 2 out of thin air. And of course my the left term has the constant and n in reverse order. So I feel like somewhere I've missed something important.



    EDIT- on a side note, it appears to me that as n goes to infinity one way or another I'm going to have infinity minus infinity, which presumably is zero, so the series associated with this integral should converge, right?
     
  6. Oct 2, 2016 #5

    Ray Vickson

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    If all you want to do is to show that ##I = \int_a^{\infty} dn /(n^2-4)## converges for ##a \geq 3##, you are doing much more work than needed. Just use the fact that ##f(n) = 1/(n^2-4)## satisfies ## 0 < f(n) < 3/n^2## for ##n \geq 3## to conclude that ##0 < I < \int_a^{\infty} 3 dn/n^2 = 3/a##. (The upper bound of ##3/n^2## is not mysterious: clearly, ##n^2 - 4## is smaller than ##n^2##, so ##1/(n^2-4)## is larger than ##1/n^2##---but not too much larger when ##n \geq 3##. Putting ##f(n) < k/n^2, \: (k > 1)##, experiment with some convenient values of ##k## that work for ##n = 3##--and so for ##n > 3## as well.)

    In fact, if you want to use the integral to show that ##\sum_{n \geq 3} 1/(n^2-4) ## converges, that, too is a waste of time: just use the fact that the series ##\sum 3/n^2## converges, which follows right away from the ratio test.

    However, if after all that you still want to do the integral ##I##, just use partial fractions to write ##f(n) = A/(n-2) + B/(n+2)## for some ##A,B## (that you can determine), then do two simple integrations. Furthermore, since ##n \geq 3## in the integration region, you can eliminate the ##|\cdots|## signs inside the logarithms, and use the facts that ##\ln(A) + \ln(B) = \ln(AB)## and ##\ln(A) - \ln(B) = \ln(A/B)##, to help you evaluate the ##n \to \infty## limit needed in evaluating the improper integral ##I##.
     
  7. Oct 2, 2016 #6

    vela

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    A problem you have here is the limits. Unless you want to deal with the complex plane, the sine function only work if ##u## takes on values between -1 and 1, which isn't the case here.

    What I would do is try the substitution ##n=2\sec \theta## in the original integral.
     
  8. Oct 3, 2016 #7
    I appreciate the added insight. However I was deliberately trying to do this problem this way for practice. The text book I am getting this from actually asks the reader to use the integral test, and as for not doing partial fractions, I just wanted to see if I could do it this way just to practice the technique (and clearly I need more practice!). ;)


    Of course the way you suggest is really appealing, given how sneaky it seems.



    That is probably a good idea. To be honest I am not very familiar at all with using trig substitution for integrals, and to be even more honest, the sad truth is my familiar with trigonometry itself is limited due to my over reliance upon sine and cosine. It's just a personality quark I have. For example, I prefer to stick with meters for everything regarding distance and just use scientific notation rather than deal with km, nm, etc. *shrugs*

    But thanks for that suggestion. I'll try it that way.
     
  9. Oct 3, 2016 #8
    Try expressing (1-n/2) as (2-n)/2...You can then use ln(a/b) = ln(a)-ln(b)
    Also remember that |a-b| = |b-a|

    That should get you there.
     
  10. Oct 3, 2016 #9

    Ray Vickson

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    The way I suggested is not at all "sneaky"; it is a 100% standard approach in such a problem. Back long ago when I wan learning this material I was taught to try partial fractions first (when the integrand was appropriate) and to go with something like trigonometric substitutions, etc., only when other, more straightforward approaches failed.
     
  11. Oct 4, 2016 #10
    Thanks. Wow I can't believe I failed to see that... this whole time I thought I made some grave error along the way, and it turns out I just was blind to the fact that I already had the right answer just expressed differently.

    [tex]\frac{1}{4}\ln{|(1-\frac{n}{2})|}\mbox{ }-\frac{1}{4}\ln{|1+\frac{n}{2}|}[/tex]
    [tex]\frac{1}{4}\ln{|(\frac{2-n}{2})|}\mbox{ }-\frac{1}{4}\ln{|\frac{2+n}{2}|}[/tex]
    [tex]\frac{1}{4}(\ln{|2-n|} - \ln{2})\mbox{ }-\frac{1}{4}(\ln{|2+n|}- \ln{2})[/tex]
    [tex]\frac{1}{4}(\ln{|n-2|} - \ln{2})\mbox{ }-\frac{1}{4}(\ln{|n+2|}- \ln{2})[/tex]
    [tex]\frac{1}{4}\ln{|n-2|}\mbox{ }-\frac{1}{4}\ln{|n+2|}[/tex]



    Again thank you everyone for helping me work this out.
     
  12. Oct 4, 2016 #11

    Ray Vickson

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    Forget the ##| \cdots |## signs; they just get in the way and obscure what you are trying to do. You said before that ##n \geq 3##, so surely ##n-2 > 0## and ##n+2 > 0##. I already said that in post #7!
     
  13. Oct 4, 2016 #12

    vela

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    But you still have an error along the way when you used the substitution ##u=\sin\theta##. The upper limit was ##u=\infty##. What's the upper limit once you change to ##\sin\theta##?
     
  14. Oct 5, 2016 #13
    Yeah you are correct.

    I thought that since I went back to the original substitution that that didn't matter. Is that not correct?
     
  15. Oct 5, 2016 #14

    vela

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    No, you can't use an invalid step in the middle of a logical argument and "undo" it later.
     
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