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Math Help - Improper integrals of type I: convergent or divergent?

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    Improper integrals of type I: convergent or divergent?

    I've got two improper integrals, and need to determine whether or not they are convergent.

    1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
    I said that this function was =< 1/x, and was hoping to say that the integral from 1 to infinity of 1/x converged, therefore my given function would... but that's not the case Can someone point me in the right direction? I know that it's supposed to converge, but is there an easy way to see this?

    2) (x^2)e^(-x^2) from 0 to infinity
    No idea where to go with this one, any hints? Something I should look for?

    Thanks!!
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    Quote Originally Posted by mistykz View Post
    I've got two improper integrals, and need to determine whether or not they are convergent.

    1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
    I said that this function was =< 1/x, and was hoping to say that the integral from 1 to infinity of 1/x converged, therefore my given function would... but that's not the case Can someone point me in the right direction? I know that it's supposed to converge, but is there an easy way to see this?

    Thanks!!
    Hint)

    \displaystyle{\int\limits_1^\infty\frac{dx}{x\sqrt  {x+3}}<\int\limits_1^\infty\frac{dx}{x\sqrt{x}}}=\  lim\limits_{b\to\infty}\int\limits_1^b\frac{dx}{x^  {3/2}}=\left.{-2\lim\limits_{b\to\infty}\frac{1}{\sqrt{x}}}\right  |_1^b=-2\lim\limits_{b\to\infty}\!\left(\frac{1}{\sqrt{b}  }-1\right)=2}
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    Ah, that makes sense! Thanks!!
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by mistykz View Post
    I've got two improper integrals, and need to determine whether or not they are convergent.

    1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
    I said that this function was =< 1/x, and was hoping to say that the integral from 1 to infinity of 1/x converged, therefore my given function would... but that's not the case Can someone point me in the right direction? I know that it's supposed to converge, but is there an easy way to see this?

    2) (x^2)e^(-x^2) from 0 to infinity
    No idea where to go with this one, any hints? Something I should look for?

    Thanks!!
    \displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}=-\frac{1}{2}\int\limits_0^\infty{x\,d(e^{-x^2})}=\left.{-\frac{1}{2}xe^{-x^2}} \right|_{x=0}^{x\to\infty}+\frac{1}{2}\int\limits_  0^b{e^{-x^2}\,dx}=}

    \displaystyle{=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{x}{e^{x^2  }}+\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{(x)'}{(e^  {x^2})'}+\frac{\sqrt{\pi}}{4}=}

    \displaystyle{=-\frac{1}{4}\lim\limits_{x\to\infty}\frac{1}{xe^{x^  2}}+\frac{\sqrt{\pi}}{4}=-\frac{1}{4}\cdot0+\frac{\sqrt{\pi}}{4}=\frac{\sqrt  {\pi}}{4}}


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    Last edited by DeMath; November 26th 2010 at 06:36 PM.
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    Senior Member DeMath's Avatar
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    Thanks! Although I believe that seems to be outside the scope of our course, it definitely works lol
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by mistykz View Post
    2) (x^2)e^(-x^2) from 0 to infinity
    No idea where to go with this one, any hints? Something I should look for?
    Thanks!!
    Quote Originally Posted by DeMath View Post
    \displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}=-\frac{1}{2}\int\limits_0^\infty{x\,d(e^{-x^2})}=\left.{-\frac{1}{2}xe^{-x^2}} \right|_{x=0}^{x\to\infty}+\frac{1}{2}\int\limits_  0^b{e^{-x^2}\,dx}=}

    \displaystyle{=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{x}{e^{x^2  }}+\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{(x)'}{(e^  {x^2})'}+\frac{\sqrt{\pi}}{4}=}

    \displaystyle{=-\frac{1}{4}\lim\limits_{x\to\infty}\frac{1}{xe^{x^  2}}+\frac{\sqrt{\pi}}{4}=-\frac{1}{4}\cdot0+\frac{\sqrt{\pi}}{4}=\frac{\sqrt  {\pi}}{4}}
    This solution is better

    \displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}=-\frac{1}{2}\int\limits_0^b{x\,d(e^{-x^2})}=\left.{-\frac{1}{2}xe^{-x^2}}\right|_{x=0}^{x\to\infty}+\frac{1}{2}\int\li  mits_0^\infty{e^{-x^2}\,dx}}

    \displaystyle{\left.{-\frac{1}{2}xe^{-x^2}}\right|_{x=0}^{x\to\infty}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{x}{e^{x^2  }}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{(x)'}{(e^  {x^2})'}=-\frac{1}{4}\lim\limits_{x\to\infty}\frac{1}{xe^{x^  2}}=-\frac{1}{4}\cdot0=0}

    \displaystyle{\int\limits_0^\infty{e^{-x^2}\,dx}=\int\limits_0^1{e^{-x^2}}\,dx}+\int\limits_1^\infty{e^{-x^2}\,dx}}

    The integral \displaystyle{\int\limits_0^1{e^{-x^2}}\,dx} is own, that is convergent.

    \displaystyle{\int\limits_1^\infty{e^{-x^2}\,dx}<\int\limits_1^\infty\frac{dx}{x^2}=\left  .{-\lim\limits_{b\to\infty}\frac{1}{x}}\right|_1^b=-\lim\limits_{b\to\infty}\!\left(\frac{1}{b}-1\right)=1}

    So, the integral \displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}} is convergent.
    Last edited by DeMath; November 26th 2010 at 07:28 PM. Reason: Add!
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    Quote Originally Posted by mistykz View Post
    I've got two improper integrals, and need to determine whether or not they are convergent.

    1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
    all you gotta do is to bound it and conclude namely that \displaystyle\int_1^\infty\dfrac{dx}{x^n}<\infty if n>1.


    Quote Originally Posted by mistykz View Post
    2) (x^2)e^(-x^2) from 0 to infinity
    play with inequalities, just analise it for x\ge1, then -x^2\le-x so e^{-x^2}\le e^{-x}, which implies that x^2e^{-x^2}\le x^2e^{-x}, and the one of the right side is known to converge.
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    Senior Member DeMath's Avatar
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    Note that I added a line in my last post.
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