integral of x^x

Discussion in 'Physics & Math' started by ddovala, Jun 27, 2004.

  1. John Connellan Valued Senior Member

    Its kinda weird alright! It seems to have a minimum (for x > 0) at x = ~0.368
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  3. shmoe Registred User Registered Senior Member

    Or at exactly x=1/e. It's graph for x>0 isn't really that exciting.
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  5. Rick Valued Senior Member

    There's no way to find an Integral such as this,we were never able to during our schooling years...

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  7. Dunnoyet Registered Senior Member

    I just put together a TI-83+ program that uses the mapping z=>z^z on the complex plane to find escapist points. Quite an interesting shape came out. Now I need to learn Fractint's script language so I can have it apply its powers to it and produce awesome pictures. The TI-83 only has 13 digits of precision, so the one picture I've made had some odd outlines that reminded me of "artifacts of the limitations of the computer" that showed up in some of the images in The Beauty of Fractals.

    Whee!!! Complex dynamics are fun...

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  8. Joey_21 Registered Member

    The antiderivative for x<sup>x</sup> isn't elegant, but it can be done. This has recently caught my mathematical attention also (for fun).

    Here is what I came up with:
    ∫ x<sup>x</sup> dx = Sum(k=0 to infinity, (k+1)<sup>-1</sup> * x<sup>k+1</sup> * Sum(n=0 to k, (-k-1)<sup>-n</sup> * (log(x))<sup>k-n</sup> / (k-n)!))
    Let's call this xTxInteg(x)

    log(x) = ln(x) and for the case where x=1 and k-n=0, log(x)<sup>k-n</sup> = log(1)<sup>0</sup> = 0<sup>0</sup> is defined to be 1 for the simple sake that it's easier to write the sum as that instead of having to include an extra conditional term each time I have log(1) to a power of zero.

    Since we have x<sup>x</sup> 's antiderivative, we know that:
    <sub>0</sub>∫ <sup>t</sup> x<sup>x</sup> dx = xTxInteg(t) - xTxInteg(0) (xTxInteg(0) = 0), therefore
    <sub>0</sub>∫ <sup>t</sup> x<sup>x</sup> dx = xTxInteg(t)
    Last edited: Sep 9, 2004
  9. Joey_21 Registered Member


    Let u = x<sup>2</sup>, therefore we have exp(u) or e<sup>u</sup>

    Take the power series for e<sup>x</sup>:
    e<sup>x</sup> = Sum(n=0 to infinity, x<sup>n</sup> / n!)

    Therefore e<sup>u</sup>=
    Sum(n=0 to infinity, (x<sup>2</sup>)<sup>n</sup> / n!)
    Sum(n=0 to infinity, x<sup>2n</sup> / n!)

    This expression is simple to integrate.

    ∫ e<sup>x<sup>2</sup></sup> dx = Sum(n=0 to infinity, x<sup>2n+1</sup> / (n! * (2n+1)))

    In fact, this expansion is easily generalizable to any exponential argument z:
    ∫ e<sup>x<sup>z</sup></sup> dx =
    Sum(n = 0 to infinity, x<sup>zn+1</sup> / (n! * (zn+1)))
    Last edited: Sep 9, 2004
  10. mifter Registered Member

    Hi, I'm having the same trouble and it's really cool that you've found a formula for it (because nowhere else I've looked has it), however, could you please explain what exactly xTxInteg(t) means or what values of n and k I should use. I'm sorry for my ignorance but I'm only a high school student and would urgently appreciate some further explanation. Thankyou
  11. Dunnoyet Registered Senior Member

    I mentioned fractals made with this formula earlier. I've put together a small web page talking about my experiments at I hope it's somewhat useful, or at least interesting to y'all.

    I think he has answered your question in an indirect way:

    The generalization that he ends with, e.g. ∫ e<sup>x<sup>z</sup></sup> dx = Sum(n = 0 to infinity, x<sup>zn+1</sup> / (n! * (zn+1))), works. Simply plug 1 in for z. Joey's method is what I call "Taylor-MacLaurin Black Magic" because I have never quite understood the full power of series. You will see (or are seeing ;D) the power of series in Calculus 2. Calculus BC, the high school course that I took, covered it. I'll have to tinker some more with my TI-83+ (oh, I wish there was an easier way that was free) to verify Joey's results, but looking at the calculator that he has written (and the advanced math skills that it shows), I'll give him credit.

    p.s. I'm fresh out of high school.
    p.s.s. Joey, your calculator is now in my Start menu, tier 1. Thanks!

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  12. Dunnoyet Registered Senior Member

    I'm not sure that anyone is looking at this thread anymore, but I'm posting anyway. I've got my (almost) final two cents to put in.

    After looking around for quite a while, it seems that James R, shmoe, Joey_21 and Crisp are right on target. Perhaps too vague or garbled in some instances, but on target. Thanks!

    x<sup>x</sup> is a special case of the Power Tower, as described on the Wolfram MathWorld website. Integrals are mentioned on that page, and it seems that the conclusions reached by those listed above are correct. Check it out--lots of pretty graphs and hard information.

    Thanks, all!

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