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Math Help - Are all functions integrable?

  1. #1
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    Are all functions integrable?

    Hi all,

    Are all functions 'Integrable'? Is there any function existing which is not integrable? can anyone give me an example??
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    Sorry, you'll have to define what you mean FAR better than that.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sreejithpdas View Post
    Hi all,

    Are all functions 'Integrable'? Is there any function existing which is not integrable? can anyone give me an example??
    No, a function is not defined if it has an infinite amount of discontiunities on the interval it is being integrated on..

    Also I could interpret this as

    Neither

    \int_0^{1}\frac{dx}{x^2} or \int_1^{\infty}\frac{dx}{\sqrt{x}}

    are "not integrable" because they diverge...


    it all depends on how you define integrable
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    Quote Originally Posted by sreejithpdas View Post
    Hi all,

    Are all functions 'Integrable'? Is there any function existing which is not integrable? can anyone give me an example??
    No. Dirichlet's function is an example.

    Quote Originally Posted by Mathstud28 View Post
    No, a function is not defined if it has an infinite amount of discontiunities on the interval it is being integrated on..
    That is wrong. A function can have infinitely many discontinouties and still be integrable.

    it all depends on how you define integrable
    Your examples with improper integrals are bad examples because "integrable" is defined on closed bounded intervals.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    No. Dirichlet's function is an example.


    That is wrong. A function can have infinitely many discontinouties and still be integrable.
    I thought that...hmm...well...I guess you cannot believe everything you hear! Thanks TPH!
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    That is wrong. A function can have infinitely many discontinouties and still be integrable.
    This is the case with Laplace Transforms, right (the functions must be piecewise continuous in order to apply the transform)?

    Also, I believe \int e^xtan(x)\,dx is another integral that can't be solved...

    That integral came up in my Differential Equations class when solving y^{//}+y=tan(x) (this was the correct equation). He gave us y^{//}-y=tan(x), and we wasted 45 minutes trying to figure that one out... :'(
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    Quote Originally Posted by Chris L T521 View Post
    This is the case with Laplace Transforms, right (the functions must be piecewise continuous in order to apply the transform)?
    I do not know the theory behind Laplace transforms so I cannot say. But I am assuming that the necessacity of "piecewise continous" can be relaxed and you can use "integrable" instead. The reason why books do not use that is because they are engineering books and engineering students do not know what "integrable" means. Also, Laplace Transforms are not a good example for the above question because the poster was asking about proper integrables, i.e. integrals over finite closed intervals.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    This is the case with Laplace Transforms, right (the functions must be piecewise continuous in order to apply the transform)?

    Also, I believe \int e^xtan(x)\,dx is another integral that can't be solved...

    That integral came up in my Differential Equations class when solving y^{//}+y=tan(x) (this was the correct equation). He gave us y^{//}-y=tan(x), and we wasted 45 minutes trying to figure that one out... :'(
    Well

    \int{e^{x}\tan(x)dx}=\int\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\cdot\sum_{n=0}^{  \infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}x^{2n+1}}{(2n+2)!}\bigg]dx

    Multiply out and integrate termwise
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    Quote Originally Posted by Chris L T521 View Post
    Also, I believe \int e^xtan(x)\,dx is another integral that can't be solved...
    Quote Originally Posted by Mathstud28 View Post
    Well

    \int{e^{x}\tan(x)dx}=\int\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\cdot\sum_{n=0}^{  \infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}x^{2n+1}}{(2n+2)!}\bigg]dx

    Multiply out and integrate termwise
    I think Chris meant that you cannot find the integral in closed form. But "integrable" does not mean having a closed form. In fact, "integrable" is used only for definite integrals. Integrable means that the Riemann Sums approximate the function well. So we can take a limiting value and call that to be the integral. Though not always it is possible.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Well

    \int{e^{x}\tan(x)dx}=\int\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\cdot\sum_{n=0}^{  \infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}x^{2n+1}}{(2n+2)!}\bigg]dx

    Multiply out and integrate termwise
    argh...series comes to the rescue again...

    But that looks nasty to integrate termwise...
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I think Chris meant that you cannot find the integral in closed form. But "integrable" does not mean having a closed form. In fact, "integrable" is used only for definite integrals. Integrable means that the Riemann Sums approximate the function well. So we can take a limiting value and call that to be the integral. Though not always it is possible.
    I have always wondered this

    Is this integrable?

    \int_0^{0}\bigg(sign(x)\bigg)'dx?

    EDIT: I call the signum function sign(x) just in case that was unclear
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    Quote Originally Posted by Mathstud28 View Post
    I have always wondered this

    Is this integrable?

    \int_0^{1}\bigg(sign(x)\bigg)'dx?

    EDIT: I call the signum function sign(x) just in case that was unclear
    You cannot write that. The sign function is not differenciable at 0. So the derivative there makes no sense.
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  13. #13
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I have always wondered this

    Is this integrable?

    \int_0^{0}\bigg(sign(x)\bigg)'dx?

    EDIT: I call the signum function sign(x) just in case that was unclear
    Well, wouldn't that give you zero? Since the upper and lower bounds are the same??
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You cannot write that. The sign function is not differenciable at 0. So the derivative there makes no sense.
    But I thought sign(0) was a point on the unit circle in the complex plane...why is it not differentiable
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Well, wouldn't that give you zero? Since the upper and lower bounds are the same??
    Assuming you could integrate it (which apparently you cant)

    it would give

    sign(0)-sign(0)

    which apparently I have come to know is synonomous with

    \pm-\pm

    but that cant be right
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