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I have improper integral $$\displaystyle \int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}}$$ and I need to find out if it converges/diverges. Now I don't if my thinking is correct:

$\displaystyle \int \frac{dx}{\sqrt{\ln(x)}}$ goes to $0$ slower than $\displaystyle \int \frac{dx}{\sqrt{x}}$, thus from integral comparison: $$\displaystyle \int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}} < \int_{1}^{2} \frac{dx}{\sqrt{x}}$$ it diverges. Is this correct? Thanks.

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  • $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – pax
    Sep 4, 2013 at 11:16
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    $\begingroup$ Also, I'm not sure I can follow your logic. The integral comparison test would only apply if you "2" was infinity $\endgroup$
    – pax
    Sep 4, 2013 at 11:21
  • $\begingroup$ My bad. I'll try to calculate it. $\endgroup$ Sep 4, 2013 at 11:29

2 Answers 2

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I solve the improper integral firstly by take it apart at $3/2$:

$$\int_1^{3/2}f(x)dx+\int_{3/2}^2f(x)dx$$

Now in the first integral we see that $$\lim_{x\to 1^+}\sqrt{x-1}f(x)=1<\infty$$ And for the second one we get $$\lim_{x\to 2^-}\sqrt{2-x}f(x)=1/\sqrt{\ln 2}<\infty$$ Therefore, both are converge and so the summation converges.

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Bounding above the integral by a convergent one does not prove the integral diverges.

Actually it converges. The integrand is non-negative and

  • at $2$, we have $\frac 1{(2-x)\log (x)}\sim \frac 1{(2-x)\log 2}$, and the integral $\int_{3/2}^{2}\frac{\mathrm dx}{\sqrt{2-x}}$ is convergent;
  • at $1$, using $\log(1+t)\overset{t\to 0}{\sim}t$, we have $\frac 1{(2-x)\log (x)}\sim \frac 1{\log(x-1+1)}\sim \frac 1{x-1}$.
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  • $\begingroup$ I've got few questions: (1) you split the integral it into intervals $[1,3/2]$ and $[3/2,2]$, right ? (2) why at $x=2$ you didn't put $2$ into $(2-x)$? $\endgroup$ Sep 4, 2013 at 13:03
  • $\begingroup$ (1) yes. (2) What do you mean? $\endgroup$ Sep 4, 2013 at 13:05
  • $\begingroup$ (2): when $x=2$ you wrote $\displaystyle \frac{1}{(2-x)log(x)}\sim \frac{1}{(2-x)(log2)}$. Why not $\displaystyle \frac{1}{(2-x)log(x)}\sim \frac{1}{(2-2)(log2)}$ (I know that it would equal $0$, but what's the main point of it?). $\endgroup$ Sep 4, 2013 at 13:08
  • $\begingroup$ And what would be the meaning of the last expression? I need to compare with something whose convergence of the integral is easily determinable, so I have to keep a function of $x$. $\endgroup$ Sep 4, 2013 at 13:12
  • $\begingroup$ I didn't know it was possible to do it this way. Thanks once again. $\endgroup$ Sep 4, 2013 at 13:15

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