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Math Help - Today's integral

  1. #1
    Eater of Worlds
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    Today's integral

    Here is an integral that looks ominous.

    Give it a whirl if you so desire.

    \int_{e^{e}}^{e^{e+1}}\left[\frac{1}{ln(x)\cdot ln(ln(x))}+ln(ln(ln(x)))\right]dx

    Mathstud?. Krizarama?. anyone?.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Here is an integral that looks ominous.

    Give it a whirl if you so desire.

    \int_{e^{e}}^{e^{e+1}}\left[\frac{1}{ln(x)\cdot ln(ln(x))}+ln(ln(ln(x)))\right]dx

    Mathstud?. Krizarama?. anyone?.
    Hmmm...well I have fiddled around with the indefinite integral. I tried a bunch of tricks but to no avail. So I am going to say that this has no closed form answer. But for the definite part, I am not completely done yet, but letting u=\ln(\ln(x)) then using differentation under the integral sign is looking promising.

    EDIT: Also, based upon human nature I am going to guess that a sub of u=\ln(x) will be in here somewhere, I got this based upon human being's propensity for nice solutions, e.g. look at the limits of integration
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  3. #3
    Eater of Worlds
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    I was thinking along your lines, mathstud. But I got sidetracked at work and was not able to look into it further.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    I was thinking along your lines, mathstud. But I got sidetracked at work and was not able to look into it further.
    Just out of curiosity, is this a real world integral or a fabricated human made one?
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by Mathstud28 View Post
    EDIT: Also, based upon human nature I am going to guess that a sub of u=\ln(x) will be in here somewhere, I got this based upon human being's propensity for nice solutions, e.g. look at the limits of integration
    If you did that repeatedly for two or three times, then it looks like you would have a bunch of nested exponential functions. Like Galactus, I am at work, so I cannot look into this one further at the moment.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by colby2152 View Post
    If you did that repeatedly for two or three times, then it looks like you would have a bunch of nested exponential functions. Like Galactus, I am at work, so I cannot look into this one further at the moment.
    Well, I have to go tutor, I will see if I can think of something. I am guessing that if I get this it will be differentiation under the integral sign with a prior substitution.

    Mathstud
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  7. #7
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    Recognition. (It is an application of the product rule.)
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  8. #8
    Super Member wingless's Avatar
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    \int_{e^{e}}^{e^{e+1}}\left[\frac{1}{ln(x)\cdot ln(ln(x))}+ln(ln(ln(x)))\right]dx

    Let f(x)=\ln \ln \ln x,
    then f'(x) = \frac{1}{x \ln x \ln \ln x}

    So the integral is \int_{e^{e}}^{e^{e+1}}\left[ xf(x)\right ]'~dx

    It's obvious from here
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    \int_{e^{e}}^{e^{e+1}}\left[\frac{1}{ln(x)\cdot ln(ln(x))}+ln(ln(ln(x)))\right]dx

    Let f(x)=\ln \ln \ln x,
    then f'(x) = \frac{1}{x \ln x \ln \ln x}

    So the integral is \int_{e^{e}}^{e^{e+1}}\left[ xf(x)\right ]'~dx

    It's obvious from here
    Thats what I had done originally and I had an answer, but the Mathworld's integrator said there was no solution, and there was no way I was going to contest that thing.
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  10. #10
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Thats what I had done originally and I had an answer, but the Mathworld's integrator said there was no solution, and there was no way I was going to contest that thing.
    Can't you stop relying on software ?

    I can say it too : "I had the answer, it was obvious ! But when I used my Ti-89, it didn't give me the result".



    Sorry, just tickling...
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Can't you stop relying on software ?

    I can say it too : "I had the answer, it was obvious ! But when I used my Ti-89, it didn't give me the result".



    Sorry, just tickling...
    I am using it less and less, but seriously, hate me or not, you cannot deny that if you really did have an answer and the online integrator said there was no solution you would disagree with it. I admit I did not see the solution till Dystopia said product rule, but nonetheless. Stop making it sound like I cannot do anything, that all my answers are from my calculator, I almost never use it anymore. Your complete and utter faith in my mathematical abilities is flattering Moo.
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  12. #12
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    I am using it less and less, but seriously, hate me or not, you cannot deny that if you really did have an answer and the online integrator said there was no solution you would disagree with it. I admit I did not see the solution till Dystopia said product rule, but nonetheless. Stop making it sound like I cannot do anything, that all my answers are from my calculator, I almost never use it anymore. Your complete and utter faith in my mathematical abilities is flattering Moo.
    I was certainly not denying your abilities

    I'm just saying that you have to rely more on yourself, that calculators/softwares only are tools... Don't you agree ?
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  13. #13
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Thats what I had done originally and I had an answer, but the Mathworld's integrator said there was no solution, and there was no way I was going to contest that thing.
    Mathworld's integrator is not able to find an anti-derivative of x\mapsto |\sin x| but it can find an anti-derivative of x\mapsto\sqrt{\sin^2x} ...
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  14. #14
    Super Member wingless's Avatar
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    Mathematica (which is made by Wolfram, just like Mathworld) gives the correct answer.
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  15. #15
    Eater of Worlds
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    I had a chance to look at it and what I saw happens to be along wingless lines, it would appear.

    Take note that the derivative of xln(ln(ln(x)))=\text{our integral}

    So, we get:

    \int(\frac{d}{dx}\left[xln(ln(ln(x)))\right])dx

    Which we just end up with:

    (e^{e+1})ln(ln(ln(e^{e+1})))\approx{11.226}

    Now, when e^{e} we get 0, so we are done.

    I had this integral written down amongst some papers and it was not finished. I do not know where it came from.
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