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A convergence and divergence test and a couple integrals

  1. Jun 13, 2010 #1
    Here are five separate problems.

    Show that the series 1/3^(ln(n)) converges and that the series 1/2^(ln(n)) diverges.

    integral (sqrt(x)*e^-sqrt(x)) dx

    integral (x/(sqrt(x-1)+2)) dx

    integral (1/(2+sin(x)+cos(x)

    integral (1-cos(x))^(5/2)) dx

    There are no other relevant equations.

    My attempts at solution lead nowhere and are too long-winded to post. I believe I am just missing one important premise I need to solve these. Any help would be appreciated.
     
  2. jcsd
  3. Jun 13, 2010 #2
    Here are some top-level hints:

    OK, for the series, try using the integral test.

    For integral #1, make the obvious u substitution, then integrate by parts twice.
    For integral #2, make the obvious u substitution, then grind through the rational polynomials integrations.
    For integral #3, use the tangent half-angle substitution.
    For integral #4, do two u substitutions.
     
  4. Jun 14, 2010 #3
    Thanks for the help.

    I made a slight parentheses mistake with the last two integrals. They should be:

    integral (1/(2+sin(x)+cos(x)) dx

    integral ((1-cos(x))^(5/2)) dx

    Does this change your hints at all? I tried manipulating the tangent half-angle formula, but I just turns into mess.

    With the 4th integral it seems that no matter what I substitute I end up with nasty, unmanageable square roots when I try to get rid of the sin(x) that comes with the dx.

    I'd be grateful for any more help.
     
  5. Jun 14, 2010 #4
    For the third integral, we make the substitution [itex] t = \tan \frac{x}{2} [/itex]. Then:

    [tex] \sin x = \frac{2t}{1+t^2}, \cos x = \frac{1 - t^2}{1 + t^2}, dx = \frac{2}{1 + t^2} dt [/tex].

    OK, so plug all that in and simplify; you wind up with this:

    [tex] \int \frac{2 dt}{3 + t^2 + 2t} [/tex]

    Now just do a little rearranging/substitution to get it into a familiar inverse trig form, and everything is spiffy.

    For the fourth integral, you have:

    [tex] \int (1 - \cos x)^{\frac{5}{2}} dx [/tex]

    Make the substitution [itex] u = 1 - \cos x [/itex]. So [itex] du = \sin x dx [/itex].

    Now there isn't a [itex] \sin x [/itex] handy in your original integrand, but that's ok.

    Suppose we took [itex] 2 - u [/itex], which is [itex] 1 + \cos x [/itex] , and we multiplied it by u, which is [itex] 1 - \cos x [/itex] . That gives us [itex] 1 - \cos^2 x [/itex], which is conveniently equal to [itex] \sin^2 x [/itex] .

    So [itex] \sin x = \sqrt{u(2-u)} [/itex]. Since this is part of du, we need to divide the integrand by it. We conveniently have a factor of [itex] \sqrt{u} [/itex] already in the integrand, so we just divide out the [itex] \sqrt{2-u} [/itex] term, which gives us:

    [tex] \int \frac{u^2}{\sqrt{2-u}} du [/tex]

    I'm sure you can take it from here.
     
  6. Jun 14, 2010 #5
    Awesome. Thanks for the help!

    I have a question about the series, though. I can show that the series is always decreasing but I don't know how to integrate 1/3^(ln(n)). I tried using a u-substitution and I ended up with the integral of (e^u/3^u) du which I don't know how to deal with either.

    Somebody told me I should use the ratio test, but it ended up being the limit as n goes to infinity of 3^((n)-(n+1)) which is 1 so the result told me nothing. She also said to compare the values of ln(2) and ln(3), but those aren't even in the series. I'm still lost with this problem.
     
  7. Jun 14, 2010 #6
    I don't see how you can use the ratio test for this one, although I may be missing something.

    To integrate [itex] \int a^{- \ln x} dx [/itex], make the substitution [itex] u = a^{- \ln x} [/itex]. Now du will have an x in the denominator by the chain rule, so you'll need to find a way to express x in terms of u (hint: [itex] a^{\log_a x} = x [/itex]). Once you do one of the integrals, the other one is just the same. Then simply show that the a = 3 integral converges, while the a=2 integral diverges.

    Then, for fun, make a general statement about the series convergence for any value of a.
     
  8. Jun 14, 2010 #7
    I ended up with the integral of -(u*x)/(ln(3)*3^-ln(x)) du

    I solved for x and found x = e^log_3(u)

    After substituting I got -(e^log_3(u)*u)/(ln(3)*3^-log_3(u)) du which baffles me.
    If I wanted to I could change 3^-log_3(u) into 1/u but I still don't see where to go from there.
     
  9. Jun 14, 2010 #8
    [tex] \int a^{- \ln x} dx [/tex]
    Make the substitution:
    [tex] u = a^{- \ln x} [/tex]
    [tex] du = \frac{(\ln a) (a^{- \ln x})}{x} dx = \frac{(\ln a) u}{x} dx[/tex]

    Find x in terms of u:
    [tex] \log_a u = - \ln x [/tex]

    [tex] \frac{\ln u}{\ln a} = -\ln x [/tex]

    [tex] x = -u^{\frac{1}{\ln a}} [/tex]

    So now:
    [tex] du = \frac{(\ln a) (a^{- \ln x})}{x} dx = \frac{(\ln a) u}{-u^{\frac{1}{\ln a}}} dx[/tex]

    [tex] du = -(\ln a) u^{1 - \frac{1}{\ln a}} dx[/tex]

    OK, now back in to the original integral. To make the du work, we need to divide the original integral by [itex] -(\ln a) u^{1 - \frac{1}{\ln a}} [/itex]. So:

    [tex] \int a^{- \ln x} dx = - \int \frac{u}{(\ln a)(u^{1 - \frac{1}{\ln a}})} du = - \int \frac{u^{\frac{1}{\ln a}}}{\ln a} du [/tex]

    OK, that's the tricky part. The rest is pretty easy.
     
  10. Jun 14, 2010 #9
    Sweet. Then the indefinite integral should be:

    (a^(-ln(x)*1/(ln(3))+1))/(ln(a) +1)

    The bad news is that evaluating this at infinity and 1 (for the integral test) gives -.59062 for a = 2 and -.47651 for a =3 which means that both of the series converge, which can not be true as the problem tells me that one converges and one diverges.

    What's going on?
     
    Last edited: Jun 14, 2010
  11. Jun 14, 2010 #10

    Dick

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    (1/2)^(log(x)) is actually x^p for some constant p. Can you find that p? (1/2)^(log(x))=e^(log(1/2)*log(x))=(e^(log(x)))^log(1/2). What do you know about the convergence of p-series?
     
    Last edited: Jun 14, 2010
  12. Jun 14, 2010 #11
    From here it goes:
    [tex] - \int \frac{u^{\frac{1}{\ln a}}}{\ln a} du = - \frac{u^{(1 -\frac{1}{\ln a})}}{(1 - \frac{1}{\ln a}) \ln a} [/tex]

    Substituting back for u:

    [tex] -\frac{{(a^{-\ln x})}^{(1 -\frac{1}{\ln a})}}{\ln a - 1} = \frac{x a^{- \ln x}}{1 - \ln a} [/tex] (+C)
     
  13. Jun 14, 2010 #12

    Dick

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    Aren't you making this a bit harder than it needs to be? a^(ln(x))=x^(ln(a)), isn't it?
     
  14. Jun 14, 2010 #13
    Yes, much harder. :(

    OP, pay attention to what Dick says.
     
  15. Jun 14, 2010 #14
    That's so beautiful yet simple. That was the clue I needed.

    Thanks to both of you for your help.
     
  16. Jun 22, 2010 #15
    Bob Busby:

    you can use Cauchy's Condensation Test for the first part of question 1:
    f(n) =< (2^n)*f(2^n)

    Also can be done by the Ye'Olde p Test :)
    ~ 1/(n^1.1)
     
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