I need to find the limit of the sequence :

$$I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$$

The only thing I have done is : take substitution $\tan x = z$

this gives

$$dx = \dfrac{1}{1 + z^2}\,dz$$

So, integral becomes :

$$I_n = \int_{0}^{\infty} \frac{z^{1/n}}{1 + z^2}\,dz$$

After this I am stuck how should I proceed ?

Can someone help me here ?

Thank you.

  • 1
    $\begingroup$ Well done so far. You can now use the dominated convergence theorem to interchange limit and integration. Thus, your limit is $\int_0^{ + \infty } {\frac{{dz}}{{1 + z^2 }}}$, which you can compute explicitly. $\endgroup$ – Gary Mar 17 '20 at 12:49
  • 2
    $\begingroup$ You can actually calculate that integral explicitly. It is $$\frac{\pi/2}{\cos\left(\frac{\pi}{2n}\right)} \,.$$ $\endgroup$ – Diger Mar 17 '20 at 12:55
  • $\begingroup$ @Garry: Oh,so we need to use Dominated Convergence.Actually I have not learnt that theorem yet ,so I thinking of using sandwich theorem or reduction to evaluate this, anyway thanks for the help. $\endgroup$ – user435638 Mar 17 '20 at 12:55

Let $ n $ be an integer greater than $ 2 $ : \begin{aligned} \int_{0}^{+\infty}{\frac{x^{\frac{1}{n}}}{1+x^{2}}\,\mathrm{d}x}&=\int_{0}^{1}{\frac{x^{\frac{1}{n}}}{1+x^{2}}\,\mathrm{d}x}+\int_{1}^{+\infty}{\frac{x^{\frac{1}{n}}}{1+x^{2}}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\frac{x^{\frac{1}{n}}}{1+x^{2}}\,\mathrm{d}x}+\int_{0}^{1}{\frac{x^{-\frac{1}{n}}}{x^{2}\left(1+\frac{1}{x^{2}}\right)}\,\mathrm{d}x} \\ &=\int_{0}^{1}{\frac{x^{\frac{1}{n}}+x^{-\frac{1}{n}}}{1+x^{2}}\,\mathrm{d}x}\\ &=\left[\left(x^{\frac{1}{n}}+x^{-\frac{1}{n}}\right)\arctan{x}\right]_{0}^{1}+\frac{1}{n}\int_{0}^{1}{\left(x^{-\frac{1}{n}}-x^{\frac{1}{n}}\right)\frac{\arctan{x}}{x}\,\mathrm{d}x}\\ I_{n}&=\frac{\pi}{2}+\frac{1}{n}J_{n}\end{aligned}

Since $ \left(\forall x\in\left[0,1\right]\right),\ \left(x^{-\frac{1}{n}}-x^{\frac{1}{n}}\right)-\left(x^{-\frac{1}{n+1}}-x^{\frac{1}{n+1}}\right)=x^{-\frac{1}{n+1}-\frac{1}{n}}\left(x^{\frac{1}{n+1}}-x^{\frac{1}{n}}\right)\left(1+x^{\frac{1}{n+1}+\frac{1}{n}}\right)\geq 0 $, the positive sequence $ \left(J_{n}\right)_{n\geq 2} $ decreases, which means $ \left(\forall n\geq 2\right),\ \left|J_{n}\right|=J_{n}\leq J_{2}=C $, and thus $ \frac{1}{n}J_{n}\underset{n\to +\infty}{\longrightarrow}0 $, which means $ I_{n}\underset{n\to +\infty}{\longrightarrow}\frac{\pi}{2} \cdot $


While the estimation done above is sufficient, it might be interesting to see how to obtain the analytical expression by elementary principles from which the result is immediately seen. Rewrite the integral using the symmetry \begin{align} I_n &= \int_0^1 \frac{x^{1/n} + x^{-1/n}}{x^2+1} \, {\rm d}x \\ &=\sum_{k=0}^\infty (-1)^k \int_0^1 \left(x^{1/n} + x^{-1/n}\right) x^{2k} \, {\rm d}x \\ &=\sum_{k=0}^\infty (-1)^k \left\{ \frac{n}{2nk+n-1} + \frac{n}{2nk+n+1} \right\} \\ &=\sum_{k=0}^\infty \left\{ \frac{n}{4nk+n-1} + \frac{n}{4nk+n+1} - \frac{n}{4nk+3n-1} - \frac{n}{4nk+3n+1} \right\} \\ &=\sum_{a\in \{-1,1\}} \sum_{k=0}^\infty \left\{ \frac{n}{4nk+n-a} - \frac{n}{4nk+3n+a} \right\} \\ &=\sum_{a\in \{-1,1\}} \sum_{k=0}^\infty \frac{\frac{n+a}{2n}}{(2k+1)^2-\left(\frac{n+a}{2n}\right)^2} \, . \tag{0} \end{align}

Now let's analyse the function $$\sum_{k=0}^\infty \frac{x}{(2k+1)^2-x^2}$$ in an analytic sense. It has simple poles at the odd integers $x=\pm (2k+1)$ and converges for all other $x\in \mathbb{C}$. Therefore we should write it as $$\sum_{k=0}^\infty \frac{x}{(2k+1)^2-x^2} = \frac{f(x)}{\cos\left(\frac{\pi x}{2}\right)} \tag{1}$$ with some analytic function $f(x)$ not vanishing at the odd integers. Calculating the residues on both sides at $x=2n+1$ it follows $$-\frac{1}{2} = (-1)^{n+1} \frac{2}{\pi} \, f(2n+1) \quad \Rightarrow \quad f(2n+1)=(-1)^n \frac{\pi}{4} \, .$$ We can thus write $$f(x)=\frac{\pi}{4} \, \sin\left(\frac{\pi x}{2}\right) \, g(x)$$ for some analytic function $g(x)$ which obeys $g(2n+1)=1$.

Now besides $g(x)$ being analytic, $g(x)$ is also bounded by (1) and the fact that $|\tan(\pi x/2)|$ is bounded $\forall x\in \mathbb{C} \setminus \mathbb{Z}$ by some constant $C>0$, and from Liouvilles theorem it follows it must be a constant $c$. To obtain that constant we go back to (1) and calculate $$\lim_{x\rightarrow 0} \sum_{k=0}^\infty \frac{1}{(2k+1)^2-x^2} = \lim_{x\rightarrow 0} c \, \frac{\pi}{4} \, \frac{\tan\left(\frac{\pi x}{2}\right)}{x} \\ \frac{\pi^2}{8} = c \, \frac{\pi^2}{8} \, ,$$ hence $c=1$.

Back into (0) it follows $$\sum_{a \in \{-1,1\}} \frac{\pi}{4} \, \tan\left(\frac{\pi(n+a)}{4n}\right) = \frac{\pi}{2\cos\left(\frac{\pi}{2n}\right)}$$ by the addition theorem of the tangens.


We could even solve the problem via DCT without your substitution: $\lim_{n\to\infty}\tan^{1/n}x=1$ for all $x\in(0,\,\tfrac{\pi}{2})$, so $\lim_{n\to\infty}I_n=\int_0^{\pi/2}dx=\frac{\pi}{2}$. As @DIger noted, $I_n$ is computable via Beta functions, viz. $\int_0^{\pi/2}\tan^{2s-1}xdx=\frac{\pi}{2}\csc\pi s$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy