Use the Comparison Theorem to determine whether the integral is convergent/divergent: $\displaystyle \int_{0}^{1} \frac{sec^{2}x}{x\sqrt{x}} dx$

I dont know which basic function to compare it to. I tried comparing it to 1/x but that didnt work out.

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- Aug 25th 2010, 02:08 PMSyNtHeSiSComparison Theorem
Use the Comparison Theorem to determine whether the integral is convergent/divergent: $\displaystyle \int_{0}^{1} \frac{sec^{2}x}{x\sqrt{x}} dx$

I dont know which basic function to compare it to. I tried comparing it to 1/x but that didnt work out. - Aug 25th 2010, 02:48 PMAckbeet
The secant function is bounded on the interval, so the convergence doesn't depend on that function. Close to zero, the numerator looks like 1. So I would compare the integrand to the function $\displaystyle x^{-3/2}.$