# How to evaluate if $\int_2^\infty {\frac{1}{\log(x)\cdot \sqrt{x^2+1}}}dx$ converges?

I am asked to evaluate if $$\int_2^\infty {\frac{1}{\log(x)\cdot \sqrt{x^2+1}}}dx$$ converges.

How can that be done? Even Wolframalpha/Mathematica 8.0 does not return a value.

Can this be done with improper integral convergence tests I,II or III?

EDIT:

Improper integral convergence test I (I guess it is also called comparison test I, at least in my native language):

If there exists $g(x)$ such that $0\leq f(x)\leq g(x)\quad \forall x \in [a;=\infty)$

and $\int_a^\infty{g(x)dx}$ converges, then $\int_a^\infty{f(x)dx}$ also converges.

Improper integral convergence test II would be the ratio test.

• Welcome to math.SE. As a new user, please see this post on how to ask a good question on this site, particularly the part about including context: meta.math.stackexchange.com/questions/9959/… – Carl Mummert Nov 19 '14 at 13:04
• @Carl I am inclined to give the OP the benefit of the doubt here - although the question could be better, the last two lines show that they have thought about it a bit. Although I have no idea what "improper integral convergence tests I,II or III" are. Perhaps the OP could clarify?... – user1729 Nov 19 '14 at 13:29

Since $$\int_{2}^{M}\frac{dx}{x\log x}=\int_{\log 2}^{\log M}\frac{dt}{t}=\log\log M-\log\log 2$$ and for any $x\geq 2$: $$\frac{1}{\log(x)\cdot\sqrt{x^2+1}}\geq\frac{1}{x\log x\cdot\sqrt{\frac{5}{4}}}=\frac{2}{\sqrt{5}}\cdot\frac{1}{x\log x}$$ the integral is divergent.

• I understand the first integral, but how did you came up with $\frac{1}{\log{x}\sqrt{x^2+1}}\geq \frac{1}{x \log{x}\cdot \sqrt{\frac{5}{4}}}$? – Kristians Kuhta Nov 19 '14 at 13:45
• @KristiansKuhta: Since $x\geq 2$, $$\sqrt{x^2+1}=x\cdot\sqrt{1+\frac{1}{x^2}}\leq x\cdot\sqrt{1+\frac{1}{4}}.$$ – Jack D'Aurizio Nov 19 '14 at 13:47
• Ok, I see now, thank you! – Kristians Kuhta Nov 19 '14 at 13:51
• Simpler: $\sqrt{x^2+1}<\sqrt{x^2+x^2}=x\sqrt2$. – David Nov 19 '14 at 20:13

Hint. First investigate $$\int_2^\infty \frac{dx}{x\log x}\ .$$

• So I evaluate this ($\int_2^\infty{\frac{dx}{x\log{x}}}=\infty$), then concider $g(x)=\frac{dx}{x\log{x}}$ and calculate $lim_{x->\infty}{\frac{f(x)}{g(x)}}$ and if its a real number, then integral diverges, right? – Kristians Kuhta Nov 19 '14 at 13:24
• That will do it (as long as the limit is not zero). – David Nov 19 '14 at 20:10