Calculate limit of this sequence $I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$ 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.
 A: 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 $
A: We first convert the integral into a beta function.
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}}(\tan x)^{\frac{1}{n}} d x &=\int_0^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x \\
&=\frac{1}{2} B\left(\frac{1}{2}+\frac{1}{2 n}, \frac{1}{2}-\frac{1}{2 n}\right) \\
&=\frac{\pi}{2} \csc \left(\frac{1}{2}+\frac{1}{2 n}\right) \pi
\end{aligned}
$$
Then \begin{align}
\lim _{n \rightarrow \infty} I_n=\frac{\pi}{2} \csc \frac{\pi}{2}=\frac{\pi}{2}
\end{align}
A: $$\lim_{n\to \infty} \int_{0}^{\infty} \frac{z^{1/n}}{1 + z^2}\,dz
= \int_{0}^{\infty} \frac{\lim_{n\to \infty} z^{1/n}}{1 + z^2}\,dz= \int_{0}^{\infty} \frac{1}{1 + z^2}\,dz=\frac\pi2
$$
A: 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.
A: 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. $\int_0^{\pi/2}\tan^{2s-1}xdx=\frac{\pi}{2}\csc\pi s$.
