Studying convergence of improper integral

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?

• 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.

• 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? Commented Jan 20, 2021 at 17:10
• @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 $(0,1)$. Commented Jan 20, 2021 at 17:31
• Ah okay, yeah it did say to consider a neighbourhood of $+\infty$. Thanks a lot, you solved my doubt :) Commented Jan 20, 2021 at 17:37

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