1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence of the integral ln(sin(x))

  1. Aug 4, 2005 #1
    I have to study the convergence of the following integral:

    [tex]\int_0^{\frac {\pi} {2}}ln(sin(x)).dx[/tex]

    My first attempt was to do the following:

    [tex]\int_0^{\frac {\pi} {2}}ln(sin(x)).dx\leq\int_0^{\frac {\pi} {2}}ln(x).dx\leq\int_0^{\frac {\pi} {2}}x.dx[/tex]

    But then i realized it was not an integral of positive terms, so my previous work was wrong.(This integral has all negative terms, right????)

    I also realized that it is an improper integral, since [tex]ln(sin(0))[/tex] doesnt exist.

    I really dont know what to do, maybe trying to express the series of ln, and sin, but i would really appreciate if you could point me into the right direction.

    Thankfully Paul.
     
  2. jcsd
  3. Aug 4, 2005 #2

    lurflurf

    User Avatar
    Homework Helper

    [tex]\log(\sin(x))=\log(x)+\log(\frac{\sin(x)}{x})[/tex]
    log(sin(x)/x) will have no problems so focus attention on
    [tex]\int_0^\frac{\pi}{2} \log(x) dx[/tex]
    or for that matter
    [tex]\int_0^h \log(x) dx[/tex]
    for any h s.t. 0<h<pi
    show
    [tex]\int_0^{0^+} \log(x) dx=0[/tex]
     
    Last edited: Aug 4, 2005
  4. Aug 4, 2005 #3

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    This seems to be an example of the general case:

    If:

    [tex] \int_a^b f(x)dx[/tex]

    converges, and:

    [tex]a\leq g(x)\leq b\quad\text{for}\quad a\leq x\leq b[/tex]

    Can we say that:

    [tex]\int_a^b f[g(x)]dx[/tex]

    also converges? I believe so but as yet cannot come up with a proof. Note that:

    [tex]\int_0^1 ln(x)dx[/tex]

    Converges and:

    [tex]0\leq Sin[x] \leq 1\quad \text{for} \quad x\in [0,\pi/2][/tex]
     
  5. Aug 4, 2005 #4
    Let me see if i understand: You said that log(sin(x)/x) doesnt have problems because it is a continuous and bounded function between 0 and pi/2 , and therefore converges, right???
    On the second part of the integral:
    [tex]\int_0^\frac{\pi}{2} \log(x) dx[/tex]

    i dont understand why you substituted pi/2 by h on the next step, instead of substituting 0 by h which is what i would have done.

    Thanks you very Much, Paul.
     
  6. Aug 4, 2005 #5

    lurflurf

    User Avatar
    Homework Helper

    It will work out the same since
    [tex]\int_0^\frac{\pi}{2} \log(x) dx=\int_0^h \log(x) dx+\int_h^\frac{\pi}{2} \log(x) dx[/tex]
    I like to focus on (0,h) because that is where the problem is. Others like to focus on (h,pi/2) since that is where there are not problems.
    For your way just show that the limit h->0 exist.
    and yes log(sin(x)/x) is good because it is continous and bounded.
     
  7. Aug 4, 2005 #6

    lurflurf

    User Avatar
    Homework Helper

    The problem is we have impropper integrals, so that will not be true in general
    [tex]\int_0^1 \log(x) dx=-1[/tex]
    conveges
    0<exp(-1/x)<1
    if x>0
    but
    [tex]\int_0^1 \log(e^{-\frac{1}{x}}) dx=-\int_0^1 \frac{1}{x} dx[/tex]
    diverges
     
  8. Aug 4, 2005 #7
    i get it

    Thank you so much, i now understand this problem, and i could get
    [tex]\int_0^{0^+} \log(x) dx=0[/tex] using parts integration.
    Saltydog's generalization seems pretty interesting, at least to me, but i think he overused the same variables a and b for different meanings, am i right????
    I will try to prove it later, with my little knowledge, but probably i will come back crying for help.

    I really appreciate all your explanations.
    Grateful, Paul.

    edit: It seems I am so slow writing my messages that everytime i finish one, there is another one that i couldnt read.
    Obviously i am not going to try to prove saltydog`s generalization now.
     
    Last edited by a moderator: Aug 4, 2005
  9. Aug 5, 2005 #8

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Alright Lurflurf. Thank you. I yield. May I have some defense: You used a composite of three functions, the second was the inverse of the first and the third was a function known to diverge in the interval. That is:

    [itex]\int_a^b f[f^{-1}(g(x))]dx[/tex]

    This is equivalent to:

    [tex]\int_a^b g(x)dx[/tex]

    which diverges.

    Thus, perhaps a better phrasing might be if:

    [tex]\int_a^b f(x)dx[/tex]

    converges and g(x) is NOT the inverse of f(x), then the proposition holds. Might you suggest a counter-example for such?

    Thanks. :smile:

    Edit:
    Perhaps a better wording would be:

    If:

    [tex]\int_a^b f(x)dx[/tex]

    converges, then under what suitable restriction imposed on g(x) will:

    [tex]\int_a^b f[g(x)]dx[/tex]

    also converge?

    Yes, I like that better and will spend some time on it.
     
    Last edited: Aug 5, 2005
  10. Aug 5, 2005 #9

    lurflurf

    User Avatar
    Homework Helper

    That was just a simple example. The problem is we are dealing with improper integrals. That is functions unbounded in the interval. Composing functions the way you suggest has a possibility of destryoying convergence.
    say we are on (0,1) f is integrable but f diverges to infinity at 0. If we compose f with g where g is small often we can destry the convergence.
    What you state will work if the functions are continuous and bounded, but not under conditions as weak as you desire.
     
  11. Aug 5, 2005 #10

    lurflurf

    User Avatar
    Homework Helper

    The question was one of convergence, but this is one of those famous lets find the value for fun integrals.
    [tex]I=\int_0^\frac{\pi}{2}\log(\sin(x))dx=\int_0^\frac{\pi}{2}\log(\cos(x))dx=\frac{1}{2}\int_0^\frac{\pi}{2}\log(\frac{\sin(2x)}{2})dx[/tex]
    change variable u=2x in the last of these and breaking int(0,pi) into 2int(0,pi/2)
    [tex]I=\frac{1}{2}\int_0^\frac{\pi}{2}\log(\frac{\sin(2x)}{2})dx=\frac{1}{2}\int_0^\frac{\pi}{2}\log(\frac{\sin(u)}{2})du[/tex]
    then
    [tex]I=\int_0^\frac{\pi}{2}\log(\frac{\sin(x)}{2})dx-\int_0^\frac{\pi}{2}\log(\sin(x))dx=\int_0^\frac{\pi}{2}\log(\frac{1}{2})dx[/tex]
    so
    [tex]I=\int_0^\frac{\pi}{2}\log(\sin(x))dx=-\frac{\pi}{2}\log(2)[/tex]
     
    Last edited: Aug 5, 2005
  12. Aug 5, 2005 #11

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Ok. I'm beginning to see that now and how it's related to the convergence of improper integrals. How very interesting. I can imagine a set of functions:

    [tex]S=\{g_i(x)\}[/tex]

    and the integral:

    [tex]\int_a^b f[g_i(x)]dx[/tex]

    for which the integral of f(x) converges. I can envision a range, a spectrum, for which convergence becomes slower and slower (the integral grows larger and larger) until convergence is abruptly lost or lost gradually? I'd be interested in studing how convergence and the loss of such is affected by S. That is, it's as if S is an operator operating on the integral. How then is convergence affected by different S? Suppose it's a question for Functional Analysis. Lots of questions but I better get out now before I interefere too much with real homework questions.

    Thanks Lurflurf. :smile:
     
    Last edited: Aug 5, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Convergence of the integral ln(sin(x))
Loading...