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I tried to use the absolute convergence test. $$\int_1^2\left|\frac{\sin{x}}{\log{x}}\right| \, dx<\int_1^2\left|\frac{1}{\log{x}}\right| \, dx$$ but I'm not sure whether $\frac{1}{\log{x}}$ converges or not.

Thanks in advance!

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  • $\begingroup$ What is $\log x$ at $x = 1$? $\endgroup$ – Gregory Apr 28 '17 at 14:59
  • $\begingroup$ 0 so it must diverge but then the comparison test doesn't work.. Any other way? $\endgroup$ – user21312 Apr 28 '17 at 15:00
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    $\begingroup$ Just because $\log(1) = 0$ doesn't mean $\int_1^2 \frac{1}{\log}$ doesn't converge. It doesn't converge because at 1 $\log$ goes to 0 as $\frac{1}{x}$. $\endgroup$ – Solomonoff's Secret Apr 28 '17 at 15:03
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    $\begingroup$ What Solomonoff said, @Gregory. The denominator of an integrand being $0$ at an endpoint of integration isn't a sufficient condition for the integral diverging. For example, $\int_0^1 \frac1{\sqrt x} \, dx = 2$. $\endgroup$ – tilper Apr 28 '17 at 15:28
  • $\begingroup$ Very true guys. $\endgroup$ – Gregory Apr 29 '17 at 19:09
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No: for $x > 1$, $\log{x} < x-1$, so $\frac{1}{\log{x}} > \frac{1}{x-1}$. $\sin{1}$ is not zero, and $1<\pi/2$, so sine is increasing for $x$ near $1$. Therefore on the interval $(1,\epsilon)$, $$ \frac{\sin{x}}{\log{x}} > \frac{\sin{1}}{x-1}, $$ and the right-hand side does not have a convergent integral. Hence by comparison, neither does the left-hand side.

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