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Is this result trivial? (improper integral representations of real functions))

  1. Aug 10, 2012 #1
    I'm a college Sophomore majoring in math and over the summer I've been playing around with improper integrals, specifically integrals with limits at infinity because they've always fascinated me. The highest calculus course I've taken is Calc II, so I might be missing something here.

    Anyways, what really got me started on this was my discovery of the Pi function (Gamma -1) accidentally.

    I was doing integrals that I made up off the top of my head and I found that
    [itex] \int_{0}^{\infty}t^ne^{-t}dt =n! [/itex]
    I then realized this integral could actually represent the factorial function for any value of n.

    So I asked myself, is there a way to represent ANY real function as an improper integral?
    Is there a function [itex]\omega (t,x) [/itex] such that any real function and a fixed lower bound can be represented by [itex]f(x)=\int_{a}^{\infty}\omega (t,x)dt[/itex] ?

    I played around with a lot more integrals until I found that [itex]\int_{0}^{\infty}e^{-t/x}dt=x [/itex] for x>0
    Since and x>0 will give back any x, any f(x)>0 should give back f(x).

    So, this is what I propose can represent any real function f(x) as an improper integral
    [itex] f(x)=
    \begin{Bmatrix}
    \int_{0}^{\infty}e^{-t/f(x)}dt, f(x)>0 & \\
    0, f(x)=0 & \\
    \int_{0}^{\infty}-e^{t/f(x)}dt, f(x)<0 &
    \end{Bmatrix}
    [/itex]
    The minus sign on the last row is supposed to be outside the integral.

    So, an example
    [itex] sin(x)=
    \begin{Bmatrix}
    \int_{0}^{\infty}e^{-t/sin(x)}dt, 2n\pi<x<2n\pi +\pi & \\
    0, x=n\pi & \\
    \int_{0}^{\infty}-e^{t/sin(x)}dt, 2n\pi-\pi<x<2n\pi &
    \end{Bmatrix}
    [/itex]

    I seem to have found a way to represent any real function as an improper integral as long as I know when it’s greater, equal to, and less than 0.
    But my question to you guys is, is this result simply trivial? Can this be used to represent real functions the way I want to? I’m not a very experienced mathematician and I’m just asking your advice.
     
    Last edited: Aug 10, 2012
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  3. Aug 10, 2012 #2

    Simon Bridge

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    Lets see if I understand you:
    You are trying to discover if, for any function f which maps R->R can be expressed in terms of some other functon g which maps R2 -> R by a definite integral over the extra coordinate so your question becomes: "given f(x), can we find a g(x,y) that satisfies: [itex]f(x)=\int_a^b g(x,y)dy[/itex] ?


    if g(x,y)=yf(x) then [itex]\int g(x,y)dy = f(x) + c[/itex] and any limits will give you just f(x) out the end.

    So the answer to your question is - yes - it is trivial to find a g(x,y) which, when integrated, will give you a specific f(x).
    Remember - an integration is a fancy way of doing a sum ... you can always make a function out of a sum of other functions.

    BTW: your examples look a lot like a fourier transform don't they :)
    if [itex]g(x,y)=h(y)e^{2\pi i xy}[/itex] and the integral is done over all space, then h is the fourier transform of f.

    So what is interesting is the specifics - how you choose g(x,y) to satisfy some other criteria.
    Good exploration btw.

    As well as fourier transforms, you may want to look at the gamma function.
     
    Last edited: Aug 10, 2012
  4. Aug 11, 2012 #3

    Mute

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    There need to be some normalization constants in there, of course. For finite limits this example would be

    [tex]g(x,y) = \frac{2y}{b^2-a^2} f(x)[/tex]

    in order for [itex]f(x) = \int_a^b dy~g(x,y)[/itex].

    To represent f(x) as an integral with infinite limits over a function g(x,y) of the form h(y)f(x), h(y) would have to integrate to 1 over the integrate limits, so [itex]h(y) \propto y[/itex] is of course not a valid choice for infinite limits.

    This does help one see that there are pretty much infinitely many ways one could write some function f(x) as the integral of some function involving f(x).

    Similarly, the Gamma function is not the only function that can generalize n!; however, it is the only log-convex (and continuous?) function that does so. Typically you need extra desirable conditions to be able to choose a unique integral representation.

    The question is, why do you want to represent functions this way? Just for curiosity's sake?
     
    Last edited: Aug 11, 2012
  5. Aug 11, 2012 #4
    Interesting find if not a useful one.
    Usually people like to represent functions with others that share some common proprety.
    For example periodic functions with integrals of trigs. (Fourier transform)
    You can find all sorts of funny looking identities if you have an idea where to look.
     
  6. Aug 11, 2012 #5

    Simon Bridge

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    Oh yeah - I got sloppy ... so the conclusion is correct but not for the reason I specified. Thanks.
     
  7. Aug 11, 2012 #6
    Hello Sigma057 !

    I don't know if your identity is original or not.
    The fact is that simple changes of variable in integrals are likely to lead to various identities.
    For example, in attachment, your identity (with the function exp) is compared to some other obtained on the same manner. Of course, in each case, f(x) has to satisfy some conditions so that the integral be convergent : The conditions are not specified for a clearer présentation.
    I cannot say if your particular identity with exp has special useful properties and/or is of particular interest.
     

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    Last edited: Aug 11, 2012
  8. Aug 11, 2012 #7

    AlephZero

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    Your first example is interesting, because it gives a definition of a "factorial" function for all real numbers, not just integers. (It's usually called the gamma function and the definition is slightly different, but that's just a detail.).

    But I'm not sure the later examples lead anywhere, since in the integrals x and f(x) are effectively just constants for each particular value of x.

    BTW there's a mistake in JJacquelin's last example. When you change the variable the limits of the integral also change, unless the limits are 0 to ##\infty##.
    $$\int_0^{\pi/2} \sin (t/a)\, dt \ne a.$$
     
  9. Aug 11, 2012 #8
    Hi AlephZero !

    You are right. I made an horrible mistake in the last example. Shame on me ! So I delete it from my first post.
    Thank you very much for the remark.
     
  10. Aug 11, 2012 #9
    Thanks for the help everyone!
    Yes, it's basically just for curiosity's sake. I just wanted to know whether it could be done or not. I thought it might lead to some interesting results like my first example, but if it doesn't, at least it was fun =)

    I'll definitely look int the Gamma function's properties, as well as Fourier Transforms and Laplace transforms as well.
     
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