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I am learning about ways to test if an integral converges or diverges and I am stuck with this one: $\displaystyle{\int{{\rm d}x \over \sin\left(\, x\right)}}$ between $0$ and $1$.

The tests I know are:

  1. The Direct Comparison Test.
  2. The 2 Way Limit Comparison Test.
  3. The 1 way Limit Comparison Test.
  4. and The Dirichlet Test.

Am I right in thinking that this integral diverges ?. And is there a way to prove it using those tests ?. Thank you !.

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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – user137731 Nov 24 '14 at 16:17
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    $\begingroup$ Try to do a comparison with $\int_0^1 \frac 1x$. $\endgroup$ – user137731 Nov 24 '14 at 16:22
  • $\begingroup$ $\sin(x)$ ~ $x$ when $x$ is small. So since $\int_0^1 \frac{1}{x}$ diverges, you can say the same about the $\sin$ one $\endgroup$ – AnalysisStudent0414 Nov 24 '14 at 16:46
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A comparison works:

For $0 < x \leq 1$, $0 < \sin x < x$ and $\displaystyle \frac{1}{\sin x} > \frac{1}{x}$. Thus

$$\int_\epsilon^1 \frac{1}{\sin x} dx \ \ > \ \ \int_\epsilon^1 \frac{1}{x} dx$$

The second integral diverges to $\infty$ as $\epsilon \rightarrow 0^+$ and hence so does the first.

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    $\begingroup$ Importantly, the last integral diverges to $+\infty$. The comparison would not have been useful if it diverged by oscillation, or to "the wrong infinity". This is probably clear, but still warrants stating for formality. $\endgroup$ – Ian Nov 24 '14 at 16:23
  • $\begingroup$ @Ian, good point yet indeed it is clear as everything's positive here. $\endgroup$ – Timbuc Nov 24 '14 at 16:32
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I know that Simon S has already answered this question, but it's nice to not have to find any inequalities. The limit comparison test will apply since $\frac{1}{\sin x}$ and $\frac{1}{x}$ are positive on the interval $(0,1]$. We can also tell by the integrand that the problem occurs at $x=0$ so we will consider the following

$$ \lim_{a \to 0} \int_a^1 \frac{1}{\sin x} dx $$

We know that

$$\frac{1}{x} \sim \frac{1}{\sin x} \mbox{as } x \rightarrow 0 $$

and that

$$ \lim_{a \to 0} \int_a^1 \frac{1}{x} dx = \lim_{a \to 0} [\log x]_a^1 $$

$$ = \lim_{a \to 0} [\log 1 - \log a] $$

So we would conclude that the limit diverges to $+\infty$, meaning the second improper integral does not exist. Thus, by the limit comparison test

$$ \int_0^1 \frac{1}{\sin x} dx \mbox{ does not exist}$$

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