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Indefinite Integral of an Absolute Convergent Function

  1. Nov 8, 2008 #1
    I was wondering if a function is absolutely convergent over a certain interval, say,
    will its indefinite integral also be absolutely convergent over the same interval?

    Also, assume that f(x) is convergent for
    g(x) = \int{\int_{0}^{\infty}f(x)dx}dy &=& {\int_{0}^{\infty}}{\int{f(x)}dydx + {C}
    this function, g(x), also be convergent on the same interval as f(x)?
    Last edited: Nov 8, 2008
  2. jcsd
  3. Nov 8, 2008 #2
    Any help would be appreciated.
    Last edited: Nov 8, 2008
  4. Nov 8, 2008 #3
    It is slightly unpleasant to read somebody's posts, if one has to use latex-software to generate some pdf or ps files out of the posted source code.

    I suggest you start using the "edit"-button now :wink:

    Get those begin and end document commands out, and use [ tex ] and [ /tex ] tags for latex equations.
  5. Nov 8, 2008 #4
    Sorry I cannot make sense of your notation.
    [tex]g(y !!!) = \int{\int_{0}^{\infty}f(x)dx}dy = F(\infty)\cdot y - F(0) \cdot y +C[/tex]
  6. Nov 8, 2008 #5


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    A single function is neither "convergent" nor "uniformly convergent". Did you mean "continuous"?
  7. Nov 8, 2008 #6
    No, what I meant was if the indefinite integral of that function converges on the same interval as the original function.
  8. Nov 8, 2008 #7
    If my notation is confusing, here is what I am saying:
    Assume that f(x) is convergent for
    g(x) = \int[{\int_{0}^{\infty}f(x)dx}]dy &=& {\int_{0}^{\infty}}[{\int{f(x)}dy]dx + {C}
    this function, g(x), also be convergent on the same interval as f(x)?
  9. Nov 8, 2008 #8
    Adding brackets didn't help. The result of the integration would still be what I wrote down before. How about you demonstrate what you mean with [tex]f(x)=\frac{1}{(x+1)^2}[/tex] ?
    And tell me what g(x) is then.
  10. Nov 8, 2008 #9
    Ok. So basically what I am saying is, assume [tex] f(x)=\frac{y}{(x+1)^2} [/tex] is convergent on [tex] (0,\infty).[/tex]. Would then, [tex] g(x) = {\int}}({\int_{0}^{\infty}}\frac{y}{(x+1)^2}}dx)dy &=& {\int_{0}^{\infty}}({\int{\frac{y}{(x+1)^2}}dy)dx + {C} = {\int_{0}^{\infty}}{\frac{y^2}{{2}{(x+1)^2}}dx + {C} [/tex] also be convergent on the same interval? Is that clear enough?!
    Last edited: Nov 8, 2008
  11. Nov 8, 2008 #10
    No it is not. "g(x)" is not a function of x but of y because the x disappears when you insert the boundaries. "f(x)" is a function of x and y so it should be f(x,y). The C comes from out of nowhere, and you illegally take it out of your integration. This sloppy notation will never give accurate results.
    You don't specify what you mean by a convergent function so I assumed you meant a the integral to converge, but I don't know. And finally you should be able to resolve the last integral with high school calculus, so you would see that the integral converges and not assume it.

    I am sorry if this sounds harsh but if you don't clean up your notation we will get nowhere.
  12. Nov 8, 2008 #11
    Sorry about that. [tex] f(x,y) &=& \frac{y}{(x+1)^2} [/tex]. [tex] {\int}}({\int_{0}^{\infty}}\frac{y}{(x+1)^2}}dx)dy &=& {\int_{0}^{\infty}}({\int{\frac{y}{(x+1)^2}}dy)dx + {C} = {\int_{0}^{\infty}}{\frac{y^2}{{2}{(x+1)^2}}dx + {C} [/tex]

    Yes, what I meant by convergence is that the integral converges in the domain [tex] (0,\infty) [/tex]. I most definitely could not have been referring to anything in that case.
    Last edited: Nov 8, 2008
  13. Nov 8, 2008 #12
    I solved my problem. Basically, I said [tex]f(x,n) &=& {e^x}{n}[/tex]. Next, I found [tex]{\int}}{\int}}{e^x}{n}dn dx = {\int}{e^x}{nx}dn = n{{\int}{e^x}{x}dx}[/tex]. Then I found [tex]{\int}}{\int}}{e^x}{n}dx dn &=& {\int}{n}{e^x}{dn} = {nx}{e^x}+{C}[/tex]. Thus, since [tex]n{{\int}{e^x}{x}dx}[/tex] obviously does not equal [tex]{nx}{e^x}+{C}[/tex], we can thus conclude that [tex]{\int}}{\int}}{f(x,n)}dndx[/tex] does not equal [tex]{\int}}{\int}}{f(x,n)}dxdn[/tex]. So the answer to my question is a resounding "NO".
    Last edited: Nov 8, 2008
  14. Nov 8, 2008 #13
    You just disproved Fubini's theorem. These calculations are false beyond repair. Sorry. Maybe some of these will cheer you up:
  15. Nov 8, 2008 #14
    Then you do it. Show me what I did wrong and disprove me, please.
  16. Nov 9, 2008 #15
    Ok. Fubini's theorem states that you may change the order of integration in normal circumstances.

    Doing infinite integrals will not work before we can do finite integrals. Doing multi variable integrals will not work before we can so single variable integrals. Doing integrals will not work before we master derivatives.

    The integral's solution is [tex]\int \int f(x,n) \, dx \,dn = \int \int ne^x \, dx \,dn = \frac{n^2}{2}e^x + c_1 x + c_2 y + c_3[/tex] with any order of integration.

    Using n is a bad choice of variables because it usually denotes constants that are not being integrated over. Your calculation shows very basic errors with integration of the simplest functions like [tex]f(x)=kx[/tex] You do not know your integration rules. I suspect you cannot derive either, because that would have enabled you to check your results. Maybe a page like http://www.calculus.org/ could help, or a good book, or a college course or a math teacher since I don't know your age. Learning these things takes a normal person a year, that is in a class with homework.
  17. Nov 9, 2008 #16
    FYI, I have taken calculus, and I got an A in it (in fact, I got the highest grade in the class). It's just that I have not yet taken multivariable calculus, which involves double integrals. It's double integrals which I find really confusing.

    Thanks for your help in preparing me for the multivariable course I will be eventually taking
  18. Nov 9, 2008 #17
    Yeah, you're right about the n thing. I should have used a better variable b/c I kept on thinking it was a constant (FAIL).

    I see my error (wow, I can't believe I made such a stupid mistake).
  19. Nov 9, 2008 #18
    Quick Question: Does Fublini's Theorem apply to the indefinite integral of an improper integral? In other words, [tex]{\int}}{\int_{0}^{\infty}}{f(x,n)}dxdn &=& {\int_{0}^{\infty}}{\int}}{f(x,n)}dndx[/tex]?
  20. Nov 9, 2008 #19
    No it doesn't. The integration constant produced by the integral over n will kill any convergence, when doing the integral over x. Furthermore I don't know if you noticed, but the double integral is not solved by simply integrating twice, you search for a function that will produce f(x,y) after deriving twice. For smooth functions it doesn't matter which derivative you do first, so I think that implies that you can exchange the order of integration if there are no boundaries, for boundaries there is Fubini saying the same thing. I have not seen integrals with and without boundaries mixed yet anyway.
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