Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\int_1^{\infty}\frac{x^6}{6x^6 − 1} dx $$

I would assume it would converge but apparently it diverges. I know it has to do with improper integrals. Can anyone explain? Thank you for your time.

share|improve this question
    
I edited your latex, please check if it is still correct –  sonystarmap Dec 17 '12 at 11:54
    
@macydanim , yes it is. Thank you very much. I'm new to posting here so I apologize for not being up to date with the formatting :-( –  Ceelos Dec 17 '12 at 11:55
2  
There’s a MathJax tutorial with some useful links here; it will give you at least a start on formatting. –  Brian M. Scott Dec 17 '12 at 11:58

4 Answers 4

up vote 12 down vote accepted

The reason why this integral diverges is the following. For large $x$ the fraction reaches a constant limit. \begin{align} \lim_{x\rightarrow \infty} \frac{x^6}{6x^6-1} = \frac{1}{6} \end{align}

That means that we integrate a functions that asymptotically behaves like $f(x)\equiv \frac{1}{6}$ which has a diverging integral. See also Wolfram Alpha for a plot of the integrand.

Concerning Brian M. Scott regards. He is right. The more correct argumentation would be: \begin{align} \frac{x^6}{6x^6-1} \geq \frac{x^6}{6x^6}=\frac{1}{6} \, \forall x \geq 1 \end{align} So \begin{align} \int_1^{\infty}\frac{x^6}{6x^6-1} \, dx \geq \int_1^{\infty}\frac{1}{6} \, dx = \infty \end{align}

share|improve this answer

Since we know that $\dfrac{1}{x^6}>0$ for all $x\geq1$ , we have $\dfrac{1}{6-\dfrac{1}{x^6}}>\dfrac{1}{6}$ and we know that RHS diverges. Hence, by comparison theorem , LHS diverges.

share|improve this answer

The integral turns out to be doubly bad. The behaviour when $x$ is large is the more obvious badness. We show that there is also fatal badness at $1$, by showing that $\displaystyle\int_1^2 \dfrac{x^6\,dx}{x^6-1}$ diverges.

Note that $x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)$. When $x\ge 1$, each of the terms $x^5,x^4,x^3,x^2,x, 1$ is $\le x^6$. It follows that if $\epsilon\gt 0$, then $$I_\epsilon=\int_{1+\epsilon}^2 \dfrac{x^6\,dx}{x^6-1}\gt \int_{1+\epsilon}^2 \frac{1}{6}\cdot \frac{dx}{x-1}.$$

The change of variable $u=x-1$ shows that $$I_\epsilon\gt \frac{1}{6}\int_\epsilon^1\frac{du}{u}.$$ But it is a familiar fact that $\displaystyle\int_\epsilon^1\dfrac{du}{u}$ blows up as $\epsilon\to 0^+$.

share|improve this answer

$\frac{x^6}{6x^6-1}$ converges toward 1/6 $(n\rightarrow\infty)$. Therefore, there exists $x_0\in\mathbb{R}$ for that $\frac{x^6}{6x^6-1}\gt\frac{1}{7}\; (x\gt x_0)$ And thus, the $\int_1^\infty\frac{x^6}{6x^6-1}\ge\int_{1}^{x_0} \frac{x^6}{6x^6-1}\text{d}x+\int_{x_0}^\infty \frac{1}{7}\text{d}x$. (This obviously diverges)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.