The question is to check the convergence of the improper integral of $\int_1^2 \frac{\sqrt{x}}{\log(x)}\,\mathrm dx$. I tried using comparision test and tried to convert or compare it to the Gamma function but it led to nowhere .
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2$\begingroup$ your integral doesn't converge on the given interval $\endgroup$– Dr. Sonnhard GraubnerSep 16, 2017 at 8:18
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$\begingroup$ Your function doesn't seem to have an elementary primitive. $\endgroup$– José Carlos SantosSep 16, 2017 at 8:18
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$\begingroup$ @Dr.SonnhardGraubner Sorry made a typo in the question . Fixed it now . How did you prove it is diverging ? $\endgroup$– TejusSep 16, 2017 at 8:22
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$\begingroup$ The integral is divergent. $\endgroup$– Archis WelankarSep 16, 2017 at 8:23
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$\begingroup$ @ArchisWelankar Yes but how though ? $\endgroup$– TejusSep 16, 2017 at 8:24
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2 Answers
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$$\int_1^2 \frac{\sqrt{x}}{\log(x)}\,\mathrm dx\geq\int_1^2 \frac{1}{x\log(x)}\,\mathrm dx\to+\infty$$
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1$\begingroup$ One relevant detail here is that $\log \log x$ is the antiderivative of $1/(x \log x)$, and that $\log \log x \to - \infty$ as $x \downarrow 1$. $\endgroup$– shalopSep 16, 2017 at 8:42
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$\begingroup$ @RobertZ It's realy $-\log\log x\to+\infty$, Furthermore when the integrand is positive how it's integral will be negative! $\endgroup$– NosratiSep 16, 2017 at 8:57
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We compare the integrand with the function $1/(x-1)$. Note that $$\lim_{x\to 1^+}\frac{\sqrt{x}(x-1)}{\ln(x)}=\lim_{t\to 0^+}\frac{t}{\ln(1+t)}=1$$ which implies that there is $r\in (1,2]$ such that for all $x\in (1,r]$ $$\frac{\sqrt{x}(x-1)}{\ln(x)}\geq \frac{1}{2}\implies \frac{\sqrt{x}}{\ln(x)}\geq \frac{1}{2(x-1)}.$$ Therefore $$\int_1^2\frac{\sqrt{x}}{\ln(x)}\, dx\geq \frac{1}{2}\int_1^r\frac{1}{x-1}\,d x=\frac{1}{2}\left(\ln(r-1)-\lim_{x\to 1^+}\ln(x-1)\right)=+\infty.$$
