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

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?

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.
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.
• But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal? – GambitSquared Nov 27 '18 at 12:03
• By the same approach $\int_{-\infty}^{+\infty}x\,\mathrm dx=0$. – José Carlos Santos Nov 27 '18 at 12:04