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Math Help - 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


    \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

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

    Tonio
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  3. #3
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    Hello, nvwxgn!

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

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

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

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

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

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

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

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


    \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 \le the degree of the denominator.

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

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