Does this integral diverge? $\int_{-\pi/2}^{\pi/2}\csc{x}dx$

Question

According to my calculusbook the following integral diverges: $$\int_{-\pi/2}^{\pi/2}\csc{x}dx$$

This is the case since $\csc{x}\ge\frac{1}{x}$ on $(0,\pi/2]$

My question is: I would have guessed the integral would be zero, since $\csc{x}$ is an odd function. Why is this not the case?

On $(0,\pi/2]$ the integral goes to $\infty$ and on $[-\pi/2,0)$ the integral goes to $-\infty$ so we have $\infty-\infty$, which is usually undefined, but doesn't the fact that $\csc{x}$ is odd imply that $\infty-\infty=0$ in this case?

Answer 1

The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $\int_{\epsilon <|x|<\frac {\pi} 2} \csc (x) dx$ as $\epsilon \to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $\frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.

Answer 2

No, it doesn't imply that. By definition, an improper integral $\int_a^bf(x)\,\mathrm dx$, where $f$ is a map from $[a,b]\setminus\{c\}$ into $\mathbb R$, converges if both integrals $\int_c^bf(x)\,\mathrm dx$ and $\int_a^cf(x)\,\mathrm dx$ converge (and, if they do, then $\int_a^b f(x)\,\mathrm dx$ is the sum of those two integrals).

In your case, none of the integrals converges. The fact that $\csc$ is odd is irrelevant.