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Thread: Improper Integrals, Convergence, Sequence Convergence? Ahh!

  1. #1
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    Improper Integrals, Convergence, Sequence Convergence? Ahh!

    Hey there! Having problems with a few of the exercises on some homework!

    Do the integrals converge or diverge? If they converge, what is their value?
    a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
    b) integral from 0 to 2 of ln(x)dx
    c)integral from zero to infinity of dx/(2x-1)

    Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

    Thank you!
    My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by matt.qmar View Post
    Hey there! Having problems with a few of the exercises on some homework!

    Do the integrals converge or diverge? If they converge, what is their value?
    a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
    b) integral from 0 to 2 of ln(x)dx
    c)integral from zero to infinity of dx/(2x-1)

    Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

    Thank you!
    My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
    a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx

    $\displaystyle \int\limits_1^\infty {\frac{{{x^3} + x}}
    {{{x^4} + 2{x^2} + 2}}dx} = \frac{1}
    {4}\int\limits_1^\infty {\frac{{4{x^3} + 4x}}
    {{{x^4} + 2{x^2} + 2}}dx} = \frac{1}
    {4}\int\limits_1^\infty {\frac{{d\left( {{x^4} + 2{x^2} + 2} \right)}}
    {{{x^4} + 2{x^2} + 2}}} =$

    $\displaystyle = \left. {\frac{1}{4}\ln \left( {{x^4} + 2{x^2} + 2} \right)} \right|_1^\infty = \frac{1}{4}\mathop {\lim }\limits_{x \to \infty } \ln \left( {{x^4} + 2{x^2} + 2} \right) - \frac{{\ln 5}}{4} = \infty .$

    So, this integral is divergent.
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  3. #3
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    One additional Question

    Also, in part a)
    Does the integral still diverge if the denominator is (x^4 + 2*x^2 + 2)^1/2?

    Thank you so much!
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by matt.qmar View Post
    Hey there! Having problems with a few of the exercises on some homework!

    Do the integrals converge or diverge? If they converge, what is their value?
    a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
    b) integral from 0 to 2 of ln(x)dx
    c)integral from zero to infinity of dx/(2x-1)

    Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

    Thank you!
    My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
    b) integral from 0 to 2 of ln(x)dx

    $\displaystyle \int\limits_0^2 {\ln xdx} = \left. {x\ln x} \right|_0^2 - \int\limits_0^2 {dx} = \left. {\left( {x\ln x - x} \right)} \right|_0^2 = 2\left( {\ln 2 - 1} \right) - \mathop {\lim }\limits_{x \to 0} \left( {x\ln x - x} \right).$

    $\displaystyle \mathop {\lim }\limits_{x \to 0} \left( {x\ln x - x} \right) = \mathop {\lim }\limits_{x \to 0} x\ln x - \mathop {\lim }\limits_{x \to 0} x = \mathop {\lim }\limits_{x \to 0} x\ln x = \left\{ \begin{gathered}
    x = \frac{1}
    {t}, \hfill \\
    x \to 0, \hfill \\
    t \to \infty \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \frac{1}
    {t}}}
    {t} = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{d}
    {{dt}}\ln \frac{1}
    {t}}}
    {{\frac{{dt}}
    {{dt}}}} = - \mathop {\lim }\limits_{t \to \infty } \frac{1}
    {t} = 0.$

    Thus, the integral $\displaystyle \int\limits_0^2 {\ln xdx}$ converges, because it has a finite value and is equal to $\displaystyle 2\left( {\ln 2 - 1} \right).$
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    Progress!

    I think I've solved Question (2)!
    let N=3/(2*epsilon)
    We can show abs(a sub n - (5/2)) < epsilon?
    (this is just the beginning and end of the proof of course!)
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by matt.qmar View Post
    Also, in part a)
    Does the integral still diverge if the denominator is (x^4 + 2*x^2 + 2)^1/2?

    Thank you so much!
    Similarly

    $\displaystyle \int\limits_1^\infty {\frac{{{x^3} + x}}
    {{\sqrt {{x^4} + 2{x^2} + 2} }}\,dx} = \frac{1}
    {4}\int\limits_1^\infty {\frac{{4{x^3} + 4x}}
    {{\sqrt {{x^4} + 2{x^2} + 2} }}\,dx} = \frac{1}
    {4}\int\limits_1^\infty {\frac{{d\left( {{x^4} + 2{x^2} + 2} \right)}}
    {{\sqrt {{x^4} + 2{x^2} + 2} }}} =$

    $\displaystyle = \left. {\frac{1}{2}\sqrt {{x^4} + 2{x^2} + 2} } \right|_1^\infty = \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \sqrt {{x^4} + 2{x^2} + 2} - \frac{{\sqrt 5 }}{2} = \infty .$
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by matt.qmar View Post
    Hey there! Having problems with a few of the exercises on some homework!

    Do the integrals converge or diverge? If they converge, what is their value?
    a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
    b) integral from 0 to 2 of ln(x)dx
    c)integral from zero to infinity of dx/(2x-1)

    Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

    Thank you!
    My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
    c) integral from zero to infinity of dx/(2x-1)

    $\displaystyle \int\limits_0^\infty {\frac{{dx}}
    {{2x - 1}}} = \frac{1}
    {2}\int\limits_0^\infty {\frac{{d\left( {2x - 1} \right)}}
    {{2x - 1}}} = \left. {\frac{1}
    {2}\ln \left| {2x - 1} \right|} \right|_0^\infty = \frac{1}
    {2}\mathop {\lim }\limits_{x \to \infty } \ln \left| {2x - 1} \right| = \infty .$

    So, this integral is divergent.
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