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Upper bound for e^(-x^2)

  1. Apr 15, 2012 #1
    Can anyone suggest an upper bound for [itex]e^{-x^2}[/itex] that can be integrated easily?
     
  2. jcsd
  3. Apr 15, 2012 #2

    LCKurtz

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    On what interval? Just guessing maybe ##[0,\infty)##? This is really easy. What have you thought of so far?
     
  4. Apr 15, 2012 #3
    I've just realised it's less than 1 for all x.
     
  5. Apr 15, 2012 #4

    LCKurtz

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    You haven't answered the question about what interval, or told us what problem you are trying to solve.
     
  6. Apr 15, 2012 #5
    It doesn't matter now; I know how to do it!
     
  7. Apr 15, 2012 #6

    LCKurtz

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    Not if you were trying a comparison test on ##[0,\infty)##.
     
  8. Apr 16, 2012 #7
    It's actually on a finite interval [itex][-n,n][/itex], but yes the integral of 1 would diverge on [itex][0,\infty)[/itex]
     
  9. Apr 16, 2012 #8

    HallsofIvy

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    Are you asserting that "1 times infinity" is a finite number?
     
  10. Apr 16, 2012 #9
    All I need to show is that [itex]\int_{-n}^n e^{-x^2} dx[/itex] converges as [itex]n\to\infty[/itex]. This is the last part of a bigger question where I'm trying to use the Monotone Convergence Theorem to show that [itex]e^{-x^2}[/itex] is Lebesgue integrable over [itex]\mathbb{R}[/itex].

    Since [itex]0 \leqslant e^{-x^2} \leqslant 1[/itex] and [itex]\int_{-n}^n 1 \; dx =0 \to 0[/itex] as [itex]n\to\infty[/itex] it follows from the comparison test that [itex]\int_{-n}^n e^{-x^2} dx[/itex] converges as [itex]n\to\infty[/itex].

    (I'm not asked to find [itex]\int_{\mathbb{R}} e^{-x^2}[/itex], just to prove it is Lebesgue integrable using the MCT.)
     
  11. Apr 16, 2012 #10

    D H

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    No, it isn't.
     
  12. Apr 16, 2012 #11
    Oh yeah, it equals [itex]2n[/itex] which diverges as [itex]n\to\infty[/itex].

    But doesn't that contradict the comparison test?

    I'm trying to show that [itex]\int_{-n}^n e^{-x^2}dx[/itex] converges, but the comparison test implies it diverges.
     
  13. Apr 16, 2012 #12
    Yep, it diverges, assuming you mean as [itex]\displaystyle\lim_{n\to\infty}[/itex]. Why do you think it converges?
     
  14. Apr 16, 2012 #13
    Well, OK. Let's start from scratch.

    I'm trying to prove [itex]f(x) = e^{-x^2} \in L^1(\mathbb{R})[/itex] ; that is, that [itex]f[/itex] is Lebesgue integrable over [itex]\mathbb{R}[/itex].

    Let [itex]f(x) = e^{-x^2} \chi_{(-\infty , \infty)}(x)[/itex] and [itex]f_n = f \chi_{[-n,n]}[/itex].

    Since [itex]f\geqslant 0[/itex], [itex](f_n)[/itex] is an increasing sequence of functions which converges everywhere to [itex]f[/itex].

    We want to show that [itex]f[/itex] is integrable so it suffices to show [itex]\int f_n[/itex] converges as [itex]n\to\infty[/itex]. It then follows from the Monotone Convergence Theorem that [itex]f\in L^1 (\mathbb{R})[/itex] (and [itex]\int f = \lim_{n\to\infty} \int f_n[/itex]).

    We use the given hint ("don't try to find [itex]\int f[/itex] ; integrate a simpler upper bound instead").

    We see that [itex]f_n[/itex] is bounded by... The integral of this upper bound converges, so by comparison test [itex]\int f_n[/itex] converges.
     
  15. Apr 16, 2012 #14

    D H

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    Of course not. Given functions [itex]f(x)[/itex] and [itex]g(x)[/itex] such that [itex]0\le f(x) \le g(x)[/itex] for all x and such that [itex]\lim_{s\to\infty}\int_{-s}^s g(x) dx[/itex] diverges says nothing about the convergence or divergence of [itex]\lim_{s\to\infty}\int_{-s}^s f(x) dx[/itex].

    There are plenty of positive definite functions [itex]f(x)[/itex] that are bounded from above by [itex]g(x)=1[/itex] whose integral over the real number line does converge. [itex]f(x)=\exp(-x^2)[/itex] is one such function.
     
  16. Apr 16, 2012 #15
    Oh yes, it's the other way around for divergence isn't it: if [itex]\int f(x) dx[/itex] diverges then so does [itex]\int g(x) dx[/itex].

    OK, so I need an upper bound whose integral converges then. Any suggestions?
     
  17. Apr 16, 2012 #16

    D H

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    No. That would be solving your homework for you.

    Huge hint: if [itex]0\le a \le 1[/itex] then [itex]\sqrt a \ge a[/itex], and if [itex]a\ge1[/itex] then [itex]\sqrt a \le a[/itex].
     
    Last edited: Apr 16, 2012
  18. Apr 16, 2012 #17

    Mentallic

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    What about something with [tex]xe^{-x^2}[/tex]
     
  19. Apr 16, 2012 #18
    But if [itex]x<0[/itex] then [itex]e^{-x^2} > xe^{-x^2}[/itex]
     
  20. Apr 16, 2012 #19

    D H

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    Do you care? exp(-x^2) is symmetric, so if the integral from 0 to infinity exists, then so does the integral from -infinity to infinity.
     
  21. Apr 16, 2012 #20
    Bingo! If only I'd drawn a graph first...
     
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