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:
Am I right in thinking that this integral diverges ?. And is there a way to prove it using those tests ?. Thank you !.
On
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
$$ \int_0^1 \frac{1}{\sin x} dx \mbox{ does not exist}$$
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.