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Infinite graph with finite area?

  1. Jan 18, 2005 #1

    kreil

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    Is it possible for the graph of a function to extend to infinity in either direction but have a finite area under the curve?

    It seems to me that since series exist that have infinite terms but are convergent, the same holds true here.

    If so, does anyone have an example function I could play around with?

    Thanks
     
  2. jcsd
  3. Jan 18, 2005 #2

    HallsofIvy

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    Yes, of course. The function would have to go to 0, rapidly, in both directions. One important example is the "error function",
    [tex]\frac{1}{2\pi}e^{-x^2}[/tex]
    which has area under the curve of 1.
     
  4. Jan 18, 2005 #3

    dextercioby

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    How about [itex] e^{-kx} [/tex] with 'k" and "x" strictly positive,equivalently,"k" and "x" strictly negative ???

    Daniel.
     
  5. Jan 18, 2005 #4

    matt grime

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    |kx| you mean?
     
  6. Jan 18, 2005 #5

    dextercioby

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    If you prefer,i was referring to this integral

    [tex]\int_{0}^{+\infty} e^{-kx} dx [/tex]

    ,k>0.

    Then it occured to me that
    [tex] \int_{-\infty}^{0} e^{-kx} dx [/tex]

    k<0

    is the same thing...

    Daniel.

    P.S.I don't like modulus under the integral sign... :yuck:
     
  7. Jan 18, 2005 #6

    Galileo

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    [tex]\int \limits_{-\infty}^{\infty}\frac{dx}{1+x^2}[/tex]

    Easily integrated with a beautiful result.
    This is one of my favourite integrals... (boy, that sounded nerdish).
     
  8. Jan 18, 2005 #7

    kreil

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    Thanks for all the equations

    Would someone take the time to show me the algebra for

    [tex]\int_{-\infty}^{\infty}(\frac{1}{2\pi}e^{-x^2})dx=1[/tex]

    I don't know how to solve integrals at infinity, the only thing I know how to use to evaluate integrals right now is the fundamental theorem (I'm a high school senior in AP calc AB, but I plan to be a math major).
     
  9. Jan 18, 2005 #8

    dextercioby

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    OMG,i saw it just now...HALLS,what did u do???????????? :surprised :surprised :surprised Where is the square root?????????????????And where did that 2 come from????????????

    Daniel.

    P.S.O.P.1-st question:Do you know Fubini's theorem??If u don't,are u willing to accept that the product of 2 improper integrals on R is equivalent with a double improper integral on R^{2}??
     
    Last edited: Jan 18, 2005
  10. Jan 18, 2005 #9

    Galileo

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    Integrals at infinity are defined by a limit, just like sequences or series.
    Similarly, if the integral limit exists, the integral is called convergent.

    [tex]\int_a^{\infty}f(x)dx=\lim_{t \to \infty} \int_a^t f(x)dx[/tex].

    For [itex]\int \limits_{-\infty}^{\infty}\exp(-x^2)dx[/itex] you need a little trick which uses a double integral.
     
  11. Jan 18, 2005 #10

    matt grime

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    come on! 2pi=1, so root(2pi)=2pi

    The OP says he only knows the fundamental theorem of calc so i'm guessing fubini's theorem is not on the syllabus.

    For the OP. It can't be done by the elementary means he knows, since it isn't an elementary integral, ie the integrand isn't the derivative of something nice.
     
  12. Jan 18, 2005 #11

    dextercioby

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    Okay,Matt,how about the DEFINITION OF ERF??? :wink: Or is that the trully "relative" thing...??

    Daniel.

    EDIT:That should be actually:

    [tex] 2\pi\cdot n\cdot i =1 ;n\in Z^{*} [/tex]

    :tongue2: :wink:
     
    Last edited: Jan 18, 2005
  13. Jan 18, 2005 #12

    Curious3141

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    Isn't it actually [itex]2\pi i[/tex] that's actually 1 ? :rofl:

    EDIT : it's actually zero, d'oh ! :yuck:
     
    Last edited: Jan 18, 2005
  14. Jan 18, 2005 #13

    matt grime

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    If 2pi were 1 then no one would make a mistake in fourier series again.... apart from the other mistakes that is.
     
  15. Jan 18, 2005 #14

    HallsofIvy

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    I'll have to remember that! Yes, I dropped the square root.

    kreil: to integrat [itex]\int_{-\infty}^\infty e^{-x^2}dx[/itex], first write
    [itex]I= \int_{-\inty}^\infty e^{-x^2}dx[/itex]. Because of the symmetry,
    [itex]\frac{I}{2}= \int_0^\infty e^{-x^2}dx[/itex]. It is also true that [itex]\frac{I}{2}= \int_0^\infty e^{-y^2}dy[/itex] since that's just a change in dummy variable.

    Multiply the two together: [itex]\frac{I^2}{4}= \int_0^\infty e^{-x^2}dx \int_0^\infty e^{-x^2}dx[/itex]. Fubini's theorm, that dextercioby referred to, basically says that that product is the same as the double or repeated integral (which Fubini's theorem also says are the same) [itex]\int_{x=0}^{\infty}\int_{y=0}^{\infty}e^{-(x^2+ y^2)} dydx[/itex]. Think of that as an integral over the first quadrant in the xy-plane and convert to polar coordinates: x2+ y[/sup]2[/sup] becomes r2 and dydx become r dr dθ. See that "r" in the differential? That's the trick that makes this possible!
    In order to cover the first quadrant r must go from 0 to [itex]\infty[/itex] and θ from 0 to [itex]\frac{\pi}{2}[/itex]. In polar coordinates, we have [itex]\frac{I^2}{4}= \int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}{\infty}e^{-r^2}rdrd\theta[/itex].

    Since there is no θ in the integrand, we have immediately [itex]\frac{I^2}{4}= \frac{\pi}{2}\int_{r=0}^{\infty}re^{-r^2}dr[/itex].

    Let u= r2. Then du= 2rdr so rdr= du/2 (see why we needed that "r"?). Of course, when r=0,u= 0 and when t= [itex]\infty[/itex], so does u. In terms of the variable u, [itex]\frac{I^2}{4}= \frac{\pi}{4}\int_{u=0}^{\infty}e^{-u}du[/itex]. That's easy to integrate and we have [itex]\frac{I^2}{4}= \frac{\pi}{4}[/itex] so [itex]I^2= \frac{\pi}[/itex] and [itex]I= \sqrt{\pi}[/itex].

    I forgot the square root, as several people pointed out!
     
    Last edited: Jan 18, 2005
  16. Jan 18, 2005 #15
    How about [tex]\int_{- \infty}^{+ \infty} x dx[/tex]
     
  17. Jan 18, 2005 #16
    You can use the basic definition of the improper integral over the reals as a limit to show that this expression diverges (represents no value in the reals).
     
  18. Jan 18, 2005 #17

    Galileo

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    That's an interesting case.

    [tex]\lim_{\epsilon_{1,2} \to _{\pm}\infty}\int_{\epsilon_1}^{\epsilon_2}xdx=\frac{1}{2}\lim_{\epsilon_{1,2} \to _{\pm}\infty}\left(\epsilon_2^2-\epsilon_1^2\right)[/tex]

    The limit doesn't exist if [itex]\epsilon_1[/itex] and [itex]\epsilon_2[/itex] approach infinity in an arbitrary manner.

    But if [itex]\epsilon_1=\epsilon_2=\epsilon[/itex]:

    [tex]\lim_{\epsilon \to \infty}\int_{-\epsilon}^{+\epsilon}xdx[/tex]
    does exist.
    It's called the Cauchy principal value of the improper integral.
     
    Last edited: Jan 18, 2005
  19. Jan 18, 2005 #18

    kreil

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    Just read all the posts, got stuck with detention for being late to school, lots of information to work on...

    Galileo-


    If you could give me a simple example I would be able to fully grasp this.



    Thank you halls of ivy for showing that to me, I will have to work on it a bit.
     
  20. Jan 18, 2005 #19

    Galileo

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    Okay. A very simple case:

    [tex]\int_{1}^{\infty}\frac{1}{x^2}dx=\lim_{t \to \infty} \int_1^t\frac{1}{x^2}dx=\lim_{t \to \infty}\left(-\frac{1}{x}\right]^t_1=\lim_{t \to \infty}\left(1-\frac{1}{t}\right)=1[/tex]

    Another one, using [itex]\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}[/itex]:

    [tex]\int_{-\infty}^{+\infty}\frac{1}{1+x^2}dx=2 \int_{0}^{\infty}\frac{1}{1+x^2}dx=2\lim_{t \to \infty} (\arctan(t)-\arctan(0))=\pi[/tex]
     
  21. Jan 18, 2005 #20

    t!m

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    I think Gabriel's Horn ties in nicely in this discussion. For those who don't know, it is the shape generated for the revolution of 1/x around the x-axis for x>= 1. In doing the improper integrals, it has a finite volume of pi, but an infinite surface area. Blows my mind.
     
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