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Math Help - Improper Integrals

  1. #1
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    Improper Integrals

    Use the comparison test to determine whether the improper integral converges or diverges:
     \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}

    So, since this is true... \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)} <  \int_{0}^{\infty} \frac{1}{\sqrt{x}} I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

     \int_{0}^{\infty} x^\frac{-1}{2} which equals:

     \lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2} equals

     \lim_{t \to \infty} \frac {x^\frac{1}{2}}{2} from 0 -> \infty

    So this just turns out to be  \infty - \frac {0^\frac{1}{2}}{2} so this is equal to...
     \infty - 0 which =  \infty

    So because this integral DIVERGES TO \infty, this tells us nothing? So what do I do now?
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    Quote Originally Posted by larson View Post
    Use the comparison test to determine wehter the improper integral converges or diverges:
     \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}

    So, since this is true... \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)} <  \int_{0}^{\infty} \frac{1}{\sqrt{x}} I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

     \int_{0}^{\infty} x^\frac{-1}{2} which equals:

     \lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2} equals

     \lim_{t \to \infty} \frac {x^\frac{1}{2}}{2} from 0 -> \infty

    So this just turns out to be  \infty - \frac {0^\frac{1}{2}}{2} so this is equal to...
     \infty - 0 which =  \infty

    So because this integral DIVERGES TO \infty, this tells us nothing? So what do I do now?
    All that you have said is that the integrals values is LESS THAN or equal to infinity. to show divergence you would need to show that it is greater than or equal to infinity.

    Try this comparison

    \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}=\int_0^1 \frac {dx}{\sqrt{x}(1 + x)} +\int_1^\infty \frac {dx}{\sqrt{x}(1 + x)}

     < \int_{0}^{1}\frac{dx}{\sqrt{x}}+\int_1^{\infty}\fr  ac{dx}{x^{3/2}}
    Last edited by TheEmptySet; April 20th 2008 at 04:35 PM.
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    Quote Originally Posted by larson View Post
    Use the comparison test to determine whether the improper integral converges or diverges:
     \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}
    Note it does converges by my above post now, but also it can be integrated with this sub

    u=\sqrt{x} \mbox{ then } du=\frac{dx}{\sqrt{x}}

    Note that the limits of int don't change so we get

    \int_{0}^{\infty}\frac{du}{1+u^2}=\tan^{-1}u|_{0}^{\infty}= \frac{\pi}{2}-0=\pi/2

    I hope this helps.
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  4. #4
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    Quote Originally Posted by larson View Post
    Use the comparison test to determine whether the improper integral converges or diverges:
     \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}

    So, since this is true... \int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)} <  \int_{0}^{\infty} \frac{1}{\sqrt{x}} I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

     \int_{0}^{\infty} x^\frac{-1}{2} which equals:

     \lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2} equals

     \lim_{t \to \infty} \frac {x^\frac{1}{2}}{2} from 0 -> \infty

    So this just turns out to be  \infty - \frac {0^\frac{1}{2}}{2} so this is equal to...
     \infty - 0 which =  \infty

    So because this integral DIVERGES TO \infty, this tells us nothing? So what do I do now?
    ...maybe I misread the past posts but \int_0^{\infty}\frac{1}{\sqrt{x}(1+x)}=2arctan(\sq  rt{x})|_0^{\infty}=\pi-0=\pi
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ...maybe I misread the past posts but \int_0^{\infty}\frac{1}{\sqrt{x}(1+x)}=2arctan(\sq  rt{x})|_0^{\infty}=\pi-0=\pi
    Oh and to make how this integral was begotten let me write it in a better form 2\int_0^{\infty}\frac{\frac{1}{2\sqrt{x}}}{(1^2+(\  sqrt{x})^2)}dx
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    Hmm... now I'm just confused...
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by larson View Post
    Hmm... now I'm just confused...
    What is confusing?
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    Is what I posted at the first post correct? It's just I should have rather of found an integral that was less than the original integral? Because I didn't really prove anything in the end, right?
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  9. #9
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    Quote Originally Posted by larson View Post
    Is what I posted at the first post correct? It's just I should have rather of found an integral that was less than the original integral? Because I didn't really prove anything in the end, right?
    Try \frac{1}{x^{\frac{3}{2}}} instead
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    Try \frac{1}{x^{\frac{3}{2}}} instead
    Converges to 0 correct?
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by larson View Post
    Converges to 0 correct?
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    Quote Originally Posted by Mathstud28 View Post
    Finally I get something correct... lol. Thanks.
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    It is double improper

    Quote Originally Posted by Mathstud28 View Post
    Try \frac{1}{x^{\frac{3}{2}}} instead
    I am sorry to say that this diverges see post #2

    \int_0^{\infty}\frac{dx}{x^{(3/2)}} \lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\frac{dx}{x^{3/2}}

    \lim_{a \to 0} \lim_{b \to \infty} -\frac{2}{\sqrt{x}}|_a^{b}=\lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\left( \frac{-2}{\sqrt{b}}+\frac{2}{\sqrt{a}}\right)=0+\infty=\i  nfty

    So it diverges

    It is double improper
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I am sorry to say that this diverges see post #2

    \int_0^{\infty}\frac{dx}{x^{(3/2)}} \lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\frac{dx}{x^{3/2}}

    \lim_{a \to 0} \lim_{b \to \infty} -\frac{2}{\sqrt{x}}|_a^{b}=\lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\left( \frac{-2}{\sqrt{b}}+\frac{2}{\sqrt{a}}\right)=0+\infty=\i  nfty

    So it diverges

    It is double improper
    Oops...forgot the integral...but you get the idea use one that when you integrate you get a degree less than one
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  15. #15
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    I still believe it converges... Mathstud pointed me to use this...  \frac {1}{x^\frac{3}{2}} which is greater than the original function. I proved that this  \frac {1}{x^\frac{3}{2}} converges, so the original much converge...
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