Does the integral $$ \int_0^{2\pi} (\sin^3(x) + 2 \ln |\sin(x)| + \tan^5 (x)) dx$$ coverge/have a definite value?

I am confused regarding the last term. $\int_0^{\pi/2} \tan^5(x) dx$ does not converge or is rather undefined. However, using Queen's rule of Definite Integration:

(i.e. $\int_0^{2a} f(x) dx = \int_{0}^{a}f(x)dx + \int_{0}^{a}f(2a-x)dx$ ), $\int_0^{2\pi}\tan^5 (x) dx$ turns out to be $0$.

So should I take the value of $\int_0^{2\pi}\tan^5 (x) dx$ as $0$ or as undefined ?

  • $\begingroup$ $\tan^5 x$ blows up like $\frac{1}{(\frac{\pi}{2} - x)^5}$ as $x \to \frac{\pi}{2}$, hence the integral does not converge. $\endgroup$ – MathematicsStudent1122 Feb 26 '17 at 8:46

Queen's rule only applies to integrals that exist in the first place (if you look it up in your book it should say as much). What the rule tells you is basically the same as $$ \lim_{a\to0^+}\int_a^{\pi/2-a}\tan^5(x)dx=0 $$ which is clearly true. However that is not what we're after. We want $$ \int_0^{\pi/2}\tan^5(x)dx=\lim_{a, b\to0^+}\int_a^{\pi/2-b}\tan^5(x)dx $$which turns out to be undefined.

  • $\begingroup$ Some people in the chat room said that I should use Cauchy's principal value of integral in cases like these. Anyway I agree with your viewpoint too :) $\endgroup$ – user404484 Feb 26 '17 at 8:54

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