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Compute $$ \int_{0}^{\pi/2}\tan^n x \ dx$$

Using reduction formula I got $$I_n = \int_0^{\pi/2} \tan^n(x) dx = \int_0^{\pi/2} \tan^{n-2}(x) \sec^2(x) dx - \int_0^{\pi/2} \tan^{n-2}(x)dx$$ But I'm not sure how to proceed...

CocoaLapin
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    https://www.wolframalpha.com/input/?i=int_0%5Epi%2F2+tan+x+dx ... Wolfie says "No" ... – Donald Splutterwit Jun 11 '17 at 20:18
  • Yeah, that's why I asked. Got this as a homework from my prof., but I can't really solve it... – CocoaLapin Jun 11 '17 at 20:23
  • Is $n$ supposed to be an integer? – Daniel Fischer Jun 11 '17 at 20:32
  • It is not specified, but I'm pretty sure it should be an integer @Daniel Fischer – CocoaLapin Jun 11 '17 at 20:37
  • Change the upper limit to $\frac{\pi}{4}$ & suppose $n$ is an even positive number ... this is much more interesting. – Donald Splutterwit Jun 11 '17 at 20:38
  • Anyway, $\tan x$ behaves like $x$ near $0$, and it behaves like $\dfrac{1}{\frac{\pi}{2} - x}$ near $\pi/2$, so $\tan^n x$ is improperly Riemann integrable/Lebesgue integrable over $[0,\pi/2]$ if and only if $\lvert n\rvert < 1$. – Daniel Fischer Jun 11 '17 at 20:41
  • Yeah, I could do that, it would be a lot easier, the solution is even here already: https://math.stackexchange.com/questions/202418/compute-lim-n-to-infty-int-0-pi-4-tann-x-dx but I have to solve this problem or show that it's unsolvable using some nice arguments... @ Donald Splutterwit – CocoaLapin Jun 11 '17 at 20:41
  • If $n$ assumes real numbers, then the integral converges if and only if $|n| < 1$. In this case, the integral can be computed using beta function identity combined with the Euler reflection formula. For the regime $n \geq 1$, notice that $$\tan x \geq \frac{\sin x}{\frac{\pi}{2} - x}, \qquad x \in [0, \pi/2). $$ This can be used to show the divergence of $I_n$. The regime $n \leq -1$ follows from the identity $I_{-n} = I_n$. – Sangchul Lee Jun 11 '17 at 20:42
  • For n=1 the integral diverges to +$\infty$, then that integral looks to diverge – Matheman Jun 11 '17 at 20:45
  • Ok, then |n|<1, although it is not specified... I could use a little help with "beta function identity combined with the Euler reflection formula", if you'd be so kind @Sangchul Lee – CocoaLapin Jun 11 '17 at 20:46
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    It is definitely beyond the usual calculus course, so let me just give you links to them without explanation. Using the beta function identity and the Euler's reflection formula, we get $$ I_n = \int_{0}^{\pi/2} \sin^n \theta \cos^{-n}\theta,d\theta=\frac{1}{2}B\left(\frac{1+n}{2},\frac{1-n}{2}\right)=\frac{1}{2}\frac{\pi}{\sin(\frac{\pi-\pi n}{2})} = \frac{\pi}{2\cos(\frac{\pi n}{2})} $$ when $|n| < 1$. – Sangchul Lee Jun 11 '17 at 20:52

2 Answers2

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$$ \int_0^{\frac{\pi}{2}}\tan^nxdx= \int_0^{\frac{\pi}{4}}\tan^nxdx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\tan^nxdx > \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\tan x dx = \ln\left(\frac{\sec(\pi/2)}{\sec(\pi/4)}\right) = \infty $$

Anonymous
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \tan^{n}\pars{x} & \,\,\,\stackrel{\mrm{as}\ x\ \to\ 0^{+}}{\sim}\,\,\, x^{n} \\[5mm] \tan^{n}\pars{x} & = \cot^{n}\pars{{\pi \over 2} - x} = {1 \over \tan^{n}\pars{\pi/2 - x}} \,\,\,\stackrel{\mrm{as}\ x\ \to\ \pars{\pi/2}^{-}}{\sim}\,\,\, \pars{{\pi \over 2} - x}^{-n} \end{align}

The integral converges whenever $\ds{\verts{\Re\pars{n}} < 1}$.

Felix Marin
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