Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Hi,

    I've just started this topic and am struggling with this question.

    For which $\alpha > 0$ does the improper integral
    $\int_0^\infty \frac{x^2\cos x}{(1+x^2)^{\frac\alpha2}}\ dx$
    exists?

    I have attached some workings that I have done, but unsure if that's even the correct way around it and don't know how to finish it.

    Sorry not sure how to get it in the right format, but the question is written at the top of my workings too.
    Attached Images
     
    Offline

    15
    ReputationRep:
    (Original post by Roxanne18)
    Hi,

    I've just started this topic and am struggling with this question.

    For which $\alpha > 0$ does the improper integral
    $\int_0^\infty \frac{x^2\cos x}{(1+x^2)^{\frac\alpha2}}\ dx$
    exists?

    I have attached some workings that I have done, but unsure if that's even the correct way around it and don't know how to finish it.

    Sorry not sure how to get it in the right format, but the question is written at the top of my workings too.
    Can you work out the limit as R tends to infinity at the end of your solution? Appreciating for what alpha that limit exists is a good starting point.

    EDIT: also OP I think that exponent should be 2 - alpha not alpha - 2
    Offline

    17
    ReputationRep:
    I'm a bit concerned about the fact that the original integral is going to converge significantly "better" than the modulus version; because cos is oscillatory I'm pretty sure the integral converges if the integrand does, which isn't true for the modulus version.

    But I'm not seeing a nice way of proving the oscillatory integrand converges. I think you'd have to do some quite involved work considering +ve and -ve intervals of cos and that feels outside the scope of the question.
    Offline

    15
    ReputationRep:
    (Original post by DFranklin)
    I'm a bit concerned about the fact that the original integral is going to converge significantly "better" than the modulus version; because cos is oscillatory I'm pretty sure the integral converges if the integrand does, which isn't true for the modulus version.

    But I'm not seeing a nice way of proving the oscillatory integrand converges. I think you'd have to do some quite involved work considering +ve and -ve intervals of cos and that feels outside the scope of the question.
    Yes, it's hard to answer without knowing what context/level of knowledge this question is being asked in.

    The answer is that

    the Lebesgue integral exists when alpha exceeds three;
    only the improper Lebesgue integral exists when alpha is three;
    otherwise the integral doesn't exist.

    But I don't know how helpful that is to the OP, not knowing how much integration theory they've met.
    Offline

    17
    ReputationRep:
    (Original post by RichE)
    Yes, it's hard to answer without knowing what context/level of knowledge this question is being asked in.

    The answer is that

    the Lebesgue integral exists when alpha exceeds three;
    only the improper Lebesgue integral exists when alpha is three;
    otherwise the integral doesn't exist.
    I don't recall the Lebesgue definitions for improper etc (one thing I've always felt Oxford does better than Cambridge), but I'm pretty sure that

    \displaystyle \lim_{L \to \infty} \int_0^L \dfrac{x^2 \cos x}{(1+x^2)^{\alpha/2}}\,dx converges for \alpha > 2.

    (For example when \alpha = 2.5 the integral is surely going to behave like \int_1^\infty \dfrac{\cos x}{\sqrt{x}} \,dx which is a (convergent) Fresnel integral).

    Edit: although I think we both suspect that the OP isn't supposed to worry about all this, but I don't think you can find the exact range without quite a bit of theory/grind.
    Offline

    15
    ReputationRep:
    (Original post by DFranklin)
    I don't recall the Lebesgue definitions for improper etc (one thing I've always felt Oxford does better than Cambridge), but I'm pretty sure that

    \displaystyle \lim_{L \to \infty} \int_0^L \dfrac{x^2 \cos x}{(1+x^2)^{\alpha/2}}\,dx converges for \alpha > 2.

    (For example when \alpha = 2.5 the integral is surely going to behave like \int_1^\infty \dfrac{\cos x}{\sqrt{x}} \,dx which is a (convergent) Fresnel integral).

    Edit: although I think we both suspect that the OP isn't supposed to worry about all this, but I don't think you can find the exact range without quite a bit of theory/grind.
    Yes you're right. So I should have written that the improper Lebesgue integral exists for alpha more than 2 and up to 3. Those improper integrals will exist in a manner akin to the Leibniz test. (For a proper Lebesgue integral to exist the integral of the absolute integrand will also exist.)

    EDIT: in fact thinking about things here, the Leibniz test may be the best, more elementary way forward for showing the improper integrals exist when alpha exceeds 2.
    Offline

    15
    ReputationRep:
    (Original post by DFranklin)
    ...
    Thanks for the rep. BTW as someone who keeps their hand in with maths, don't know whether

    https://www.amazon.co.uk/Towards-Hig...er+mathematics

    would be of any interest to you...
    Offline

    11
    ReputationRep:
    (Original post by Roxanne18)
    Hi,

    I've just started this topic and am struggling with this question.

    For which $\alpha > 0$ does the improper integral
    $\int_0^\infty \frac{x^2\cos x}{(1+x^2)^{\frac\alpha2}}\ dx$
    exists?

    Sorry not sure how to get it in the right format, but the question is written at the top of my workings too.
    Wrap [tex]...[/tex] tags around the latex (which I've cleaned up a bit).

    For which \alpha > 0 does the improper integral \displaystyle \int_0^\infty \frac{x^2\cos x}{(1+x^2)^{ \alpha/2}}\ dx exist?
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RichE)
    Yes, it's hard to answer without knowing what context/level of knowledge this question is being asked in.

    The answer is that

    the Lebesgue integral exists when alpha exceeds three;
    only the improper Lebesgue integral exists when alpha is three;
    otherwise the integral doesn't exist.

    But I don't know how helpful that is to the OP, not knowing how much integration theory they've met.
    Hi, thanks for the reply.
    I have just started year 2 of bachelors Mathematics degree. The title of this module is improper integrals, and I have never heard of Lebesgue integrals.
    Offline

    15
    ReputationRep:
    (Original post by Roxanne18)
    Hi, thanks for the reply.
    I have just started year 2 of bachelors Mathematics degree. The title of this module is improper integrals, and I have never heard of Lebesgue integrals.
    If you've met Leibniz's alternating series test, then I'd suggest splitting the integral up into intervals (n pi/2, (n+2)pi/2) where n is odd will help.
    Offline

    17
    ReputationRep:
    (Original post by Roxanne18)
    Hi, thanks for the reply.
    I have just started year 2 of bachelors Mathematics degree. The title of this module is improper integrals, and I have never heard of Lebesgue integrals.
    You don't really need to know about Lebesgue for this.

    I assume you're familiar with the alternating series (Leibniz) test, and in particular that

    \displaystyle \sum \frac{1}{n^\alpha} diverges for \alpha \leq 1, but \displaystyle \sum \frac{(-1)^n}{n^\alpha} doesn't diverge unless \alpha \leq 0

    Obviously, if you were to take the modulus during the analysis of the 2nd series, you'd end up with the first, and therefore think it diverged in cases where it doesn't.

    You have a similar problem here, but there's not really an integral version of the alternating series test, so you need to improvise. (In general terms, you form a series based on the integral that you can show converges, and then show that the integral can't behave very differently from the series).

    It's not exactly hard, but it's involved enough that I do question whether you're supposed to do it.
    • Thread Starter
    Offline

    9
    ReputationRep:
    [QUOTE=DFranklin;74233896]You don't really need to know about Lebesgue for this.

    I assume you're familiar with the alternating series (Leibniz) test, and in particular that

    \displaystyle \sum \frac{1}{n^\alpha} diverges for \alpha \leq 1, but \displaystyle \sum \frac{(-1)^n}{n^\alpha} doesn't diverge unless \alpha \leq 0

    Obviously, if you were to take the modulus during the analysis of the 2nd series, you'd end up with the first, and therefore think it diverged in cases where it doesn't.

    You have a similar problem here, but there's not really an integral version of the alternating series test, so you need to improvise. (In general terms, you form a series based on the integral that you can show converges, and then show that the integral can't behave very differently from the series).



    So why are we talking about when it converges, is that only when the integral exists ?
    Offline

    17
    ReputationRep:
    (Original post by Roxanne18)
    So why are we talking about when it converges, is that only when the integral exists ?
    It's the same thing. I think pure mathematicians are going to prefer to use converges (particularly once discussing the technical details of why a particullar integral exists or not).
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 24, 2017
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.