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?