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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.

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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

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

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  • $\begingroup$ 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 $9684.69$. $\endgroup$ Jul 22 '16 at 16:26
  • $\begingroup$ @WillFisher How does one know that the integrand is less than $\frac1{x^2}$? $\endgroup$
    – Mark Viola
    Jul 22 '16 at 16:28
  • $\begingroup$ 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. $\endgroup$ Jul 22 '16 at 16:29
  • $\begingroup$ @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. $\endgroup$
    – Mark Viola
    Jul 22 '16 at 16:31
  • $\begingroup$ @WillFisher The value of the integral is roughly $143$. $\endgroup$
    – Mark Viola
    Jul 22 '16 at 16:36
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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 $-2$).

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