Struggling with the integrability of $\int_{\frac{\pi}{2}}^{\pi}(\tan(x))^{\frac{1}{3}}\text{d}x$

Question

I know quite a lot tools to determine the integrability of functions, but in this case I really don't know where to start: $$\int_{\frac{\pi}{2}}^{\pi}(\tan(x))^{\frac{1}{3}}\text{d}x$$

Answer 1

On the integration range $\tan x$ is negative, hence I assume that $x^{1/3}$ is defined as $-(-x)^{1/3}$ for negative $x$. Given that, by setting $x=\pi-z$, then $z=\arctan t$, we have: $$ I = -\int_{0}^{\pi/2}\tan(z)^{1/3}\,dz = -\int_{0}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt$$ and the last integral is convergent, since: $$0\leq \int_{0}^{1}\frac{t^{1/3}}{1+t^2}\,dt \leq \int_{0}^{1}t^{1/3}\,dt = \frac{3}{4}$$ and: $$0\leq \int_{1}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt \leq \int_{1}^{+\infty}t^{-5/3}\,dt = \frac{3}{2}.$$ We may also compute it exactly: $$ \int_{0}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt=\int_{0}^{+\infty}\frac{3u^3}{1+u^6}\,du = \frac{\pi}{\sqrt{3}}.$$

Answer 2

HINT:

$$\tan x=\frac{\sin x}{\cos x}=O\left((x-\pi/2)^{-1}\right)$$

and the integral

$$\int_0^1 \frac{dx}{x^a}$$

converges for $a<1$.