# Study the convergence or divergence of the improper integral

I am stuck in solving the following exercise, please help me with the improper integral

$$\int_2^\infty \frac{1}{\sqrt x\cdot \ln x} dx.$$ I am asked to determine whether it is divergent or convergent.

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Hint: $\ln x<\sqrt x$ for sufficiently large $x$. – David Mitra Mar 14 '14 at 9:25
this means it diverges by D.C.T am i correct? – MathDisease Mar 14 '14 at 9:31
Depends on what D.C.T. is – 5xum Mar 14 '14 at 9:35
by comparison test this integral is gretear than 1/x and 1/x diverges by p-integral so this is my way is it correct? – MathDisease Mar 14 '14 at 9:38
Almost. There is an $N$ so that ${1\over x\ln x}>{1\over x}$ for $x\ge N$. Since $\int_N^\infty 1/x\,dx$ diverges, so does your integral. I think you can take $N=2$, actually; so your argument is ok (up to being badly phrased). – David Mitra Mar 14 '14 at 9:42

Setting $t=\sqrt{x}$, we have $$\int_2^\infty\frac{1}{\sqrt{x}\ln x}\,dx=\int_{\sqrt{2}}^\infty\frac{2t}{t\ln t^2}\,dt=\int_{\sqrt{2}}^\infty\frac{1}{\ln t}\,dt \ge\int_{\sqrt{2}}^\infty\frac{1}{t}\,dt=\infty.$$