find the limit: $\lim_{n\to\infty}\int_{0}^{\infty} \frac{\sqrt x}{1+ x^{2n}} dx$ Calculate the following limit
$$\lim_{n\to\infty}\int_{0}^{\infty} \frac{\sqrt x}{1+ x^{2n}} dx$$
I tried to apply dominated convergence theorem but I could not find the dominating function. even I broke down the integration from $0$ to $1$ and $1$ to infinity. then found only integration from $0$ to $1$ is possible. Do you have any ideas? 
 A: We have
$$\int_0^\infty \frac{x^{1/2}}{1+x^n}\,dx=\int_0^1 \frac{x^{1/2}}{1+x^n}\,dx+\int_1^\infty \frac{x^{1/2}}{1+x^n}\,dx$$
For $0\le x\le 1$, we have
$$\frac{x^{1/2}}{1+x^n}\le x^{1/2}$$
For $1\le x$, we have for $n\ge 1$
$$\frac{x^{1/2}}{1+x^n}\le \frac{x^{1/2}}{1+x^2}$$
Therefore, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{n\to \infty}\int_0^\infty \frac{x^{1/2}}{1+x^n}\,dx&=\int_0^1 \lim_{n\to \infty}\frac{x^{1/2}}{1+x^n}\,dx+\int_1^\infty \lim_{n\to \infty} \frac{x^{1/2}}{1+x^n}\,dx\\\\
&=2/3+0\\\\
&=2/3
\end{align}$$
A: A dominating function is $\sqrt x \cdot \chi_{[0,1]} + x^{-3/2}\cdot\chi_{[1,\infty)},$ which is in $L^1([0,\infty)).$ Since $f_n(x) \to \sqrt x \cdot\chi_{[0,1]}$ pointwise a.e., the DCT shows the limit is $\int_0^1\sqrt x \, dx = 2/3.$
A: Hint: you only care about $n$ big and $x$ big. 
A: No problem when $x$ is $0$ (check by yourself, the function is well defined for $x \geq 0$). So your function is integrable in any set $[0, a]$, with $a>0$.
When $x$ goes to infinity, then: 
$$\frac{\sqrt x}{(1+ x^{2n})} \sim \frac{x^\frac{1}{2}}{x^{2n}} = x^{\frac{1}{2} - 2n}$$
It is well known that the integral
$$\int_a^{+\infty}x^{-b}dx$$
converges when $b > 1$. In your case:
$$b = -\frac{1}{2} + 2n > 1 \Rightarrow n > \frac{3}{4}$$
(last part fixed thanks to the suggestion of Daniel Fischer)
A: Suppose we are interested in 
$$I = \int_0^\infty \frac{\sqrt{x}}{1+x^{2n}} dx$$
and evaluate it by integrating
$$f(z) = \frac{\sqrt{z}}{1+z^{2n}}
= \frac{\exp(1/2\log(z))}{1+z^{2n}}$$
around a keyhole  contour with the branch cut of  the logarithm on the
positive real axis and the range of the argument of the logarithm from
$0$ to $2\pi.$
Introduce $$\rho_k = \exp(\pi i/2/n + \pi i k/n)$$
where $0\le k\lt 2n$ to obtain
$$ (1-\exp(\pi i)) \times I = 
2\pi i\sum_{k=0}^{2n-1} \mathrm{Res}_{z=\rho_k} f(z).$$
We get
$$I = \pi i \sum_{k=0}^{2n-1}
\frac{\exp(1/2\log(\rho_k))}{2n \rho_k^{2n-1}}
= \pi i \sum_{k=0}^{2n-1} \rho_k
\frac{\exp(1/2\log(\rho_k))}{2n \rho_k^{2n}}
\\ = - \pi i \sum_{k=0}^{2n-1} \rho_k
\frac{\exp(1/2\log(\rho_k))}{2n}
\\ = -\frac{\pi i}{2n}
\sum_{k=0}^{2n-1} \exp(\pi i/2/n + \pi i k/n)
\exp(\pi i/4/n + \pi i k/2/n)
\\ = -\frac{\pi i}{2n} \exp(3\pi i/4/n)
\sum_{k=0}^{2n-1} \exp(3 \pi i k/2/n)
\\ = -\frac{\pi i}{2n} \exp(3\pi i/4/n)
\frac{\exp(2n \times 3\pi i/2/n)-1}{\exp(3\pi i/2/n)-1}
\\ = \frac{\pi i}{n} \exp(3\pi i/4/n)
\frac{1}{\exp(3\pi i/2/n)-1}
\\ = \frac{\pi i}{n}
\frac{1}{\exp(3\pi i/4/n)-\exp(-3\pi i/4/n)}
\\ = \frac{\pi}{2n} \frac{1}{\sin(3\pi/4/n)}.$$
It is convenient to write this as
$$\frac{2}{3} \frac{3\pi}{4n} \frac{1}{\sin(3\pi/4/n)}.$$
Now the Taylor series for $\frac{x}{\sin(x)}$
starts
$$\left.\frac{x}{\sin(x)}\right|_{x=0}
+ \left.\left(\frac{x}{\sin(x)}\right)'\right|_{x=0}
+ \frac{1}{2}\left.\left(\frac{x}{\sin(x)}\right)''\right|_{x=0}
\\ = 1 + \frac{1}{6} x^2  + \cdots$$
With $x=3\pi/4/n$ we get for the limit as
$n\rightarrow\infty$ the value
$$\frac{2}{3}.$$
Remark. The integral along the large circle vanishes because
we have by the ML bound
$$\lim_{R\rightarrow\infty } 2\pi R \frac{\sqrt{R}}{R^{2n}-1}
\rightarrow 0$$
and along the small circle enclosing the origin we get
applying ML a second time
$$\lim_{\epsilon\rightarrow 0} 2\pi\epsilon \sqrt{\epsilon}
\rightarrow 0.$$
Remark II. Actually we do not need the Taylor series
of $\frac{x}{\sin(x)}$ as we can obtain the constant term
with L'Hopital to get
$$\lim_{x\rightarrow 0} \frac{x}{\sin(x)}
= \lim_{x\rightarrow 0} \frac{1}{\cos(x)} = 1.$$
