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Math Help - Converging integrals

  1. #1
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    Converging integrals

    Show whether or not each of the following integrals converge:
    i)

    ii)
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\int\limits_1^\infty  {e^{ - \left( {3x^2  + 2} \right)} dx = e^{ - 2} } \int\limits_1^\infty  {e^{ - 3x^2 } dx} <br /> <br />

    now it's easy to verify that:

    <br />
e^{ - 3x^2 }  \leqslant e^{ - 3x} \quad \forall 1 \leqslant x<br />

    thus:

    <br />
\int\limits_1^\infty  {e^{ - 3x^2 } } dx < \int\limits_1^\infty  {e^{ - 3x} } dx\quad \forall 1 \leqslant x<br />

    now:

    <br />
\int\limits_1^\infty  {e^{ - 3x} } dx = \left. { - \frac{1}<br />
{3}e^{ - 3x} } \right|_1^\infty   = \frac{1}<br />
{3}e^{ - 3}  < \infty <br />

    thus according to the comparison test we conclude that the integral converges.

    2.

    note that the integrand is an even function, thus:

    <br />
\int\limits_{ - \infty }^\infty  {e^{ - x^2 } dx}  = 2\int\limits_0^\infty  {e^{ - x^2 } dx} <br />

    now following the lines of the previous exercise we note that:

    <br />
e^{ - x^2 }  \leqslant e^{ - x} \quad \forall x \geqslant 0<br />

    ....
    Last edited by Peritus; May 4th 2008 at 10:35 AM.
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  3. #3
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    Krizalid's Avatar
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    Quote Originally Posted by matty888 View Post

    ii)
    Since e^x\ge x+1,\,\forall\,x\in\mathbb R we have e^{x^{2}}\ge x^{2}+1\implies e^{-x^{2}}\le \frac{1}{x^{2}+1}.

    Hence \int_{-\infty }^{+\infty }{e^{-x^{2}}\,dx}\le \int_{-\infty }^{+\infty }{\frac{1}{1+x^{2}}\,dx}=\pi . The given integral converges.
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  4. #4
    Senior Member Peritus's Avatar
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    Quote Originally Posted by Krizalid View Post
    Since e^x\ge x+1,\,\forall\,x\in\mathbb R we have e^{x^{2}}\ge x^{2}+1\implies e^{-x^{2}}\le \frac{1}{x^{2}+1}.

    Hence \int_{-\infty }^{+\infty }{e^{-x^{2}}\,dx}\le \int_{-\infty }^{+\infty }{\frac{1}{1+x^{2}}\,dx}=\pi . The given integral converges.
    nice
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