find the limit: $\lim_{n\to\infty}\int_{0}^{\infty} \frac{\sqrt x}{1+ x^{2n}} dx$

Question

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?

Answer 1

Hint: you only care about $n$ big and $x$ big.

Answer 2

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)

Answer 3

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

$$\begin{align} \lim_{n\to \infty}\int_0^\infty \frac{x^{1/2}}{1+x^n}\,dx&=\int_0^1 \lim_{n\to \infty}\frac{x^{1/2}}{1+x^n}\,dx+\int_1^\infty \lim_{n\to \infty} \frac{x^{1/2}}{1+x^n}\,dx\\\\ &=2/3+0\\\\ &=2/3 \end{align}$$

Answer 4

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 $\int_0^1\sqrt x \, dx = 2/3.$

Answer 5

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 $$\lim_{x\rightarrow 0} \frac{x}{\sin(x)} = \lim_{x\rightarrow 0} \frac{1}{\cos(x)} = 1.$$