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Integral of e^-mod(x^n)

  1. Oct 29, 2009 #1
    OK, this is going back to a problem I studied back in Calculus class; it's been a few years since I graduated, so bear with me please... :)

    It's easy to integrate [tex]\int^{\infty}_{-\infty}\\e^{-mod(x)}\\dx[/tex] (giving a value of 2), and with a bit of grunt-work, the same can be done for [tex]\int^{\infty}_{-\infty}\\e^{-mod(x^2)}\\dx[/tex] (giving the value of [tex]\sqrt{\pi}[/tex]). But I was curious about [tex]\int\\e^{-mod(x^n)}[/tex] for values of n>2.
    Would it be considered valid to take the power series of the function and integrate it, then evaulate that at the end points?

    For example, for the general case of x to an nth power... (oh heck...I tried to used the forum code but it got so tangled up...here's one I MSPainted earlier instead).

    My question is, is this integration and evaluation valid? Have I integrated it properly? Does it count? (I never did check if the integrated power series converges). It certainly appears to work when I evaluated the partial sums of the series with x tending towards infinity, for the cases of m=1 and m=2, but is it legal?

    Attached Files:

  2. jcsd
  3. Oct 30, 2009 #2
    It would work for a finite sum, and I think it works for an infinite sum if the series converges (which it does for your conditions on n)... but I cannot give you an absolute answer, only that I think you are safe here.

    You've switched m and n around from your introduction in the attachment!
  4. Oct 30, 2009 #3
    Ah, it's good to know I haven't completely lost my touch at calculus then. :)

    I did switch them? *looks* O bother! That's what I get for using a picture that I created quite some time ago...:tongue2:
  5. Oct 30, 2009 #4

    Gib Z

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    Convergence is not a sufficient condition to interchange an integral and a summation: You must have uniform convergence.

    I don't see how the final series can have a value? It doesn't converge! If the upper limit was some fixed value rather than infinity we could see from the alternating series test that the series converges because the terms decrease monotonically towards zero (after a certain term, at least) .

    I don't understand what the "mod(x^2)" means? Could you clarify what it means, and how you got the value of 2 for when m=1 ?

    It so happens that [tex]\int_0^{\infty} e^{-x^n} dx = \Gamma \left(1+\frac{1}{n}\right)[/tex].
  6. Oct 30, 2009 #5
    Huh, that was meant to be [tex]-mod(x)^2[/tex] - I'm not very good at writing neatly in this BulletinBoard maths code.

    OK, I'll do this for m=1.

    [tex]\int^{\infty}_{-\infty}\\e^{-mod(x)} = 2\int^{\infty}_{0}\\e^{-x}[/tex]

    [tex]= 2\int^{\infty}_{0}\sum^{\infty}_{n=0}\\\frac{-x^n}{n!}[/tex]

    [tex]= 2[\sum^{\infty}_{n=1}\frac{(-1)^n}{n+1}\frac{x^{n+1}}{n!}]\\\|^{\infty}_{x=0}[/tex]

    [tex]= 2[\sum^{\infty}_{n=1}\frac{(-1)^n\;x^{n+1}}{(n+1)!}\\\|^{\infty}_{x=0}[/tex]

    [tex]= -2\\e^{-x}\\\|^{\infty}_{x=0}[/tex]

    [tex]= -2\\e^{-\infty} - -2\\e^{-0}[/tex]

    [tex]= 0 + 2\1 = 2[/tex]

    But for m=2, that is, for the integral of [tex]e^{-mod(x)^2}[/tex], you can't convert the integrated power series back into a simple function, but evaluating that sum for any sufficiently large x (as x --> infinity) does apparently produce [tex]\sqrt{\pi}[/tex] (I ran it through on Excel once).

    I didn't know that it could be expressed in terms of the Gamma function, that's interesting to know (my professor never mentioned that).
  7. Oct 30, 2009 #6
    So, mod mean modulus here?

  8. Oct 30, 2009 #7
    So that's how to write it! Heh, I just had a brain freeze about how to write the modulus symbol in forum code, I can't believe that.

    Yes, that is what I meant, [tex]|x|[/tex].

    *Slaps "n00b" sign on forehead* :p
  9. Oct 30, 2009 #8
    this implies at once

  10. Oct 30, 2009 #9
    Huh. For some reason, my class was never taught that identity, and our professor said that


    was not solvable at all for [tex]n\geq3[/tex]. Hence my trying to construct another way. O well, waste of my time. :smile:
  11. Oct 30, 2009 #10

    Gib Z

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    Probably meant not solvable in terms of elementary constants and the such. For integer and half integer arguments the Gamma function yields a "nice" value but no such expression has been found for other values.

    If you want to get ahead or impress your teacher, try proving the relation.
  12. Oct 30, 2009 #11
    Heh, I couldn't impress him anyway. :smile: I graduated years ago and haven't kept in contact with anyone from uni, faculty or fellow students.

    I've also been working on/off on trying to find the general derivative of the function [tex]y = \\^{n}\\x[/tex] (tetration, positive integer n only), and it seems to be defined recursively and not in a nice easy way either (involving more than one summation sign).
    Although differentiating [tex]y = \\^{\infty}\\x[/tex] is much easier, as it essentially boils down to differentiating the equation [tex]y=x^y[/tex]. It's a slightly messy answer:
    [tex]y' = \frac{y^2}{x-x\ln(y)}[/tex]
    if I recall correctly, but by far easier than for finite power towers of x.
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