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Prove $$\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} \,dx = \frac{\pi}{2} \sec \left(\frac{\pi}{2n}\right)$$

for all natural numbers $n \ge 2$.

There are several answers (e.g. A1 A2) to this integral. But, they all involve the gamma function, or the beta function, or contour integration etc. Can one solve it using only 'real' 'elementary' techniques?

For $n = 2$ and $n = 3$, it can be solved using elementary substitutions and partial fractions. Can it be done for any $n$?

Quanto
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Vue
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    You may try setting $t^n = \tan x$ and do something like https://math.stackexchange.com/q/4459448 – Gary Jun 29 '22 at 02:54
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    @Gary Could you explain more about how that would work? I understand I can factorize $x^{2n}+1$ but I'm having trouble making it into partial fractions (how to calculate each numerators). – Vue Jun 29 '22 at 03:16
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    The way I could compute it uses the Beta Integral. – robjohn Jun 29 '22 at 03:52
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    The Beta Integral is usually handled with real methods. I would consider it elementary. – robjohn Jun 29 '22 at 04:01
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    @robjohn Later in the solution, we would need to use the reflection formula of gamma function. Is there a real proof for that? Please clarify. – Laxmi Narayan Bhandari Jun 29 '22 at 06:50
  • @LaxmiNarayanBhandari: I posted my answer. I have removed the Beta function and Reflection Formula and folded them into the identity which can be used to prove the Reflection Formula. The cited answer is involved, but totally real. – robjohn Jun 29 '22 at 07:17

3 Answers3

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$$ \begin{align} \int_0^{\frac\pi2}\sqrt[n]{\tan(x)}\,\mathrm{d}x\ &=\int_0^\infty\frac{u^{1/n}\,\mathrm{d}u}{1+u^2}\tag{1a}\\ &=\frac12\int_0^\infty\frac{v^{\frac{1-n}{2n}}}{1+v}\,\mathrm{d}v\tag{1b}\\ &=\frac\pi2\csc\left(\pi\frac{n+1}{2n}\right)\tag{1c}\\[6pt] &=\frac\pi2\sec\left(\frac\pi{2n}\right)\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: set $x=\tan^{-1}(u)$
$\text{(1b)}$: set $u=v^{1/2}$
$\text{(1c)}$: apply $(2)$ below
$\text{(1d)}$: $\csc(\pi/2+x)=\sec(x)$

Here is the argument from $(3)$ of this answer with more explanation: $$ \begin{align} \int_0^\infty\frac{x^{\alpha-1}}{1+x}\,\mathrm{d}x &=\int_0^1\frac{x^{-\alpha}+x^{\alpha-1}}{1+x}\,\mathrm{d}x\tag{2a}\\ &=\sum_{k=0}^\infty(-1)^k\int_0^1\left(x^{k-\alpha}+x^{k+\alpha-1}\right)\mathrm{d}x\tag{2b}\\ &=\sum_{k=0}^\infty(-1)^k\left(\frac1{k-\alpha+1}+\frac1{k+\alpha}\right)\tag{2c}\\ &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\alpha}\tag{2d}\\[6pt] &=\pi\csc(\pi\alpha)\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: break the integral into two parts: $[0,1]$ and $(1,\infty)$
$\phantom{\text{(2a):}}$ substitute $x\mapsto1/x$ in the second part
$\text{(2b)}$: apply the series for $\frac1{1+x}$
$\text{(2c)}$: evaluate the integrals
$\text{(2d)}$: write as a principal value sum
$\text{(2e)}$: apply $(8)$ from this answer

robjohn
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  • I think it's bad form to write $x\mapsto\tan^{-1}(x)$. Better $x\mapsto\tan^{-1}(u)$. Also, you have two equations 1d, and no 1c. – Gerry Myerson Jun 30 '22 at 04:27
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    Since the scope of each of these variables is its individual integral, as long as the substitution is done uniformly, there is no problem. However, I have added new variable names. – robjohn Jun 30 '22 at 09:40
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An elementary solution: \begin{align} \int_{0}^{\frac{\pi}{2}} &\sqrt[n]{\tan x}dx = \frac12 \int_{0}^{\frac{\pi}{2}} \overset{y^n=\tan x}{ \sqrt[n]{\tan x}+\sqrt[n]{\cot x}}\ {dx} \>\>\>\>\>\>\>\> a_k=\small{\frac{(2k-1)\pi}{2n}}\\ &= \int_0^\infty\frac{n(y^n+y^{n-2})}{2(1+y^{2n})}dy = \int_0^\infty \sum_{k=1}^n \frac{(-1)^{k+1}\sin a_k \cos a_k}{y^2-2y\cos a_k +1 }dy\\ &=- \sum_{k=1}^n (-1)^{n-k}a_k\cos a_k =-\frac{d}{dt}\bigg(\sum_{k=1}^n (-1)^{n-k}\sin a_k t\bigg)_{t=1}\\ &=- \frac{d}{dt}\bigg(\frac{\sin \pi t}{2\cos\frac{\pi t}{2n}}\bigg)_{t=1} = \frac\pi2 \sec\frac\pi{2n} \end{align}

Quanto
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\begin{aligned} \int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} d x&=\int_{0}^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x\\ &= \int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{2 n}+\frac{1}{2}\right)-1} x \cos^ {2\left(-\frac{1}{2n}+\frac{1}{2}\right)-1 }x d x\\ &=\frac{1}{2} B\left(\frac{1}{2 n}+\frac{1}{2},-\frac{1}{2 n}+\frac{1}{2}\right)\\ &=\frac{1}{2} \pi \csc \left[\pi\left(\frac{1}{2 n}+\frac{1}{2}\right)\right]\\ &=\frac{\pi}{2} \sec \frac{\pi}{2 n} \end{aligned} where the second last line using the property of Beta function:

$$B(x, 1-x)=\pi \csc (\pi x) \quad x \notin \mathbb{Z}.$$

Lai
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