# why does this integral diverge?

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

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.

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I edited your latex, please check if it is still correct – k1next 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
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

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}

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$\frac{x^6}{6x^6-1}$ 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)

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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.

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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 $\epsilon\to 0^+$.

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