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Math Help - Improper Integration

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    Improper Integration

    The question is evaluating the integral from negative infinity to positive infinity, of (30-x^4) dx. I think that the answer is negative infinity because I got a very large negative number, and that would be divergent, anyone think the same?
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    Quote Originally Posted by elocin View Post
    The question is evaluating the integral from negative infinity to positive infinity, of (30-x^4) dx. I think that the answer is negative infinity because I got a very large negative number, and that would be divergent, anyone think the same?
    You got a very large negative number from where? As an approximation? Don't trust such methods if you want proof.

    The integral is divergent. To demonstrate it, split the interval into two and show that either integral diverges by letting one of the limits of integration tend to infinity or -infinity. That is, show that \lim_{b\to\infty}\int_0^b\left(30 - x^4\right)\,dx = \lim_{b\to\infty}\left(30b - \frac{b^5}5\right) = -\infty, which would imply that \int_0^\infty\left(30 - x^4\right)\,dx diverges, so the integral over the whole real line also diverges.
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    Quote Originally Posted by elocin View Post
    The question is evaluating the integral from negative infinity to positive infinity, of (30-x^4) dx. I think that the answer is negative infinity because I got a very large negative number, and that would be divergent, anyone think the same?

     \int_{-\infty}^{\infty} 30 - x^4 \, dx = \lim_{t \to \infty} \int_{-t}^{t} 30 - x^4 \, dx = \lim_{t \to \infty} (30 x - \frac{x^5}5)\bigg{|}_{-t}^t =\lim_{t \to \infty} (60 t - \frac{2t^5}5)

    Now finding its limit and formally proving that the integral diverges to negative infinity should do the job.
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    Quote Originally Posted by Isomorphism View Post
     \int_{-\infty}^{\infty} 30 - x^4 \, dx = \lim_{t \to \infty} \int_{-t}^{t} 30 - x^4 \, dx = \lim_{t \to \infty} (30 x - \frac{x^5}5)\bigg{|}_{-t}^t =\lim_{t \to \infty} (60 t - \frac{2t^5}5)
    One problem with your method: does \int_{-\infty}^{\infty} x^3\,dx converge or diverge?

    Unless you want the Cauchy principal value, you need to treat the endpoints separately by splitting the integral or using a double limit.
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    Oh, okay, so integrating the integral gives you 30b-(b^5/5). So how do you know that it goes to negative infinity? This is probably a dumb question, but I don't really understand how you can tell as b goes to infinity, the integral goes to negative infinity? Please help me understand how to figure that out! Thanks!!
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    Quote Originally Posted by elocin View Post
    Oh, okay, so integrating the integral gives you 30b-(b^5/5). So how do you know that it goes to negative infinity? This is probably a dumb question, but I don't really understand how you can tell as b goes to infinity, the integral goes to negative infinity? Please help me understand how to figure that out! Thanks!!
    \int_{-\infty}^\infty(30-x^4)\,dx = \int_{-\infty}^0(30-x^4)\,dx + \int_0^\infty(30-x^4)\,dx

    The original integral diverges if either or both of these integrals diverge.

    \int_0^\infty(30-x^4)\,dx = \lim_{b\to\infty}\int_0^b(30 - x^4)\,dx = \lim_{b\to\infty}\left(30x - \frac{x^5}5\right)\Bigg|_0^b

    = \lim_{b\to\infty}\left(30b - \frac{b^5}5\right) = \lim_{b\to\infty}\frac{150b - b^5}5.

    =\lim_{b\to\infty}\frac{150 - b^4}{5/b}

    =\frac{-\infty}0=-\infty (note that \frac\infty0 is not an indeterminate form; it diverges to \infty).

    For a rough formal proof, note that \forall M < 0,\text{ choosing }N > 0\text{ to be } N = 150 - M\text{ gives }

    b > N\Rightarrow b > 150 - M\Rightarrow b - 150 > -M

    \Rightarrow 150 - b < M\Rightarrow 150b - b^5 < M (this follows from the axioms of order, since b > N\Rightarrow b>150\Rightarrow b>1).
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    Ohh I see! Thank yoU!
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    Quote Originally Posted by Reckoner View Post
    One problem with your method: does \int_{-\infty}^{\infty} x^3\,dx converge or diverge?
    Nice!

    However I have a doubt. Am I right in saying that my method is a sufficient condition for the integral to diverge?

    Unless you want the Cauchy principal value, you need to treat the endpoints separately by splitting the integral or using a double limit.
    Never heard of it

    So I will have to learn what "Cauchy principal value" is .... I will do that.

    Thank you
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    \int_{-\infty}^{\infty} 30 - x^4 \, dx = \lim_{t \to \infty} \int_{-t}^{t} 30 - x^4 \, dx
    From Isomorphism is not right. As Reckoner said,
    \int_{-\infty}^\infty(30-x^4)\,dx = \int_{-\infty}^0(30-x^4)\,dx + \int_0^\infty(30-x^4)\,dx
    is right, but the general form is \int_{-\infty}^\infty(30-x^4)\,dx = \int_{-\infty}^a(30-x^4)\,dx + \int_a^\infty(30-x^4)\,dx with a as a finite number.
    It can be shown that if
    \int_{-\infty}^{\infty} f(x) dx exists, then lim as n tends to +\infty of \int_{-n}^{n}f(x)dx exists.
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    Quote Originally Posted by Isomorphism View Post
    However I have a doubt. Am I right in saying that my method is a sufficient condition for the integral to diverge?
    Yes, it is sufficient, but not necessary. If the improper integral exists, then the Cauchy principal value exists and is equal to it, so by the contrapositive, the divergence of the principal value implies that the improper integral diverges.

    Quote Originally Posted by arbolis View Post
    but the general form is \int_{-\infty}^\infty(30-x^4)\,dx = \int_{-\infty}^a(30-x^4)\,dx + \int_a^\infty(30-x^4)\,dx with a as a finite number.
    Thank you, I should have been clearer about that. You can indeed choose any real value at which to split the interval. I chose 0 only because it was convenient to do so.
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