# Limit and convergence/divergence of an integral

I was working on a problem concerning the function $$f(x) = \frac{x^2}{\ln(x)^\sqrt{x}}$$ asking for the value of $$\lim_{x \to \infty}f(x)$$ and for the convergence/divergence of $$\int_2^\infty f(x) \ dx$$

I wasn't sure if my approach for the limit was entirely valid (or maybe too complicated): $$\lim_{x \to \infty} f(x) = e^{\lim_{x \to \infty} \ln(f(x))} = e^{\lim_{x \to \infty} 2\ln(x) - \sqrt x \ln(\ln(x))}$$ $$\sqrt x > \ln(x)$$ Hence: $$\lim_{x \to \infty} \ln(f(x)) = -\infty \ ; \lim_{x \to \infty} f(x) = e^{-\infty} = 0 \ .$$

As for the integral, I started out by taking the "sum comparison approach" and tried out the root and ratio tests, but those came out inconclusive. I suspect I could use the Limit Comparison Test somehow, but I couldn't find a sufficient "g(x)" to compare to.

Could someone validate/correct my solution for the limit and/or point me in the right direction on the integral question?

Thank you kindly.

• the limit $0$ is correct Jul 22 '16 at 15:44

Let $f(x)=\frac{x^2}{\log^{\sqrt{x}}(x)}$. Then, we have

\begin{align} \int_2^\infty f(x)\,dx&=\int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx \end{align}

Enforcing the substitution $x\to x^2$ yields

\begin{align} \int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx&=2\int_\sqrt{2}^\infty \frac{x^5}{\log^x(x^2)}\,dx\\\\ &=2\int_\sqrt{2}^\infty \frac{x^5}{2^x\log^x(x)}\,dx\\\\ &\le \frac2{\log^\sqrt 2(\sqrt 2)}\int_0^\infty x^5e^{-\log(2)x}\,dx\\\\ &= \frac{240}{\log^\sqrt 2(\sqrt 2)\log^6(2)}\\\\ &<\infty \end{align}

• Nice work. But I feel like a simple comparison is better. Comparing it to $1/x^2$ bounds the actual value between $0$ and $1/2$. This bounds it between $0$ and $9684.69$. Jul 22 '16 at 16:26
• @WillFisher How does one know that the integrand is less than $\frac1{x^2}$? Jul 22 '16 at 16:28
• See my response. It doesn't quite bound it by $1/2$ cause we need $x$ mildly large. But I would still think it gets a closer approximation. Jul 22 '16 at 16:29
• @WillFisher I read it. But it uses the "for sufficiently large $x$" argument. Although it does not impact convergence, the bound you provided does not hold over the entire domain of integration. This answer does and the bound is not so bad. Jul 22 '16 at 16:31
• @WillFisher The value of the integral is roughly $143$. Jul 22 '16 at 16:36

For the limit, notice that $f (x)$ is bounded below by $y=0$. Next realize that for $x$ sufficiently large we have that $\ln x>x^{1/3}$ hence for large $x$ we have $$0 <f (x)<g (x)=\frac {x^2}{x^{\sqrt {x}/3}}=x^{2-\sqrt {x}/3}$$ Then notice that for $x>81$ we have that $2-\sqrt {x}/3<-1$ (it equals $-1$ at $x=81$ and is decreasing from there on out). Hence for large $x$ we have $g (x)<1/x$. We conclude that for large $x$ we have $$0 <f (x)<\frac {1}{x}$$ Thus by the Squeeze Theorem $f (x)$ converges to $0$.

For the integral notice that for large $x$ we have $f (x)<1/x^2$ which converges (to see this, look at the $g (x)$ we used to bound $f (x)$ and notice that the exponent is decreasing and it reaches $-2$).