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Thread: Find a N so that an integral is convergent

  1. #1
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    Find a N so that an integral is convergent

    I can't work this out. Any help would be greatly appreciated.

    Find n so that the following integral is convergent.



    Answer: n = 1/3
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  2. #2
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    Quote Originally Posted by nvwxgn View Post
    I can't work this out. Any help would be greatly appreciated.

    Find n so that the following integral is convergent.



    Answer: n = 1/3


    $\displaystyle \frac{nx^2}{x^3+1}-\frac{1}{3x+1}=\frac{(3n-1)x^3+nx^2-1}{(x^3+1)(3x+1)}$ , and as we need to have a rational function of the form

    $\displaystyle \frac{P(x)}{Q(x)}$ , with $\displaystyle deg(Q)=deg(P)+1+\epsilon\,,\,\,\epsilon>0$ , we see that $\displaystyle n=1/3$ does the trick

    Tonio
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  3. #3
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    Hello, nvwxgn!

    $\displaystyle \text{Find }n\text{ so that the following integral is convergent.}$

    . . $\displaystyle \int^{\infty}_1\left(\frac{nx^2}{x^3+1} - \frac{1}{3x+1}\right)dx$

    We have: .$\displaystyle \int \bigg[\frac{n}{3}\left(\frac{3x^2}{x^3+1}\right) - \frac{1}{3}\left(\frac{3}{3x+1}\right)\bigg]\,dx$

    . . . . . . $\displaystyle =\;\frac{n}{3}\int\frac{3x^2\,dx}{x^3+1} - \frac{1}{3}\int\frac{3\,dx}{3x+1} $

    . . . . . . $\displaystyle =\;\frac{n}{3}\ln(x^3+1) - \frac{1}{3}\ln(3x+1) + C $

    . . . . . . $\displaystyle =\;\frac{1}{3}\bigg[n\ln((x^3+1) - \ln(3x+1)\bigg] + C$

    . . . . . . $\displaystyle =\;\frac{1}{3}\bigg[\ln(x^3+1)^n - \ln(3x+1)\bigg] + C$

    . . . . . . $\displaystyle =\;\frac{1}{3}\ln\frac{(x^3+1)^n}{3x+1} + C $


    $\displaystyle \text{We want: }\,\lim_{x\to\infty}\bigg[\frac{1}{3}\ln\frac{(x^3+1)^n}{3x+1}\bigg] \:\text{ to converge.}$

    This happens if the degree of the numerator $\displaystyle \le$ the degree of the denominator.

    Therefore: .$\displaystyle n \,\le\,\frac{1}{3}$

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