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Thread: Converging improper integrals

  1. #1
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    Converging improper integrals

    I've been asked to determine for what values of n does the integral of x^n.sin(1/x) .dx from 1 to infinity, converge. I've been given the Maclaurin series expansion of sin(x).

    I'm not sure how to tackle this problem can anyone offer any advice please?

    For what values of n is this integral converging?
    Converging improper integrals-screen-shot-2017-12-02-22.26.01.png
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  2. #2
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    Re: Converging improper integrals

    Try plugging in 1/x to the Maclaurin series and multiplying out by x^n
    Last edited by SlipEternal; Dec 2nd 2017 at 03:58 PM.
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    Re: Converging improper integrals

    I've tried its not a nice results, Maclaurin series expansion of sin(x) is given as a hint. Could I possibly replace all the x's in the series with 1/x to get the series expansion for sin(1/x) and then sub the first few terms of this expansion into the integral. Then I could multiply out the series and x^n?
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  4. #4
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    Re: Converging improper integrals

    Lol, I just changed my response to that exactly
    Thanks from CRD98
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    Re: Converging improper integrals

    Quote Originally Posted by CRD98 View Post
    I've tried its not a nice results, Maclaurin series expansion of sin(x) is given as a hint. Could I possibly replace all the x's in the series with 1/x to get the series expansion for sin(1/x) and then sub the first few terms of this expansion into the integral. Then I could multiply out the series and x^n?
    yes and note that $\displaystyle \int_1^\infty ~\dfrac{1}{x^n}~dx $ converges for $n>1$
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  6. #6
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    Re: Converging improper integrals

    Quote Originally Posted by CRD98 View Post
    I've tried its not a nice results, Maclaurin series expansion of sin(x) is given as a hint. Could I possibly replace all the x's in the series with 1/x to get the series expansion for sin(1/x) and then sub the first few terms of this expansion into the integral. Then I could multiply out the series and x^n?
    You can also plug in the indefinite integral into Wolframalpha and figure out for what values of n the antiderivative converges as x approaches infinity and the antiderivative exists at 1.

    http://m.wolframalpha.com/input/?i=i...%5B1%2Fx%5D%5D
    Thanks from CRD98
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    Re: Converging improper integrals

    Quote Originally Posted by romsek View Post
    yes and note that $\displaystyle \int_1^\infty ~\dfrac{1}{x^n}~dx $ converges for $n>1$
    So in my case the integral will converge for values for n < 0 , as this will give an integrand similar to the one you posted?
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