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I have a doubt on how to apply asymptotic comparison to this improper integral: $\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$

I'll call $f(x)=\dfrac{4x}{4x^8 + 1}$, and notice it is non negative as $x \to +\infty$.

If I understand the idea correctly, the I need to find a function $g(x)$ (also non negative as $x \to +\infty$) such that:

  • if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$ is finite and not zero: $\displaystyle\int_0^{+\infty} f(x) dx$ converges if and only if $\displaystyle\int_0^{+\infty}g(x)dx$ converges.
  • if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 0$ and $\displaystyle\int_0^{+\infty} g(x) dx$ converges: $\displaystyle\int_0^{+\infty}f(x)dx$ converges.
  • if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = +\infty$ and $\displaystyle\int_0^{+\infty} f(x) dx$ converges: $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

If I choose $g(x) = \dfrac{1}{x^7}$, I have $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 1$, so the first case applies. I know that $\displaystyle \int_0^{+\infty}\dfrac{1}{x^7}dx$ diverges, so $\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$ should diverge, but it does not..

What am I doing wrong?

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2 Answers 2

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  • if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$ is finite and not zero: $\displaystyle\int_0^{+\infty} f(x) dx$ converges if and only if $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

This is false. The reason that $\int_0^\infty \frac1{x^7}dx$ diverges is not because of the behaviour as $x\to\infty$ but rather near $x=0$.

Try splitting up your integral into two intervals, say $(0,1)$ and $(1,\infty)$. In particular, note that $\int_1^\infty \frac1{x^7}dx$ converges.

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  • $\begingroup$ Makes sense.. I wonder why my professor only gave me that formulation of the criterion, only considering integrals that diverge due to their behaviour as $x \to \infty$... So I guess choosing a comparison function in the $\frac{1}{x^{\alpha}}$ shape is only useful if, case $\alpha > 1$, I consider $\int_k^{+\infty}g(x)dx$ or if, case $alpha < 1$ I consider $\int_0^k g(x)dx$, with $k \in \mathbb{R}$, right? $\endgroup$
    – mell_o_tron
    Jan 20, 2021 at 17:10
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    $\begingroup$ @user256439 The missing part of the criterion is that $f(x)$ and $g(x)$ must be defined on some interval $[a, \infty)$. In this case, we can't choose $a=0$ because $g(0)$ is not defined. You can see that choosing $a=1$ (or any other positive number) leads to the correct conclusion; it just remains to take care of the other part of the integral, i.e., on $(0,1)$. $\endgroup$
    – Théophile
    Jan 20, 2021 at 17:31
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    $\begingroup$ Ah okay, yeah it did say to consider a neighbourhood of $+\infty$. Thanks a lot, you solved my doubt :) $\endgroup$
    – mell_o_tron
    Jan 20, 2021 at 17:37
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I think, you used some thing wrong. But it seems you have the idea.

Hint
$$\frac{4}{x^{-1}(4x^8+1)}$$ split the integral to $I_1$ and $I_2$ where $$I_1=\displaystyle\int_0^{1} \frac{4x}{4x^8 + 1}dx$$and $$I_2=\displaystyle\int_1^{+\infty} \frac{4x}{4x^8 + 1}dx$$ Now, $$I_1\leq \displaystyle\int_0^{+1} \frac{4}{x^{-1}}dx$$ and $$I_2\leq \displaystyle\int_1^{+\infty} \frac{4}{x^7}dx$$ I think, it is clear now. I hope that helps

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