Find a N so that an integral is convergent

**nvwxgn** · May 11th 2011, 06:37 PM

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

**tonio** · May 11th 2011, 06:57 PM

Originally Posted by nvwxgn

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

**Soroban** · May 11th 2011, 07:06 PM

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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