Study of a sequence of functions?

Question

I'm working on this exercise. $$f_{n}(x)=\frac{\log(x)}{\arctan(x^{\frac{1}{n}})+x^{n}}$$ for $n\ge2$ and $x \in (0,+\infty)$.

I want to verify that $f_{n}\in L^{1}(0,+\infty)$ for every $n\ge2$.

So, since $f_{n}\in C^{0}(0,+\infty)$, I can study $|f_{n}|$ in $U(0^{+})$ and in $U(+\infty)$.

Could someone give me a hint about how to compute $|f_{n}|$ evaluated in $0$ and $+\infty$? Thank you so much.

Answer 1

For small $x>0$, $\tan^{-1}(x^{1/n})$ is like $x^{1/n}$ and $|\log x|\leq \dfrac{1}{x^{1/(2n)}}$, so $|f_{n}(x)|\leq\dfrac{1}{x^{1/(2n)}}\dfrac{1}{x^{1/n}+x^{n}}\leq\dfrac{1}{x^{3/(2n)}}$ but $\displaystyle\int_{0}^{\eta}\dfrac{1}{x^{3/(2n)}}dx<\infty$ for $n\geq 2$.

For large $x>0$, $\log x\leq x^{1/2}$, so $\displaystyle\int_{M}^{\infty}f_{n}(x)dx\leq\int_{M}^{\infty}\dfrac{x^{1/2}}{x^{n}}dx=\int_{M}^{\infty}\dfrac{1}{x^{n-(1/2)}}dx<\infty$ for $n\geq 2$.

Answer 2

Apply L'Hospital rule:

Derivative of numerator= $1/x$

Derivative of denumerator= $1/nx^{(1/n-1)}/(1+x^{(2/n)} ) + n x^{(n-1)}$

Therefore

$f_n(0) =\infty$

$f_n(\infty)=0$