Convergence of $\int_0^{\pi/2}(\tan x)^p\,dx$

Question

For what values of $p$ the integral is converge/diverge?

$$\int_{0}^{\pi/2} (\tan x)^p ~{\rm d}x$$

I tried use the fact that $\displaystyle{\tan x=\frac{\sin x}{\cos x}}$ but it didn't work.

Answer 1

You can use the substitution $\tan (x) = t$ to get $$\int \limits_0^{\pi/2} (\tan(x))^p \, \mathrm{d} x = \int \limits_0^\infty \frac{t^p}{1+t^2} \, \mathrm{d} t \, .$$ Now think about how the integrand behaves for small and large values of $t$ and you should find that $-1 < p < 1$ must hold for the integral to be finite.

Answer 2

First we note the definition of the Beta function: $$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Where $\Gamma(\cdot)$ is the Gamma function.

Then we consider the integral $$I(a,b)=\int_0^{\pi/2}\sin^ax\ \cos^bx\ dx$$ Substituting $t=\sin^2x$ provides $$I(a,b)=\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$ $$I(a,b)=\frac12\int_0^1t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$ $$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$ $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ Since the identity I gave in the beginning holds for $\operatorname{Re}(a),\operatorname{Re}(b)>0$, we know that $I(a,b)$ holds for $$\operatorname{Re}(a),\operatorname{Re}(b)>-1$$ And thus your integral $$\int_0^{\pi/2}\tan^pt\ dt=I(p,-p)$$ Holds for $$\operatorname{Re}(p)>-1,\qquad \operatorname{Re}(-p)>-1\\\therefore -1<\operatorname{Re}(p)<1$$