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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?

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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.

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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.

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  • $\begingroup$ But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal? $\endgroup$ – GambitSquared Nov 27 '18 at 12:03
  • $\begingroup$ By the same approach $\int_{-\infty}^{+\infty}x\,\mathrm dx=0$. $\endgroup$ – José Carlos Santos Nov 27 '18 at 12:04
  • $\begingroup$ @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out. $\endgroup$ – SinTan1729 Nov 27 '18 at 12:04
  • $\begingroup$ @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though? $\endgroup$ – GambitSquared Nov 27 '18 at 12:22
  • $\begingroup$ @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out? $\endgroup$ – GambitSquared Nov 27 '18 at 12:23

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