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Math Help - "Super Integral" (Apparently)

  1. #1
    Bar0n janvdl's Avatar
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    "Super Integral" (Apparently)

    Saw this one on the univ's notice board...

    I don't know how hard this will be for you people, especially Krizalid, but I'm interested to see the answer.


    <br />
\int\limits_1^{100} {x^x } dx<br />
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    Eater of Worlds
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    because there is no elementary function whose derivative is x^x.

    I reckon someone should define one. That is why it is not easy by elementary means.

    I did run it through the calculator and got a huge answer.

    1.784\times{10^{199}}

    Maple nor the Integrator will give an indefinite integral for this. They just spit back the same thing.
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  3. #3
    Eater of Worlds
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    If you wanted to get a numerical answer, I would just try a Riemann sum or even Simpson's rule.
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    because there is no elementary function whose derivative is x^x.

    I reckon someone should define one. That is why it is not easy by elementary means.

    I did run it through the calculator and got a huge answer.

    1.784\times{10^{199}}

    Maple nor the Integrator will give an indefinite integral for this. They just spit back the same thing.
    Now how could I have forgotten you, Galactus? Why did I only think of Krizalid?

    Well you did get an answer...

    I'm interested to see if anyone else has some other method of solving it...
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    Eater of Worlds
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    Just for fun, I generated a graph with Maple using 1 to 10. For 1..100, the y value was too large.

    Look how huge it is just using 1 to 10, let alone 1 to 100.

    Afterall, this is quite the exponential. 2^2, 3^3, ....., 100^100.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    Just for fun, I generated a graph with Maple using 1 to 10. For 1..100, the y value was too large.

    Look how huge it is just using 1 to 10, let alone 1 to 100.

    Afterall, this is quite the exponential. 2^2, 3^3, ....., 100^100.
    I then assume this is a question for very advanced students? It's like challenging questions that they post on the notice board sometimes.

    But I still feel there must be an easier way of solving it. You know when you just have a feeling that there is a better, more elegant method?
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    Eater of Worlds
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    We could rewrite it as x^{x}=e^{xln(x)}, but I don't know what good it'll do.

    The Big K may have a trick up his sleeve.
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  8. #8
    GAMMA Mathematics
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    You could try and write the function out as a Taylor series. We have f(x)=x^x and f'(x) = x^x(\ln{x}+1). I don't have the time right now... anyone want to volunteer?

    Problem is that there are no elementary functions whose derivative is x^x.
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  9. #9
    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    The Big K may have a trick up his sleeve.
    That's what I'm thinking...
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    There is an identity with x^x, but again it is non-elementary.

    \int_0^1 x^x ~ dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n^n}.
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    Hacker got a typo, it should be \sum\limits_{n = 0}^\infty  {\frac{{( - 1)^{n + 1} }}<br />
{{(n + 1)^{n + 1} }}} or \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n }}<br />
{{n^n }}}. This identity is called as "Sophomore's Dream." One can prove it easily with galactus' trick (post #7) after series expansion and integration.

    ----

    As for the problem, I dunno, I think I would've thought in numerical methods to tackle it too.
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    GAMMA Mathematics
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    One trouble with Taylor expansion is where to center it at... 1? 50? 100?

    The derivatives of the function are rather beautiful...

    f(x)=x^x
    f'(x)=x^x(1+\ln{x})
    f''(x)=x^x(1+\ln{x}+\frac{1}{x})
    f^{(3)}(x)=x^x(1+\ln{x}+\frac{1}{x}-\frac{1}{x^2})
    f^{(4)}(x)=x^x(1+\ln{x}+\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3})

    I think you get the idea...

    Let's choose a=1 for simplicity...

    f(1)=1
    f'(1)=1
    f''(1)=2
    f^{(3)}(1)=1
    f^{(4)}(1)=3

    x^x = 1 + (x-1) + (x-1)^2 + \frac{(x-1)^3}{6} + \ldots

    This is obviously not enough of an expansion for x^x. More work needs to be done, but this is a start.
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  13. #13
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    Quote Originally Posted by Krizalid View Post
    Hacker got a typo, it should be \sum\limits_{n = 0}^\infty  {\frac{{( - 1)^{n + 1} }}<br />
{{(n + 1)^{n + 1} }}} or \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n }}<br />
{{n^n }}}. This identity is called as "Sophomore's Dream." Mr F says: This was actually an expression applied by Borwein et al (Borwein, J.; Bailey, D.; and Girgensohn Experimentation in Mathematics: Computational Paths to Discovery. Wellesley, MA: A K Peters, 2004.) ....

    One can prove it easily with galactus' trick (post #7) after series expansion and integration.

    ----

    As for the problem, I dunno, I think I would've thought in numerical methods to tackle it too.
    Actually it's \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^{{\color{red}n+1}}}}<br />
{{n^n }}} = 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + ....

    Johann Bernoulli first calculated \, \int_{0}^{1} x^x \, dx\, in a paper he published in 1697. The series converges extremely rapidly (an attribute noted by Bernoulli) .... it only takes a few terms to calculate the integral accurately to ten decimal places.

    Krizalid is quite right in what he says regarding the proof:

    x^x = 1 + x \ln x + \frac{1}{2!} \, x^2 (\ln x)^2 + \frac{1}{3!} \, x^3 (\ln x)^3 + \frac{1}{4!} \, x^4 (\ln x)^4 + ......

    and each of these integrals is simple to do. Especially if you know the reduction formula

    \int x^m (\ln x)^n \, dx = \frac{1}{m+1} \, x^{m+1} (\ln x)^n - \frac{n}{m+1}\, \int x^m (\ln x)^{n-1} \, dx
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    Quote Originally Posted by mr fantastic View Post
    Actually it's \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^{{\color{red}n+1}}}}<br />
{{n^n }}} = 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + ....
    Ahhh, time ago I don't see the identity.

    \int_0^1x^k\ln^kx\,dx can be computed via substitution u=-\ln x, it yields an integral with gamma function-type.
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    Quote Originally Posted by janvdl View Post
    Saw this one on the univ's notice board...

    I don't know how hard this will be for you people, especially Krizalid, but I'm interested to see the answer.


    <br />
\int \limits_0^{100} {x^x} \, dx<br />
    Regarding the direction this thread has taken to date: The technical infelicity is the integral limits .....

    Perhaps \int \limits_0^{100} {x^x } \, dx - \int\limits_0^{1} {x^x } \, dx .....

    Or a cunning substitution that gives the integral limits and yields the answer ..... u = \ln x doesn't seem to quite cut it ......
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