Convergence or divergence of the integral $\int_0^1 dx/\sin x $

Question

Is this Convergent or Divergent

$$\int_0^1 \frac{1}{\sin(x)}\mathrm dx $$

So little background to see if I am solid on this topic otherwise correct me please :)

To check for convergence I can look for a "bigger" function and compare if that is convergent, If yes then for sure the one in question is too. So if the "bigger" function is not convergent can we conclude that the function in question is divergent or do we have to check for divergence to? That is a "smaller" function which has to be divergent?

And for this question I have no Idea WHAT kind of function to compare with:/

Answer 1

Is this Convergent or Divergent ?

Yes! :-)

I have no Idea WHAT kind of function to compare with:/

Hint: $\sin x\sim x$ for $x\to0$.

Answer 2

You could use the following facts to find proper values of $A$ as well. Maybe it helps you for other integrands:

Let $\lim_{x\to a^+}~(x-a)^pf(x)=A$. Then:

• If $p<1$ and $A$ is finite then $\int_a^bfdx$ converges.

• If $p\ge1$ and $A\neq0$ or $A=\infty$ then $\int_a^bfdx$ diverges.

Now, you can see that why @Lucian's short hint can guide you. Note that in above points $f(x)$ is unbounded only at the lower bound $a$.

Answer 3

We have $$\frac{1}{{\sin x}}:\frac{1}{x} = \frac{x}{{\sin x}} = \frac{1}{{\sin x/x}} \to 1$$ as $x\to 0^+$. In other word, $\frac{1}{\sin x}\sim \frac{1}{x}$ as $x\to 0^+$. But, it is well known that $\int_0^{1}\frac{1}{x}dx$ is divergent. So, we conclude that $\int_0^1 {\frac{1}{{\sin x}}dx}$ is divergent.