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Convergence of the following improper integral for $\alpha>0$: $$\int_{0}^{\frac{\pi}{2}}\tan^\alpha (x)\mathrm{d}x$$

I've used the substitution $x=1/t$ to have the argument to tend to zero: $$ \int_{\frac{2}{\pi}}^{+\infty }\left |\tan^\alpha\left(\frac{1}{t}\right)\frac{1}{t^{2}}\right |dt \leqslant \int_{\frac{2}{\pi}}^{+\infty } \frac{1}{t^{\alpha +2}} $$ Which converges for $\alpha+2<1 \Rightarrow \alpha>-1$

I think it's correct, but Wolfram Alpha suggests that it converges for $\alpha<1$, so I wish to know what is wrong with my solution. Thank you.

sugofoscolo
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  • Note that $\tan(t)$ is not improperly integrable on $[0,\pi/2)$, so we cannot hope the integral converges for $\alpha \ge 1$. For $\alpha <1$, try the substitution $z=\tan(t)$; think about what $dt$ looks like in this case. – Integrand Jul 30 '20 at 19:22
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    You assume that $\tan(x)\leq x,$ but $\tan x\geq x.$ So $\tan^a x \leq x^a$ only when $a\leq 0.$ So your proof works for -1<a\leq 0.$ – Thomas Andrews Jul 30 '20 at 19:59

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Considering $$I(\alpha)=\int_{0}^{\frac{\pi}{2}}\tan^\alpha (x)\,dx$$ let $x=\tan ^{-1}(t)$ to make $$I(\alpha)=\int_{0}^{\infty}\frac{t^{\alpha }}{t^2+1}\,dt$$ Around $t=0$ there is already a problem since $$\frac{t^{\alpha }}{t^2+1}=\left(1-t^2+O\left(t^4\right)\right) t^{\alpha }\sim t^{\alpha }$$. Around infinity, the integrand is $\sim t^{\alpha-2 }$ and its integral is $\sim t^{\alpha-1 }$ if $\alpha\neq1$ or $\sim\ln t$ if $\alpha=1$.

This would give $$I(\alpha)=\frac{\pi}{2} \sec \left(\frac{\pi \alpha }{2}\right)\qquad \text{if} \qquad -1<\Re(\alpha )<1$$

FShrike
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Claude Leibovici
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