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Math Help - Improper Integral

  1. #1
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    Improper Integral

    \int_1^\infty \frac{1}{(x^4+x+1)^{\frac{1}{3}}}dx

    I don't think that I can evaluate the integral easily, so I'm looking for a comparative function. If I say that:

    x^4 + x + 1 \geq x for [1,\infty)

    then I can also say that

    \frac{1}{x^4 + x + 1} \leq \frac{1}{x}

    \frac{1}{(x^4+x+1)^{\frac{1}{3}}} \leq \frac{1}{x^{\frac{1}{3}}}

    The problem is that
    \int_1^\infty \frac{1}{x^{\frac{1}{3}}} = \infty
    which doesn't help me in determining the value of the original integral.

    Can anybody see what I am doing wrong or recommend another function to use as a comparison?

    Thanks.
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Improper Integral

    How about comparing 1/(u^1/4) with 1/(u^1/3)

    (where u = x^4 + x + 1
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  3. #3
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    Re: Improper Integral

    Thanks, but I don't see how this solves the problem. Changing the exponent from 1/3 to 1/4 still leaves me with an integral that I can't evaluate easily, and doesn't give me a function that I can show goes to infinity or to a finite number. What am I missing?

    Thanks again!
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Improper Integral

    Quote Originally Posted by joatmon View Post
    \int_1^\infty \frac{1}{(x^4+x+1)^{\frac{1}{3}}}dx

    I don't think that I can evaluate the integral easily, so I'm looking for a comparative function. If I say that:

    x^4 + x + 1 \geq x for [1,\infty)

    then I can also say that

    \frac{1}{x^4 + x + 1} \leq \frac{1}{x}

    \frac{1}{(x^4+x+1)^{\frac{1}{3}}} \leq \frac{1}{x^{\frac{1}{3}}}

    The problem is that
    \int_1^\infty \frac{1}{x^{\frac{1}{3}}} = \infty
    which doesn't help me in determining the value of the original integral.

    Can anybody see what I am doing wrong or recommend another function to use as a comparison?

    Thanks.
    For x>1 is x^{4}+x+1 > x^{4} so that is also \frac{1}{x^{4}+x+1}<\frac{1}{x^{4}} and the integral \int_{1}^{\infty} \frac{dx}{x^{\frac{4}{3}}}=3 converges , so that...

    Kind regards

    \chi \sigma
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  5. #5
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    Re: Improper Integral

    Thank you!
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