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Question: Is the following integral convergent or divergent $\int_{0}^{1} \frac{1}{\sin x}dx$? Use a comparative theorem to prove your results.

Answer attempt: I want to know if my attempt at a solution is acceptable.

$$\int_{0}^{1} \frac{1}{\sin x}dx$$

By using a variable substitution:

$t = \sin x$, $x = \arcsin t$, $dx = \frac{1}{\sqrt{1-t^2}}dt$

we get:

$$\int_{0}^{\sin 1} \frac{1}{t} \times \frac{1}{\sqrt{1-t^2}} dx$$

By using the following comparison:

$\frac{1}{t} \times \frac{1}{\sqrt{1-t^2}} \geq \frac{1}{t}$

This means that if the following integral is divergent we have proved that the original integral is also divergent:

$$\int_{0}^{\sin 1} \frac{1}{t}dx = \ln(\sin 1) - \ln(0)$$

$\ln(0)$ is undefined wich must mean that the original integral is divergent.

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    $\begingroup$ Looks good to me. $\endgroup$ – infinitylord Nov 28 '16 at 8:26
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    $\begingroup$ It's not so much that $\ln 0$ is undefined, but rather that you have an improper integral where $\lim_{b\to0}\ln x= - \infty$. $\endgroup$ – zahbaz Nov 28 '16 at 8:35
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    $\begingroup$ Why not just use $\sin x \leqslant x$ in that interval? $\endgroup$ – RRL Nov 28 '16 at 8:38
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Your attempt is fine. This is another approach, maybe simpler.

Note that $0<\sin(x)\leq x$ for $x\in (0,1]$, then for $0<t\leq 1$, $$\int_t^1\frac{dx}{\sin(x)}\geq \int_t^1\frac{dx}{x}= -\ln(t)$$ which implies that $$\int_0^1\frac{dx}{\sin(x)}=\lim_{t\to 0^+}\int_t^1\frac{dx}{\sin(x)}\geq \lim_{t\to 0^+}(-\ln(t))=+\infty.$$

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We have

$$\frac{1}{\sin(x)}\sim \frac{1}{x}\;\;(x\to 0^+)$$

and

$$\int_0^1\frac{dx}{x}$$ divergent

$$\implies \int_0^1\frac{dx}{\sin(x)}$$ divergent since the integrands are nonnegative and equivalent.

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  • $\begingroup$ You should add the criterion is valid because both functions are non-negative on $(0,1]$. $\endgroup$ – Bernard Nov 28 '16 at 9:15
  • $\begingroup$ @Bernard didn't you see $0^+$. $\endgroup$ – hamam_Abdallah Nov 28 '16 at 10:12
  • $\begingroup$ I did see it, but I think you should be more explicit on applying the criterion, as it is false if the functions do not have a constant sign. $\endgroup$ – Bernard Nov 28 '16 at 10:17

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